ideal gas law problem solving with answers

13.3 The Ideal Gas Law

Learning objectives.

By the end of this section, you will be able to:

State the ideal gas law in terms of molecules and in terms of moles.
Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume.
Use Avogadro’s number to convert between number of molecules and number of moles.

In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, N 2 N 2 , and oxygen, O 2 O 2 , are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas, but note that this discussion also applies to monatomic gases, such as helium.)

Gases are easily compressed. We can see evidence of this in Table 13.2 , where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β β . This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 13.18 . Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.

To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See Figure 13.19 .)

In many common circumstances, including, for example, room temperature air, the gas particles have negligible volume and do not interact with each other, aside from perfectly elastic collisions. In such cases, the gas is called an ideal gas, and the relationship between the pressure, volume, and temperature is given by the equation called the ideal gas law. An equation such as the ideal gas law, which relates behavior of a physical system in terms of its thermodynamic properties, is called an equation of state.

Ideal Gas Law

The ideal gas law states that

where P P is the absolute pressure of a gas, V V is the volume it occupies, N N is the number of atoms and molecules in the gas, and T T is its absolute temperature. The constant k k is called the Boltzmann constant in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product PV PV is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of V V . The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that N N is the total number of atoms and molecules, independent of the type of gas.)

Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure P P is essentially equal to atmospheric pressure, and the volume V V increases in direct proportion to the number of atoms and molecules N N put into the tire. Once the volume of the tire is constant, the equation PV = NkT PV = NkT predicts that the pressure should increase in proportion to the number N of atoms and molecules .

Example 13.6

Calculating pressure changes due to temperature changes: tire pressure.

Suppose your bicycle tire is fully inflated, with an absolute pressure of 7 . 00 × 10 5 Pa 7 . 00 × 10 5 Pa (a gauge pressure of just under 90 . 0 lb/in 2 90 . 0 lb/in 2 ) at a temperature of 18 . 0 º C 18 . 0 º C . What is the pressure after its temperature has risen to 35 . 0 º C 35 . 0 º C ? Assume that there are no appreciable leaks or changes in volume.

The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.

We know the initial pressure P 0 = 7 .00 × 10 5 Pa P 0 = 7 .00 × 10 5 Pa , the initial temperature T 0 = 18 . 0ºC T 0 = 18 . 0ºC , and the final temperature T f = 35 . 0ºC T f = 35 . 0ºC . We must find the final pressure P f P f . How can we use the equation PV = NkT PV = NkT ? At first, it may seem that not enough information is given, because the volume V V and number of atoms N N are not specified. What we can do is use the equation twice: P 0 V 0 = NkT 0 P 0 V 0 = NkT 0 and P f V f = NkT f P f V f = NkT f . If we divide P f V f P f V f by P 0 V 0 P 0 V 0 we can come up with an equation that allows us to solve for P f P f .

Since the volume is constant, V f V f and V 0 V 0 are the same and they cancel out. The same is true for N f N f and N 0 N 0 , and k k , which is a constant. Therefore,

We can then rearrange this to solve for P f P f :

where the temperature must be in units of kelvins, because T 0 T 0 and T f T f are absolute temperatures.

1. Convert temperatures from Celsius to Kelvin.

2. Substitute the known values into the equation.

The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.

Making Connections: Take-Home Experiment—Refrigerating a Balloon

Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why?

Example 13.7

Calculating the number of molecules in a cubic meter of gas.

How many molecules are in a typical object, such as air in a tire? We can use the ideal gas law to give us an idea of how large N N typically is.

Calculate the number of molecules in a cubic meter of air at standard temperature and pressure (STP), which is defined to be 0 º C 0 º C and atmospheric pressure.

Because pressure, volume, and temperature are all specified, we can use the ideal gas law PV = NkT PV = NkT , to find N N .

1. Identify the knowns.

2. Identify the unknown: number of molecules, N N .

3. Rearrange the ideal gas law to solve for N N .

4. Substitute the known values into the equation and solve for N N .

This number is undeniably large, considering that a gas is mostly empty space. N N is huge, even in small volumes. For example, 1 cm 3 1 cm 3 of a gas at STP has 2 . 68 × 10 19 2 . 68 × 10 19 molecules in it. Once again, note that N N is the same for all types or mixtures of gases.

Moles and Avogadro’s Number

It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number ( N A ) ( N A ) , in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed, and the value of Avogadro’s number is

Avogadro’s Number

One mole always contains 6 . 02 × 10 23 6 . 02 × 10 23 particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular (molar) mass, which can be calculated by multiplying the number of moles of the substance by its atomic mass. The atomic masses of elements are given in the periodic table of elements and in Appendix A

Check Your Understanding

The active ingredient in a Tylenol pill is 325 mg of acetaminophen ( C 8 H 9 NO 2 ) ( C 8 H 9 NO 2 ) . Find the molar mass of acetaminophen, and from this, the number of moles and the number of molecules of acetaminophen in a single pill.

We first need to calculate the molar mass (the mass of one mole) of acetaminophen. To do this, we need to multiply the number of atoms of each element by the element’s atomic mass.

Then we need to calculate the number of moles in 325 mg.

Then use Avogadro’s number to calculate the number of molecules.

Example 13.8

Calculating moles per cubic meter and liters per mole.

Calculate: (a) the number of moles in 1 . 00 m 3 1 . 00 m 3 of gas at STP, and (b) the number of liters of gas per mole at STP.

Strategy and Solution

(a) We are asked to find the number of moles per cubic meter, and we know from Example 13.7 that the number of molecules per cubic meter at STP is 2 . 68 × 10 25 2 . 68 × 10 25 . The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let n n stand for the number of moles,

(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain

This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.

The (average) molar weight of dry air (approximately 80% N 2 N 2 and 20% O 2 O 2 ) at STP is M = 28.8 g/mol. M = 28.8 g/mol. Thus the mass of one cubic meter of air is 1.28 kg. The density of dry room temperature air is about 10% lower. If a living room has dimensions 5 m × 5 m × 3 m, 5 m × 5 m × 3 m, the mass of air inside the room is around 90 kg, which is the typical mass of a human.

The density of air at standard conditions ( P = 1 atm ( P = 1 atm and T = 20 º C ) T = 20 º C ) is 1.20 kg/m 3 1.20 kg/m 3 . At what pressure is the density 0.60 kg/m 3 0.60 kg/m 3 if the temperature and number of molecules are kept constant?

The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation PV = NkT PV = NkT , we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and P f = 0 . 50 atm . P f = 0 . 50 atm .

The Ideal Gas Law Restated Using Moles

A very common expression of the ideal gas law uses the number of moles, n n , rather than the number of atoms and molecules, N N . We start from the ideal gas law,

and multiply and divide the equation by Avogadro’s number N A N A . This gives

Note that n = N / N A n = N / N A is the number of moles. We define the universal gas constant R = N A k R = N A k , and obtain the ideal gas law in terms of moles.

Ideal Gas Law (in terms of moles)

The ideal gas law (in terms of moles) is

The numerical value of R R in SI units is

In other units,

You can use whichever form of R R is most convenient for a particular problem.

Example 13.9

Calculating number of moles: gas in a bike tire.

How many moles of gas are in a bike tire with a volume of 2 . 00 × 10 – 3 m 3 ( 2 . 00 L ) , 2 . 00 × 10 – 3 m 3 ( 2 . 00 L ) , a pressure of 7 . 00 × 10 5 Pa 7 . 00 × 10 5 Pa (a gauge pressure of just under 90 . 0 lb/in 2 90 . 0 lb/in 2 ), and at a temperature of 18 . 0 º C 18 . 0 º C ?

Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law, PV = nRT PV = nRT , for the number of moles n n .

2. Rearrange the equation to solve for n n and substitute known values.

The most convenient choice for R R in this case is 8 . 31 J/mol ⋅ K, 8 . 31 J/mol ⋅ K, because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in Example 13.6 , but we would get the same answer if we used the final values.

The ideal gas law can be considered to be another manifestation of the law of conservation of energy (see Conservation of Energy ). Work done on a gas results in an increase in its energy, increasing pressure and/or temperature. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules.

The Ideal Gas Law and Energy

Let us now examine the role of energy in the behavior of gases. When you inflate a bike tire by hand, you do work by repeatedly exerting a force through a distance. This energy goes into increasing the pressure of air inside the tire and increasing the temperature of the pump and the air.

The ideal gas law is closely related to energy: the dimensions on both sides are those of energy, with units of joules when using SI units. The right-hand side of the ideal gas law in PV = NkT PV = NkT is NkT NkT . This term is proportional to the amount of translational kinetic energy of N N atoms or molecules at an absolute temperature T T , as we shall see formally in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature . The left-hand side of the ideal gas law is PV PV , which also has the units of joules. Pressure is force per unit area, so pressure multiplied by volume is force times displacement, or energy. The important point is that there is energy in a gas related to both its pressure and its volume. The energy can be changed when the gas is doing work as it expands—something we explore in Heat and Heat Transfer Methods —similar to what occurs in gasoline or steam engines and turbines.

Problem-Solving Strategy: The Ideal Gas Law

Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal.

Step 2 Make a list of what quantities are given, or can be inferred from the problem as stated (identify the known quantities). Convert known values into proper SI units (K for temperature, Pa for pressure, m 3 m 3 for volume, molecules for N N , and moles for n n ).

Step 3 Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful.

Step 4 Determine whether the number of molecules or the number of moles is known, in order to decide which form of the ideal gas law to use. The first form is PV = NkT PV = NkT and involves N N , the number of atoms or molecules. The second form is PV = nRT PV = nRT and involves n n , the number of moles.

Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed.

Step 6 Substitute the known quantities, along with their units, into the appropriate equation, and obtain numerical solutions complete with units. Be certain to use absolute temperature and absolute pressure.

Step 7 Check the answer to see if it is reasonable: Does it make sense?

Liquids and solids have densities about 1000 times greater than gases. Explain how this implies that the distances between atoms and molecules in gases are about 10 times greater than the size of their atoms and molecules.

Atoms and molecules are close together in solids and liquids. In gases they are separated by empty space. Thus gases have lower densities than liquids and solids. Density is mass per unit volume, and volume is related to the size of a body (such as a sphere) cubed. So if the distance between atoms and molecules increases by a factor of 10, then the volume occupied increases by a factor of 1000, and the density decreases by a factor of 1000.

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Ideal Gas Law Example Problem

Find Moles of Gas Using the Ideal Gas Law

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The ideal gas law is an equation of state the describes the behavior of an ideal gas and also a real gas under conditions of ordinary temperature and low pressure. This is one of the most useful gas laws to know because it can be used to find pressure, volume, number of moles, or temperature of a gas.

The formula for the ideal gas law is:

PV = nRT

P = pressure V = volume n = number of moles of gas R = ideal or universal gas constant = 0.08 L atm / mol K T = absolute temperature in Kelvin

Sometimes, you may use another version of the ideal gas law:

PV = NkT

N = number of molecules k = Boltzmann constant = 1.38066 x 10 -23 J/K = 8.617385 x 10 -5 eV/K

Ideal Gas Law Example

One of the easiest applications of the ideal gas law is finding the unknown value, given all the others.

6.2 liters of an ideal gas is contained at 3.0 atm and 37 °C. How many moles of this gas are present?

The ideal gas law states

Because the units of the gas constant are given using atmospheres, moles, and Kelvin, it's important to make sure you convert values given in other temperature or pressure scales. For this problem, convert °C temperature to K using the equation:

T = °C + 273

T = 37 °C + 273 T = 310 K

Now, you can plug in the values. Solve ideal gas law for the number of moles

n = PV / RT

n = ( 3.0 atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K) n = 0.75 mol

There are 0.75 mol of the ideal gas present in the system.

What Is the Ideal Gas Law?
Ideal Gas Example Problem: Partial Pressure
How to Calculate the Density of a Gas
Ideal Gas Law: Worked Chemistry Problems
How to Calculate Density of a Gas
Avogadro's Law Example Problem
Gay-Lussac's Gas Law Examples
Ideal Gas vs Non-Ideal Gas Example Problem
Charles' Law Example Problem
Ideal Gas Law Definition and Equation
Ideal Gas Law Test Questions
Ideal Gas Definition
Henry's Law Example Problem
Chemistry Definition of Gas Constant (R)
The Combined Gas Law in Chemistry
The Major Laws of Chemistry

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Ideal Gas Law Example Problem 1

The ideal gas law describes the behavior of an ideal gas, but can also be used when applied to real gases under a wide variety of conditions. This allows us to use this law to predict the behavior of the gas when the gas is subjected to changes in pressure, volume or temperature.

The Ideal Gas Law is expressed as

PV = nRT where P = Pressure V = Volume n = number of moles of gas particles T = Absolute Temperature in Kelvin and R is the Gas Constant .

The Gas Constant, R, while a constant, depends on the units used to measure pressure and volume. Here are a few values of R depending on the units.

R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·K R = 8.2057 m 3 ·atm/mol·K R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

This ideal gas law example problem shows the steps needed to use the Ideal Gas Law equation to determine the amount of gas in a system when the pressure, volume, and temperature are known.

A cylinder of argon gas contains 50.0 L of Ar at 18.4 atm and 127 °C. How many moles of argon is in the cylinder?

The first step of any Ideal Gas Law problem is to convert temperatures to the absolute temperature scale, Kelvin. At relatively low temperatures, the 273 degree difference makes a very large difference in calculations.

To change °C to K, use the formula

T = °C + 273

T = 127 °C + 273 T = 400 K

The second step is to choose the ideal gas constant value of R suitable for our units. Our example has liters and atm. Therefore, we should use

R = 0.0821 liter·atm/mol·K

Our example wants us to find the number of moles of gas.

solve for n

$Ideal Gas Law Example - Solve for moles$

plug in our values

$Ideal Gas Law Example math 2$

n = 28.0 mol

There are 28.0 moles of argon in the cylinder.

There are two important factors to keep in mind when doing this type of problem. First, the temperature is measured as absolute temperature. Second, use the correct value of R for your problem. Using the correct units of R will avoid embarrassing unit errors.

One thought on “ ideal gas law example problem ”.

Most likely, because I made the math step image from my notes and not what I actually wrote here.

The temperature was supposed to be 400 and not 300. I updated the math step to reflect the correct calculation for the problem given.

Thank you for pointing out this mistake.

Comments are closed.

Chemistry Steps

General Chemistry

The ideal gas laws show the correlation of the temperature, pressure, volume, and the amount of a gas. Although, this is not how the laws were determined, I found my students grasping the concepts a lot easier using the following model.

We are going to take a pump filled with some gas and a freely moving plunger and by changing the gas parameters determine their correlation.

Boyle’s Law

Boyle’s Law shows the correlation of the pressure and the volume of a gas. To illustrate it, we change the volume of the gas by pushing down the plunger. This decreases the volume, and because the gas molecules have less space, the pressure is increasing:

ideal gas law problem solving with answers

The observation is that the volume of a fixed quantity of gas at constant temperature is inversely proportional to its pressure.

This relationship can be written as:

\[{\rm{V}} \sim \,\frac{{\rm{1}}}{{\rm{P}}}\]

To bring in the equal sign, we introduce a constant:

PV = constant

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

Income ∼ number of cars

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income = price x number of cars

So, for the gas pressure and volume, we are not interested too much in the constant, but rather in its linkage of pressure and volume at positions 1 and 2. Because the P x V product is constant, we can write that:

P 1 V 1 = constant = P 2 V 2

This is the practical implication of the Boyle’s law that is used for solving gas problems.

For example ,

The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters.

First, write down what you gave and what needs to be determined. If nothing is mentioned about any parameter, for example, the moles and the temperature in this case, it is assumed that they are constant, so you don’t need to worry about them.

P 1 = 2.30 atm V 1 = 1.80 L V 2 = 1.20 L P 2 = ?

\[{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\]

Now, rearrange to calculate P 2 :

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{\rm{2}}{\rm{.30}}\;{\rm{atm}}\;{\rm{ \times 1}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{1}}{\rm{.20}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.45}}\;{\rm{atm}}\]

Charle’s Law

Let’s now consider what happens if we heat up the gas leaving the plunger to move freely. The gas is going to expand, and the correlation is that the volume of a gas increases with temperature:

The volume is directly proportional to the temperature of the gas:

Therefore, for different states of gas, we can write the Charle’s law as:

For example,

What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200. °C?

Write down what is given and what needs to be determined first:

V 1 = 3.50 L T 1 = 20 o C T 2 = 200. o C V 2 = ?

Now, before doing anything else, remember to always convert the temperature to Kelvin when solving a gas problem:

So, T 1 = 20 + 273 = 293 K , T 2 = 200 + 273 = 473 K

The question studies the correlation between the volume and the temperature of a gas, so we need to use the Charle’s law.

Write it down and rearrange it to get an expression of V 2 .

\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{V}}_{\rm{2}}}\; = \;\frac{{{\rm{473}}\;\cancel{{\rm{K}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}}}{{{\rm{293}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.65}}\;{\rm{L}}\]

Gay-Lussac’s law

To study the relationship between the pressure and the temperature of a gas, the barrel is held at a fixed position to prevent changing the volume, and the sample is heated up:

In this case, the pressure increases with the temperature and for different states of the gas, the Gay-Lussac’s law is written as:

A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

P 1 = 1.40 atm T 1 = 23.0 °C P 2 = 400.0 K T 2 = ?

Convert the temperature to Kelvin right away!

T 1 = 23.0 + 273 = 296 K

Write down the gas law and rearrange it to get an expression of P 2 .

\[\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{\rm{1}}{\rm{.40}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{400}}{\rm{.0}}\;\cancel{{\rm{K}}}}}{{{\rm{296}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.89}}\;{\rm{atm}}\]

Avogadro’s Law

The Avogadro’s law of ideal gases demonstrates that the volume of a gas is directly proportional to its number of moles. In other words, the more gas, the larger the volume which again is a very intuitive observation.

For this, the plunger is again left free, and pumping some gas into the system increases the volume that the gas occupies:

This is summarized in the following formula:

Talking of the volume and the moles of a gas, remember that at STP, one mole of any gas occupies a volume of 22.4 L called the molar volume , V o .

This is because in ideal gases, the size of molecules is very small compared to the intermolecular distances. So, most of the volume is almost like an empty, and therefore, it does not matter what gas it is, the volume is determined to be 22.4 L at 0 o C and 1 atm.

The Ideal Gas Law

The examples above are great to demonstrate the individual gas law, however, notice that in all experiments, we assumed or set up the experiment, such that two parameters are constant, and we study the correlation of the other two. For example, in the Boyle’s law, we study a constant amount of gas at a constant temperature and find that the pressure increase as the volume is decreased.

To combine all the laws together and have the four variables (n, P, V, T) in one place, the Ideal Gas Law equation is obtained:

The R is called the ideal gas constant . Although it has different values and units, you will mostly be using this:

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

The ideal gas law equation is used when you need to find P, V, T, or n , for a system where they do not change .

A sample of hydrogen gas is added into a 5.80 L container at 56.0 °C. How many moles of the gas is present in the container if the pressure is 6.70 atm?

Rearrange the ideal gas law to get an expression for the moles (n):

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\]

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{6}}{\rm{.70}}\;\cancel{{{\rm{atm}}}}\; \times \;{\rm{5}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ }} \times \;{\rm{329}}\;\cancel{{\rm{K}}}}}\; = \;1.44\;{\rm{mol}}\]

Remember, to change the pressure to atm when the ideal gas law equation is used! This is because the units of R contain atm when the 0.08206 value is used. And this is what most problems in this chapter use.

How do I know which gas law to use?

You are probably wondering about this question now that every gas law brings a new equation. For this, there is what is called the combined gas law and as long as you remember it, you do not need to remember all the gas laws to solve a problem.

Let’s keep it for another article because there is quite a lot of information in this one.

Gay-Lussac’s Law
Celsius or Kelvin
Ideal-Gas Laws
Combined Gas Law Equation
How to Know Which Gas Law Equation to Use
Molar Mass and Density of Gases
Graham’s Law of Effusion and Diffusion
Graham’s Law of Effusion Practice Problems
Dalton’s Law of Partial Pressures
Mole Fraction and Partial Pressure of the Gas
Gases in Chemical Reactions
Gases-Practice Problems

Boyle’s Law: The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 litters.

Boyle’s Law: After changing the pressure of a gas sample from 760.0 torr to 0.800 atm, it occupies 4.30 L volume. What was the initial volume of the gas?

Charles’s Law: What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200 °C?

Charles’s Law: The volume of a gas decreased from 2.40 L to 830. mL and the final temperature is set at 40.0 °C. Assuming a constant pressure, calculate the initial temperature of the gas in kelvins.

Gay-Lussac’s Law: A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

Chemistry: Gas Laws

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12.4: Ideal Gas Law

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learning objectives

Describe how ideal gas law was derived.

The ideal gas law is the equation of state of a hypothetical ideal gas (an illustration is offered in ). In an ideal gas, there is no molecule-molecule interaction, and only elastic collisions are allowed. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It was first stated by Émile Clapeyron in 1834 as a combination of Boyle’s law and Charles’ law.

Atoms and Modules in a Gas : Atoms and molecules in a gas are typically widely separated, as shown. Because the forces between them are quite weak at these distances, they are often described by the ideal gas law.

Empirical Derivation

Boyle’s law states that pressure P and volume V of a given mass of confined gas are inversely proportional:

\[\mathrm{P∝\dfrac{1}{V},}\]

while Charles’ law states that volume of a gas is proportional to the absolute temperature T of the gas at constant pressure

\[\mathrm{V∝T.}\]

By combining the two laws, we get

\[\mathrm{\dfrac{PV}{T}=C,}\]

where C is a constant which is directly proportional to the amount of gas, n (representing the number of moles).

The proportionality factor is the universal gas constant, R, i.e. $\mathrm{C = nR}$.

Hence the ideal gas law

\[\mathrm{PV=nRT}\]

Equivalently, it can be written as $\mathrm{PV=NkT}$,

where k is Boltzmann’s constant and N is the number of molecules.

(Since N = nN A , you can see that $\mathrm{R=N_{Ak}}$, where N A is Avogadro’s number. )

Note that the empirical derivation does not consider microscopic details. However, the equation can be derived from first principles in the classical thermodynamics (which goes beyond the scope of this Atom ).

Microscopic version

We have seen in the Atom on “Origin of Pressure” that

\[\mathrm{P=\dfrac{Nm\bar{v^2}}{3V},}\]

where P is the pressure, N is the number of molecules, m is the mass of the molecule, v is the speed of molecules, and V is the volume of the gas. Therefore, we derive a microscopic version of the ideal gas law

\[\mathrm{PV=\dfrac{1}{3}Nm\bar{v^2}}\]

An isothermal process is a change of a system in which the temperature remains constant: $\mathrm{ΔT = 0}$.

Identify conditions at which isothermal processes can occur.

An isothermal process is a change of a system in which the temperature remains constant: ΔT = 0. Typically this occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change occurs slowly enough to allow the system to adjust continually to the temperature of the reservoir through heat exchange. In contrast, an adiabatic process occurs when a system exchanges no heat with its surroundings (Q = 0). In other words, in an isothermal process, the value ΔT = 0 but Q ≠ 0, while in an adiabatic process, ΔT ≠ 0 but Q = 0.

For an ideal gas, the product PV (P: pressure, V: volume) is a constant if the gas is kept at isothermal conditions (Boyle’s law). According to the ideal gas law, the value of the constant is NkT, where N is the number of molecules of gas and k is Boltzmann’s constant.

This means that $\mathrm{p=\frac{NkT}{V}=\frac{Constant}{V}}$ holds.

The family of curves generated by this equation is shown in the graph presented in. Each curve is called an isotherm. Such graphs are termed indicator diagrams—first used by James Watt and others to monitor the efficiency of engines. The temperature corresponding to each curve in the figure increases from the lower left to the upper right.

Isotherms of an Ideal Gas : Several isotherms of an ideal gas on a PV diagram.

Calculation of Work

In thermodynamics, the work involved when a gas changes from state A to state B is simply:

\[\mathrm{W_{A \rightarrow B}=∫_{V_A}^{V_B}PdV.}\]

(This equation is derived in our Atom on “Constant Pressure” under kinetic theory. Note that $\mathrm{P = \frac{F}{A}}$. This definition is consistent with our definition of work being force times distance. )

For an isothermal, reversible process, this integral equals the area under the relevant pressure-volume isotherm, and is indicated in blue in for an ideal gas. Again, $\mathrm{P = \frac{nRT}{V}}$ applies and with T being constant (as this is an isothermal process), we have:

Work Done by Gas During Expansion : The blue area represents “work” done by the gas during expansion for this isothermal change.

\[ \begin{align} \mathrm{W_{A \rightarrow B}} & \mathrm{=∫_{V_A}^{V_B}pdV=∫_{V_A}^{V_B} \dfrac{NkT}{V}dV} \\ & \mathrm{=NkT \ln \dfrac{⁡V_B}{V_A}.}\end{align} \]

By convention, work is defined as the work the system does on its environment. If, for example, the system expands by a piston moving in the direction of force applied by the internal pressure of a gas, then the work is counted as positive. As this work is done by using internal energy of the system, the result is that the internal energy decreases. Conversely, if the environment does work on the system so that its internal energy increases, the work is counted as negative (for details on internal energy, check our Atom on “Internal Energy of an Ideal Gas”).

Constant Pressure

Isobaric processis a thermodynamic process in which the pressure stays constant (at constant pressure, work done by a gas is $\mathrm{PΔV}$).

Describe behavior of monatomic gas during isobaric processes.

Under a certain constraint (e.g., pressure), gases can expand or contract; depending on the type of constraint, the final state of the gas may change. For example, an ideal gas that expands while its temperature is kept constant (called isothermal process) will exist in a different state than a gas that expands while pressure stays constant (called isobaric process). This Atom addresses isobaric process and correlated terms. We will discuss isothermal process in a subsequent Atom.

Isobaric Process

An isobaric process is a thermodynamic process in which pressure stays constant: $\mathrm{ΔP = 0}$. For an ideal gas, this means the volume of a gas is proportional to its temperature (historically, this is called Charles’ law ). Let’s consider a case in which a gas does work on a piston at constant pressure P, referring to Fig 1 as illustration. Since the pressure is constant, the force exerted is constant and the work done is given as W=Fd, where F (=PA) is the force on the piston applied by the pressure and d is the displacement of the piston. Therefore, the work done by the gas (W) is:

\[\mathrm{W=PAd}\]

Because the change in volume of a cylinder is its cross-sectional area A times the displacement d, we see that Ad=ΔV, the change in volume. Thus,

\[\mathrm{W=PΔV}\]

(as seen in Fig 2—isobaric process ). Note: if ΔV is positive, then W is positive, meaning that work is done by the gas on the outside world. Using the ideal gas law $\mathrm{PV=NkT (P=const)}$,

Fig 2 : A graph of pressure versus volume for a constant-pressure, or isobaric process. The area under the curve equals the work done by the gas, since W=PΔV.

\[\mathrm{W=NkΔT}\]

(Eq. 1) for an ideal gas undergoing an isobaric process.

Monatomic Gas

According to the first law of thermodynamics,

\[\mathrm{Q=ΔU+W}\]

(Eq. 2), where W is work done by the system, U is internal energy, and Q is heat. The law says that the heat transferred to the system does work but also changes the internal energy of the system. Since,

$\mathrm{U=\frac{3}{2}NkT}$ for a monatomic gas, we get $\mathrm{ΔU=\frac{3}{2}NkΔT}$

(Eq. 3; for the details on internal energy, see our Atom on “Internal Energy of an Ideal Gas”). By using the Equations 1 and 3, Eq. 2 can be written as:

$\mathrm{Q=\frac{5}{2}NkΔT}$ for monatomic gas in an isobaric process.

Specific Heat

Specific heat at constant pressure is defined by the following equation:

$\mathrm{Q=ncPΔT}$

Here n is the amount of particles in a gas represented in moles. By noting that $\mathrm{N = N_An}$ and $\mathrm{R = kN_A}$ (N A : Avogadro’s number, R: universal gas constant), we derive:

$\mathrm{c_P=\frac{5}{2}kN_A=\frac{5}{2}R}$ for a monatomic gas.

Problem Solving

With the ideal gas law we can figure pressure, volume or temperature, and the number of moles of gases under ideal thermodynamic conditions.

Identify steps used to solve the ideal gas equation.

The Ideal Gas Law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It is most accurate for monatomic gases at high temperatures and low pressures.

The ideal gas law has the form:

\[\mathrm{PV=nRT,}\]

where R is the universal gas constant, and with it we can find values of the pressure P, volume V, temperature T, or number of moles n under a certain ideal thermodynamic condition . Typically, you are given enough parameters to calculate the unknown. Variations of the ideal gas equation may help solving the problem easily. Here are some general tips.

The ideal gas law can also come in the form:

\[\mathrm{PV=NkT,}\]

where N is the number of particles in the gas and k is the Boltzmann constant.

To solve the ideal gas equation:

Write down all the information that you know about the gas.
If necessary, convert the known values to SI units.
Choose a relevant gas law equation that will allow you to calculate the unknown variable.
Substitute the known values into the equation. Calculate the unknown variable.

Remember that the general gas equation only applies if the molar quantity of the gas is fixed. For example, if a gas is mixed with another gas, you may have to apply the equation separately for individual gases.

Let’s imagine that at the beginning of a journey a truck tire has a volume of 30,000 cm 3 and an internal pressure of 170 kPa. The temperature of the tire is 16 ∘ C. By the end of the trip, the volume of the tire has increased to 32,000 cm 3 and the temperature of the air inside the tire is 40 ∘ C. What is the tire pressure at the end of the journey?

Tire Pressure : Tire pressure may change significantly during the operation of the vehicle. This is mostly due to the temperature change of the air in tires.

Step 1. Write down all the information that you know about the gas: P 1 = 170 kPa and P 2 is unknown. V 1 = 30,000 cm 3 and V 2 = 32,000 cm 3 . T 1 = 16 ∘ C and T 2 = 40 ∘ C.

Step 2. Convert the known values to SI units if necessary: Here, temperature must be converted into Kelvin. Therefore, T 1 = 16 + 273 = 289 K, T 2 = 40 + 273 = 313 K

Step 3. Choose a relevant gas law equation that will allow you to calculate the unknown variable: We can use the general gas equation to solve this problem: $\mathrm{\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}}$.

Therefore, $\mathrm{P_2= \frac{P_1 \times V_1 \times T_2}{ T_1 \times V_2}}$.

Step 4. Substitute the known values into the equation. Calculate the unknown variable:

\[\mathrm{P_2=\dfrac{170 \times 30,000 \times 313}{289 \times 32,000}=173 \; kPa}\]

The pressure of the tire at the end of the journey is 173 kPa.

Note that in Step 2 we did not bother to convert the volume values to m 3 . In Step 4, pressure appears both in the numerator and denominator. In this case the conversion was not necessary.

Avogador’s Number

The number of molecules in a mole is called Avogadro’s number (N A )—defined as 6.02x 10 23 mol -1 .

Explain relationship between Avogadro’s number and mole.

When measuring the amount of substance, it is sometimes easier to work with a unit other than the number of molecules. A mole (abbreviated mol) is a base unit in the International System of Units (SI). It is defined as any substance containing as many atoms or molecules as there are in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s constant (N A ) , in recognition of Italian scientist Amedeo Avogadro.

Amadeo Avogadro : Amedeo Avogadro (1776–1856). He established the relationship between the masses of the same volume of different gases (at the same temperature and pressure) corresponds to the relationship between their respective molecular weights.

Avogadro’s number (N) refers to the number of molecules in one gram-molecule of oxygen. This indicates an amount of substance as opposed to an independent dimension of measurement. In 1811 Amedeo Avogadro first proposed that the volume of a gas (at a given pressure and temperature) is proportional to the number of atoms or molecules, regardless of the nature of the gas (i.e., this number is universal and independent of the type of gas). In 1926, Jean Perrin won the Nobel Prize in Physics, largely for his work in determining the Avogadro constant (by several different methods). The value of Avogadro’s constant, N A , has been found to equal 6.02×10 23 mol −1 .

Role in Science

Avogadro’s constant is a scaling factor between macroscopic and microscopic (atomic scale) observations of nature. As such, it provides the relation between other physical constants and properties. For example, it establishes a relationship between the gas constant R and the Boltzmann constant k,

\[\mathrm{R=kN_A=8.314472(15) J \; mol^{−1}K^{−1};}\]

and the Faraday constant F and the elementary charge e,

\[\mathrm{F=N_Ae=96485.3383(83) C \; mol^{−1}.}\]

Measuring N A

The determination of N A is crucial to the calculation of an atom’s mass, since the latter is obtained by dividing the mass of a mole of the gas by Avogadro’s constant. In his study on Brownian motion in 1905, Albert Einstein proposed that this constant could be determined based on the quantities observable in Brownian motion. Subsequently, Einstein’s idea was verified, leading to the first determination of N A in 1908 through the experimental work of Jean Baptiste Perrin.

Absolute Temperature

Absolute temperature is the most commoly used thermodyanmic temperature unit and is the standard unit of temperature.

Describe relationship between absolute temperature and kinetic energy.

Thermodynamic temperature is the absolute measure of temperature. It is one of the principal parameters of thermodynamics and kinetic theory of gases. Thermodynamic temperature is an “absolute” scale because it is the measure of the fundamental property underlying temperature: its null or zero point (“absolute zero”) is the temperature at which the particle constituents of matter have minimal motion and cannot become any colder. That is, they have minimal motion, retaining only quantum mechanical motion, as diagramed in.

Graph of Pressure Versus Temperature : Graph of pressure versus temperature for various gases kept at a constant volume. Note that all of the graphs extrapolate to zero pressure at the same temperature

At its simplest, “temperature” arises from the kinetic energy of the random motions of matter’s particle constituents such as molecules or atoms, as seen in. Therefore, it is reasonable to choose absolute zero, where all classical motion ceases, as the reference point (T=0) of our temperature system. By using the absolute temperature scale (Kelvin system), which is the most commonly used thermodynamic temperature, we have shown that the average translational kinetic energy (KE) of a particle in a gas has a simple relationship to the temperature:

Translational Motion of Helium : Real gases do not always behave according to the ideal model under certain conditions, such as high pressure. Here, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure.

\[\mathrm{\bar{KE}=\dfrac{3}{2}kT.}\]

Note that this equation would not look this elegant if the Fahrenheit scale were used instead.

The Kelvin scale

The kelvin (or “absolute temperature”) is the standard thermodyanmic temperature unit. It is one of the seven base units in the International System of Units (SI) and is assigned the unit symbol K. By international agreement, the unit kelvin and its scale are defined by two points: absolute zero and the triple point of Vienna Standard Mean Ocean Water (water with a specified blend of hydrogen and oxygen isotopes). Absolute zero, the lowest possible temperature, is defined precisely as 0 K and −273.15 °C. The triple point of water is defined precisely as 273.16 K and 0.01 °C.

Ideal gas law was derived empirically by combining Boyle’s law and Charles’ law.
Although the empirical derivation of the equation does not consider microscopic details, the ideal gas law can be derived from first principles in the classical thermodynamics.
Pressure and volume of a gas can be related to the average velocity of molecues: $\mathrm{PV=\frac{1}{3}Nm\bar{v^2}.}$
Isothermal processes typically occur when a system is in contact with an outside thermal reservoir ( heat bath), and the change occurs slowly enough to allow the system to adjust continually to the temperature of the reservoir through heat exchange.
For an ideal gas, from the ideal gas law $\mathrm{PV = NkT, PV}$ remains constant through an isothermal process. A curve in a P-V diagram generated by the equation $\mathrm{PV = const}$ is called an isotherm.
For an isothermal, reversible process, the work done by the gas is equal to the area under the relevant pressure -volume isotherm. It is given as $\mathrm{W_A \rightarrow B=NkT \ln \frac{⁡V_B}{V_A}}$.
Gases can expand or contract under a certain constraint. Depending on the constraint, the final state of the gas may change.
The heat transferred to the system does work but also changes the internal energy of the system. In an isobaric process for a monatomic gas, heat and the temperature change satisfy the following equation: $\mathrm{Q=\frac{5}{2}NkΔT}$.
For a monatomic ideal gas, specific heat at constant pressure is $\mathrm{\frac{5}{2}R}$.
Write down all the information that you know about the gas and convert the known values to SI units if necessary.
Choose a relevant gas law equation that will allow you to calculate the unknown variable, and substitute the known values into the equation. Then calculate the unknown variable.
The general gas equation only applies if the molar quantity of the gas is fixed.
Avogadro hypothesized that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules, regardless of the type of gas.
Avogadro’s constant is a scaling factor between macroscopic and microscopic (atomic scale) observations of nature. It provides the relation between other physical constants and properties.
Albert Einstein proposed that Avogadro’s number could be determined based on the quantities observable in Brownian motion. NA was measured for the first time by Jean Baptiste Perrin in 1908.
Temperature arises from the kinetic energy of the random motions of matter ‘s particle constituents such as molecules or atoms. Therefore, it is reasonable to choose absolute zero, where all classical motion ceases, as the reference point.
By international agreement, the unit kelvin and its scale are defined by two points: absolute zero and the triple point of the standardized water.
At absolute zero, the particle constituents of matter have minimal motion and cannot become any colder. They retain minimal, quantum mechanical motion.
mole : In the International System of Units, the base unit of amount of substance; the amount of substance of a system which contains as many elementary entities as there are atoms in 12 g of carbon-12. Symbol: mol.
ideal gas : A hypothetical gas whose molecules exhibit no interaction and undergo elastic collision with each other and with the walls of the container.
Avogadro’s number : the number of constituent particles (usually atoms or molecules) in one mole of a given substance. It has dimensions of reciprocal mol and its value is equal to 6.02214129·1023 mol-1
adiabatic : Occurring without gain or loss of heat.
internal energy : The sum of all energy present in the system, including kinetic and potential energy; equivalently, the energy needed to create a system, excluding the energy necessary to displace its surroundings.
the first law of thermodynamics : A version of the law of energy conservation: the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings.
specific heat : The ratio of the amount of heat needed to raise the temperature of a unit mass of substance by a unit degree to the amount of heat needed to raise that of the same mass of water by the same amount.
SI units : International System of Units (abbreviated SI from French: Le Système international d’unités). It is the modern form of the metric system.
gas constant : A universal constant, R, that appears in the ideal gas law, (PV = nRT), derived from two fundamental constants, the Boltzman constant and Avogadro’s number, (R = NAk).
Faraday constant : The magnitude of electric charge per mole of electrons.
Brownian motion : Random motion of particles suspended in a fluid, arising from those particles being struck by individual molecules of the fluid.
absolute zero : The coldest possible temperature: zero on the Kelvin scale and approximately -273.15°C and -459.67°F. The total absence of heat; the temperature at which motion of all molecules would cease.
International System of Units : (SI): The standard set of basic units of measurement used in scientific literature worldwide.
Vienna Standard Mean Ocean Water : A standard defining a standardized isotopic composition of water.

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Ideal Gas Law Calculator

Table of contents

This ideal gas law calculator will help you establish the properties of an ideal gas subject to pressure, temperature, or volume changes . Read on to learn about the characteristics of an ideal gas, how to use the ideal gas law equation, and the definition of the ideal gas constant.

We also recommend checking out our combined gas law calculator for further understanding of the basic thermodynamic processes of ideal gases.

What is an ideal gas?

An ideal gas is a special case of any gas that fulfills the following conditions:

The gas consists of a large number of molecules that move around randomly.

All molecules are point particles (they don't take up any space).

The molecules don't interact except for colliding.

All collisions between the particles of the gas are perfectly elastic (visit our conservation of momentum calculator to learn more).

The particles obey Newton's laws of motion.

Ideal gas law equation

The properties of an ideal gas are all summarized in one formula of the form:

p p p – Pressure of the gas, measured in Pa;
V V V – Volume of the gas, measured in m³;
n n n – Amount of substance, measured in moles ;
R R R – Ideal gas constant; and
T T T – Temperature of the gas, measured in kelvins.

To find any of these values, simply enter the other ones into the ideal gas law calculator.

For example, if you want to calculate the volume of 40 moles of a gas under a pressure of 1013 hPa and at a temperature of 250 K, the result will be equal to:

V = nRT/p = 40 × 8.31446261815324 × 250 / 101300 = 0.82 m³ .

Ideal gas constant

The gas constant (symbol R) is also called the molar or universal constant. It is used in many fundamental equations, such as the ideal gas law.

The value of this constant is 8.31446261815324 J/(mol·K) .

The gas constant is often defined as the product of Boltzmann's constant k (which relates the kinetic energy and temperature of a gas) and Avogadro's number (the number of atoms in a mole of substance):

You might find this air pressure at altitude calculator useful, too.

When can I use the ideal gas law?

You can apply the ideal gas law for every gas at a density low enough to prevent the emergence of strong intermolecular forces. In these conditions, every gas is more or less correctly modeled by the simple equation PV = nRT , which relates pressure, temperature, and volume.

What is the formula of the ideal gas law?

The formula of the ideal gas law is:

P — Pressure , in pascal;
V — Volume in cubic meters;
n — Number of moles ;
T — Temperature in kelvin; and
R — Ideal gas constant .

Remember to use consistent units! The value commonly used for R , 8.314... J/mol·K refers to the pressure measured exclusively in pascals.

What is the pressure of 0.1 moles of a gas at 50 °C in a cubic meter?

268.7 Pa , or 0.00265 atm . To find this result:

Convert the temperature into kelvin:

T [K] = 273.15 + 50 = 323.15 K .

Compute the product of temperature, the number of moles, and the gas constant: nRT = 0.1 mol × 323.15 K × 8.3145 J/mol·K = 268.7 J (that is, energy ).

Divide by the volume. In this case, the volume is 1 , hence:

P = 268.7 Pa .

What are the three thermodynamics laws that can be identified in the ideal gas law?

The ideal gas law has four parameters . One of them is the number of moles which is a bit outside the scope of thermodynamics. The other three are pressure, temperature, and volume . We can identify three laws by fixing, in turn, each one of the three:

Fixing the temperature, we find the isothermal transformation (or Boyle's law ): PV = k .
Fixing the volume, we find the isochoric transformation ( Charles's law ): P/T = k .
Fixing the pressure, we have the isobaric transformation ( Gay-Lussac's law ): V/T = k .

How do I calculate the temperature of a gas given moles, volume and pressure?

To calculate the temperature of a gas given the pressure and the volume, follow these simple steps:

Calculate the product of pressure and volumes. Be sure you're using consistent units: a good choice is pascals and cubic meters .

Calculate the product of the number of moles and the gas constant . If you used pascals and cubic meters, the constant is R = 8.3145 J/mol·K .

Divide the result of step 1 by the result of step 2: the result is the temperature (in kelvin ): T = PV/nR

Pressure (p)

Amount of substance (n)

Temperature (T)

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Total mass of the gas (m)

Molar mass (M)

Gas constant (R)

(40.0 mmHg) (12.3 liters) = (60.0 mmHg) (x) x = 8.20 L Note three significant figures.

(1.00 atm) ( 3.60 liters) = (2.50 atm) (x) x = 1.44 L

(400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot) x = 133 atm

(1.56 L) (1.00 atm) = (3.00 atm) (x) 0.520 L

(11.2 liters) (0.860 atm) = (x) (15.0 L) x = 0.642 atm

(745.0 mmHg) (500.0 mL) = (760.0 mmHg) (x) x = 490.1 mL

(740.0 mmHg) (350.0 mL) = (760.0 mmHg) (x)

(63.0 atm) (338 L) = (1.00 atm) (x)

(166.0 kPa) (273.15 mL) = (101.325 kPa) (x)

(18.0 mmHg) (77.0 L) = (760.0 mmHg) (x)

Volume will decrease.

It will double in size.

(0.755 atm) (4.31 liters) = (1.25 atm) (x)

(8.00 atm) (600.0 mL) = (2.00 atm) (x)

(800.0 torr) (400.0 mL) = (1000.0 torr) (x)

100 °C is to 101.3 kPa as 88 °C is to x x = 89.144 kPa

P 1 V 1 = P 2 V 2 (101.3) (2.0) = (88.144) (x) x = 2.27 L The balloon will not burst.

(1.00 atm) (2.00 L) = (x) (5.00 L) x = 0.400 atm (1.50 atm) (3.00 L) = (y) (5.00 L) y = 0.900 atm

0.400 atm + 0.900 atm = 1.30 atm

(1.00) (2.00) = n 1 RT in the first bulb moles gas = n 1 = 2.00/RT (1.50) (3.00) = n 2 RT in the second bulb moles gas = n 2 = 4.50/RT

total volume = 2.00 + 3.00 = 5.00 (P 3 ) (5.00) = (n 1 + n 2 )RT (P 3 ) (5.00) = (2.00/RT + 4.50/RT)RT (P 3 ) (5.00) = 6.50 P 3 = 6.50 / 5.00 = 1.30 atm

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6.7: The Ideal Gas Law

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Learning Objectives

Learn and apply the ideal gas law.
Learn and apply the combined gas law.

There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.

Ideal Gas Law

As with the other gas laws, we can also say that $\frac{\left( P \times V \right)}{\left( T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.

The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $R$ for the constant, the equation becomes:

\[\dfrac{P \times V}{T \times n} = R \nonumber \]

The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:

\[PV = nRT \nonumber \]

The variable $R$ in the equation is called the ideal gas constant .

Evaluating the Ideal Gas Constant

The value of $R$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $\text{kPa}$, $\text{atm}$, or $\text{mm} \: \ce{Hg}$. Therefore, $R$ can have three different values.

We will demonstrate how $R$ is calculated when the pressure is measured in $\text{kPa}$. The volume of $1.00 \: \text{mol}$ of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm)is measured to be $22.414 \: \text{L}$. We can substitute $101.325 \: \text{kPa}$ for pressure, $22.414 \: \text{L}$ for volume, and $273.15 \: \text{K}$ for temperature into the ideal gas equation and solve for $R$.

\[\begin{align*} R &= \frac{PV}{nT} \\[4pt] &= \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} \\[4pt] &= 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol} \end{align*} \nonumber \]

This is the value of $R$ that is to be used in the ideal gas equation when the pressure is given in $\text{kPa}$. The table below shows a summary of this and the other possible values of $R$. It is important to choose the correct value of $R$ to use for a given problem.

Notice that the unit for $R$ when the pressure is in $\text{kPa}$ has been changed to $\text{J/K} \cdot \text{mol}$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $\left( \text{J} \right)$.

Example $\PageIndex{1}$ Oxygen Gas

What volume is occupied by $3.76 \: \text{g}$ of oxygen gas at a pressure of $88.4 \: \text{kPa}$ and a temperature of $19^\text{o} \text{C}$? Assume the oxygen is ideal.

Example $\PageIndex{2}$: Argon Gas

A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?

Exercise $\PageIndex{1}$

A 0.0997 mol sample of O 2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?

Exercise $\PageIndex{2}$

For a 0.00554 mol sample of H 2 , P = 23.44 torr and T = 557 K. What is its volume?

One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law , and its mathematical form is

\[\frac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n \nonumber \]

This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in Kelvin.

Example $\PageIndex{3}$:

A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?

Exercise $\PageIndex{3}$

If P 1 = 662 torr, V 1 = 46.7 mL, T 1 = 266 K, P 2 = 409 torr, and T 2 = 371 K, what is V 2 ?

As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.

The ideal gas constant is calculated to be $8.314 \: \text{J/K} \cdot \text{mol}$ when the pressure is in kPa.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
The combined gas law relates pressure, volume, and temperature of a gas.

Contributors and Attributions

Marisa Alviar-Agnew ( Sacramento City College )

Henry Agnew (UC Davis)

COMMENTS

ChemTeam: Ideal Gas Law: Problems #1
Problem #1: Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP. 1) Rearrange PV = nRT to this: 2) Substitute: V = 1.19 L (to three significant figures) Problem #2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass of argon in the sample.
7.3.1: Practice Problems- Applications of the Ideal Gas Law
PROBLEM 7.3.1.10 7.3.1. 10. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN 3 ). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide. Answer.
13.3 The Ideal Gas Law
The answer lies in the large separation of atoms and molecules in gases, ... Problem-Solving Strategy: The Ideal Gas Law. Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal. ... Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio ...
Ideal Gas Law Example Problem
For this problem, convert °C temperature to K using the equation: Now, you can plug in the values. Solve ideal gas law for the number of moles. n = ( 3.0 atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K) n = 0.75 mol. There are 0.75 mol of the ideal gas present in the system. Test your knowledge and find the number of moles of an ideal gas with ...
7.2.1: Practice Problems- The Gas Laws
PROBLEM 7.2.1.11 7.2.1. 11. A high altitude balloon is filled with 1.41 × 10 4 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48 °C and the pressure is 63.1 torr? Answer. Click here to see a video solution.
Ideal Gas Law Example Problem
R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K. This ideal gas law example problem shows the steps needed to use the Ideal Gas Law equation to determine the amount of gas in a system when the pressure, volume, and temperature are known. Problem. A cylinder of argon gas contains 50.0 L of Ar at 18.4 atm and 127 °C.
Ideal Gas Laws and Practice Problems-Chemistry Steps
This is the practical implication of the Boyle's law that is used for solving gas problems. For example, The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters. First, write down what you gave and what needs to be determined.
Calculations using the ideal gas equation
A balloon contains 0.692 mol ‍ of ammonia gas, NH 3 (g) ‍ , at 280 K ‍ and a pressure of 0.810 bar ‍ . What is the volume, in liters, of the balloon? Express the answer using 3 significant figures.
Ideal gas law (practice)
AP Chemistry equations and constants. Gas equations: P V = n R T P 1 = P total × χ 1, where χ 1 = moles of gas 1 total moles P total = P 1 + P 2 + P 3 +. . . K = ° C + 273. Gas variables and constants: P = pressure V = volume T = temperature n = number of moles R = 8.314 J mol − 1 K − 1 = 0.082 06 L atm mol − 1 K − 1 = 62.36 L Torr ...
The ideal gas law (PV = nRT) (video)
Transcript. The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume). No gas is truly ideal, but the ideal gas law does provide a good approximation of real gas behavior under many conditions.
Gas Laws Problem Sets
Problem Set GL6: Combined Gas Law. Solve a two-state problem involving pressure, volume, and temperature changes. Includes 6 problems. Problem Set GL7: Ideal Gas Law 1. Use the ideal gas law to relate the pressure volume, temperature and the number of moles. Includes 6 problems. Problem Set GL8: Ideal Gas Law 2.
PDF Ideal Gas Law Problems
3) A 113L sample of helium at 27°C is cooled at constant pressure to -78.0°C. Calculate the new volume of the helium. 4) What volume of He is occupied by 2.35 mol of He at 25°C and a pressure of 0.980 atm? 5) An aerosol can contains 400.0 ml of compressed gas at 5.2 atm pressure. When the gas is sprayed into a large plastic bag, the bag ...
ChemTeam: Gas Law
Problem #10: Calculate the final pressure inside a scuba tank after it cools from 1.00 x 10 3 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm. The answer should be determined to three significant figures. Bonus Problem: Consider an ideal gas with an absolute temperature of T 1.
5.4: The Ideal Gas Law
Summary. The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known. Ideal gas equation: PV = nRT, where R = 0.08206L ⋅ atm K ⋅ mol = 8.3145 J K ⋅ mol.
Ideal Gas Law
The answer to the question, "Which law relates to the Ideal Gas Law for the variables n and V?" is Avogadro's Law. ... When solving ideal gas law problems, it is a good idea to organize the values ...
12.4: Ideal Gas Law
Step 3. Choose a relevant gas law equation that will allow you to calculate the unknown variable: We can use the general gas equation to solve this problem: P1V1 T1 = P2V2 T2 P 1 V 1 T 1 = P 2 V 2 T 2. Therefore, P2 = P1×V1×T2 T1×V2 P 2 = P 1 × V 1 × T 2 T 1 × V 2. Step 4. Substitute the known values into the equation.
What is the ideal gas law? (article)
If we want to use N number of molecules instead of n moles , we can write the ideal gas law as, P V = N k B T. Where P is the pressure of the gas, V is the volume taken up by the gas, T is the temperature of the gas, N is the number of molecules in the gas, and k B is Boltzmann's constant, k B = 1.38 × 10 − 23 J K.
Ideal Gas Law Calculator
Calculate the product of the number of moles and the gas constant. If you used pascals and cubic meters, the constant is R = 8.3145 J/mol·K. Divide the result of step 1 by the result of step 2: the result is the temperature (in kelvin ): T = PV/nR. Use the ideal gas law calculator to find the pressure, volume, and temperature of a gas.
ChemTeam: Boyle's Law Problems #1-15
2) Now, we can solve the problem using Boyle's Law: P 1 V 1 = P 2 V 2 (101.3) (2.0) = (88.144) (x) x = 2.27 L The balloon will not burst. Comment: Boyle's Law assumes that the temperature and amount of gas are constant. Since we never knew the starting temperature, we will assume it never changed as the balloon rose.
7.2: The Gas Laws (Problems)
PROBLEM 7.2.10 7.2. 10. How many grams of gas are present in each of the following cases? 0.100 L of CO 2 at 307 torr and 26 °C. 8.75 L of C 2 H 4, at 378.3 kPa and 483 K. 221 mL of Ar at 0.23 torr and -54 °C. Answer a. Answer b. Answer c.
The Ideal Gas Law
The Ideal Gas Law is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. The Ideal Gas Law is a combination of simpler gas laws such as Boyle's, Charles's, Avogadro's and Amonton's laws. The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good ….
6.7: The Ideal Gas Law
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable R for the constant, the equation becomes: P × V T × n = R. The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:
ChemTeam: Combined Gas Law
Problem #8: The pressure of a gas is reduced to 75% of its initial value and the volume is increased by 40% of its initial value. Find the final temperature, given that the initial temperature was −10 °C. Solution: Let us assign P 1 = 1, therefore P 2 = 0.75 Let us assign V 1 = 1, therefore V 2 = 1.4 . I won't bother with units on P or V.

13.3 The Ideal Gas Law

Ideal Gas Law

Example 13.6

Making Connections: Take-Home Experiment—Refrigerating a Balloon

Example 13.7

Moles and Avogadro’s Number

Avogadro’s Number

Check Your Understanding

Example 13.8

Strategy and Solution

The Ideal Gas Law Restated Using Moles

Ideal Gas Law (in terms of moles)

Example 13.9

The Ideal Gas Law and Energy

Problem-Solving Strategy: The Ideal Gas Law

Ideal Gas Law Example Problem

Ideal Gas Law Example

Ideal Gas Law Example Problem 1

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Chemistry: Gas Laws

12.4: Ideal Gas Law

Empirical Derivation

Microscopic version

Calculation of Work

Constant Pressure

Isobaric Process

Monatomic Gas

Specific Heat

Problem Solving

Avogador’s Number

Role in Science

Measuring N A

Absolute Temperature

The Kelvin scale

Ideal Gas Law Calculator

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Ideal gas law equation

Ideal gas constant

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What is the pressure of 0.1 moles of a gas at 50 °C in a cubic meter?

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Margin Size

6.7: The Ideal Gas Law

Learning Objectives

Ideal Gas Law

Evaluating the Ideal Gas Constant

Example \(\PageIndex{1}\) Oxygen Gas

Example \(\PageIndex{2}\): Argon Gas

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Example \(\PageIndex{3}\):

Exercise \(\PageIndex{3}\)

Contributors and Attributions

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