first law of thermodynamics problem solving

The first law of thermodynamics – problems and solutions

30 The first law of thermodynamics – problems and solutions

Wanted: the change in internal energy of the system

W is positive if work is done by the system

Internal energy increases by 500 Joule.

2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?

ΔU = 2000+2500

– If heat is added to the system, then the internal energy of the system increases

6.  Calculate the work done by a system when it absorbs 600 J of heat and its internal energy increases by 150 J. Solution: Using the first law of thermodynamics: \[ W = Q – \Delta U = 600 – 150 = 450\, \text{J} \]

12. Calculate the work done by the system when it loses 300 J of heat, and its internal energy decreases by 100 J. Solution: Using the first law of thermodynamics: \[ W = Q – \Delta U = -300 – (-100) = -200\, \text{J} \]

18. Calculate the work done on the system when it loses 900 J of heat, and its internal energy decreases by 500 J. Solution: Using the first law of thermodynamics: \[ W = \Delta U – Q = (-500) – (-900) = 400\, \text{J} \]

24. Calculate the work done by the system when it absorbs 1100 J of heat, and its internal energy increases by 550 J. Solution: Using the first law of thermodynamics: \[ W = Q – \Delta U = 1100 – 550 = 550\, \text{J} \]

30. Calculate the work done by the system when it loses 1600 J of heat, and its internal energy decreases by 800 J. Solution: Using the first law of thermodynamics: \[ W = Q – \Delta U = -1600 – (-800) = -800\, \text{J} \]

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The first law of thermodynamics + solved problems.

Read Chemistry May 21, 2019 Physical Chemistry

– The first law of Thermodynamics states that The total energy of an isolated system remains constant though it may change from one form to another.

Internal Energy

– A thermodynamic system containing some quantity of matter has within itself a definite quantity of energy.

– This energy includes not only the translation kinetic energy of the molecules but also other molecular energies such as rotational, and vibrational energies.

– The kinetic and potential energy of the nuclei and electrons within the individual molecules also contribute to the energy of the system.

– Internal Energy  is The total of all the possible kinds of energy of a system.

– The word (internal) is often omitted and the energy of a system always implies internal energy.

– The internal energy of a system, like temperature, pressure, volume, etc., is determined by the state of a system and is independent of the path by which it is obtained.

– Hence internal energy of a system is a state function.

– For example, we consider the heating of one mole of liquid water from 0 º to 100 º C.

– The change in energy is 1.8 kcal and is the same regardless of the form in which this energy is transferred to the water by heating, by performing work, by electrical means, or in any other way.

– Since the value of the internal energy of a system depends on the mass of the matter contained in a system, it is classed as an extensive property.

Symbol Representation of Internal Energy and Sign Conventions

– The internal energy of a system is represented by the symbol (E) (Some books use the symbol U).

– It is neither possible nor necessary to calculate the absolute value of the internal energy of a system.

– In thermodynamics, we are concerned only with the energy changes when a system changes from one state to another.

– If ΔE is the difference in energy of the initial state (E in ) and the final state (E f ), we can write

ΔE = E f – E in

 ΔE is +ve if E f is greater than E in

and ΔE is –ve if E f is less than E in .

– A system may transfer energy to or from the surroundings as heat or as work, or both.

Units of Internal Energy

– The SI unit for the internal energy of a system is the joule (J).

– Another unit of energy that is not an SI unit is the calorie, 1 cal = 4.184 J.

The first law of Thermodynamics statement

– The first law of thermodynamics is, in fact, an application of the broad principle known as the Law of Conservation of Energy to the thermodynamic system.

– First law of Thermodynamics states that: The total energy of an isolated system remains constant though it may change from one form to another.

– When a system is changed from state A to state B, it undergoes a change in the internal energy from E A to E B .

– Thus, we can write:

ΔE = E B – E A

– This energy change is brought about by the evolution or absorption of heat and/or by work being done by the system.

– Because the total energy of the system must remain constant, we can write the mathematical statement of the First Law as:

ΔE = q – w            …(1)

– where q = the amount of heat supplied to the system, w = work done by the system

– Thus the First Law may also be stated as the net energy change of a closed system is equal to the heat transferred to the system minus the work done by the system.

– To illustrate the mathematical statement of the First Law, let us consider the system ‘expanding hot gas’.

– The gas expands against an applied constant pressure by volume ΔV.

– The total mechanical work done is given by the relation:

w = P × ΔV          …(2)

– From (1) and (2), we can restate:

ΔE = q – P × ΔV

Other Definitions of the First Law of Thermodynamics

(1) Whenever energy of a particular type disappears equivalent amount of another type must be produced.

(2) Total energy of a system and surroundings remains constant (or conserved)

(3) It is impossible to construct a perpetual motion machine that can produce work without spending energy on it.

Some Special Forms of the First Law of Thermodynamics

– The mathematical statement of the First Law of Thermodynamics is:

Case (1): For a cyclic process involving isothermal expansion of an ideal gas

Case (2): For an isochoric process (no change in volume) there is no work of expansion i.e. w = 0.

Case (3): For an adiabatic process there is no change in heat gained or lost i.e. q = 0.

– In other words, the decrease in internal energy is exactly equal to the work done on the system by surroundings.

Case (4): For an isobaric process there is no change in pressure, i.e. P remains constant.

or ΔE = q – PΔV

Solved Problem

Problem (1): Find ΔE, q, and w if 2 moles of hydrogen at 3 atm pressure expand isothermally at 50ºC and reversibly to a pressure of 1 atm.

– Since the operation is isothermal and the gas is ideal

Problem (2): 1g of water at 373 K is converted into steam at the same temperature. The volume of water becomes 1671 ml on boiling. Calculate the change in the internal energy of the system if the heat of vaporization is 540 cal/g.

– As the vaporization takes place against a constant pressure of 1 atmosphere, work done for an irreversible process, w, is:

Problem (3): A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 liters to a volume of 10 liters. In doing so it absorbs 400 J thermal energy from its surroundings. Determine ΔE for the process.

Problem (4): Calculate the maximum work done when pressure on 10 g of hydrogen is reduced from 20 to one atmosphere at a constant temperature of 273 K. The gas behaves ideally. Will there be any change in internal energy? Also, calculate ‘q’.

Reference:  Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolor edition.

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Chapter 15 Thermodynamics

108 15.1 The First Law of Thermodynamics

• Define the first law of thermodynamics.
• Describe how conservation of energy relates to the first law of thermodynamics.
• Identify instances of the first law of thermodynamics working in everyday situations, including biological metabolism.
• Calculate changes in the internal energy of a system, after accounting for heat transfer and work done.

If we are interested in how heat transfer is converted into doing work, then the conservation of energy principle is important. The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is

Here[latex]\boldsymbol{\Delta{U}}[/latex]is the change in internal energy [latex]\boldsymbol{U}[/latex]of the system.[latex]\boldsymbol{Q}[/latex]is the net heat transferred into the system —that is,[latex]\boldsymbol{Q}[/latex]is the sum of all heat transfer into and out of the system.[latex]\boldsymbol{W}[/latex]is the net work done by the system —that is,[latex]\boldsymbol{W}[/latex]is the sum of all work done on or by the system. We use the following sign conventions: if[latex]\boldsymbol{Q}[/latex]is positive, then there is a net heat transfer into the system; if[latex]\boldsymbol{W}[/latex]is positive, then there is net work done by the system. So positive[latex]\boldsymbol{Q}[/latex]adds energy to the system and positive[latex]\boldsymbol{W}[/latex]takes energy from the system. Thus[latex]\boldsymbol{\Delta{U}=Q-W}.[/latex]Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. Heat engines are a good example of this—heat transfer into them takes place so that they can do work. (See Figure 2 .) We will now examine[latex]\boldsymbol{Q},\:\boldsymbol{W},[/latex]and[latex]\boldsymbol{\Delta{U}}[/latex]further.

MAKING CONNECTIONS: LAW OF THERMODYNAMICS AND LAW OF CONSERVATION OF ENERGY

The first law of thermodynamics is actually the law of conservation of energy stated in a form most useful in thermodynamics. The first law gives the relationship between heat transfer, work done, and the change in internal energy of a system.

Heat Q and Work W

Heat transfer ([latex]\boldsymbol{Q}[/latex]) and doing work ([latex]\boldsymbol{W}[/latex]) are the two everyday means of bringing energy into or taking energy out of a system. The processes are quite different. Heat transfer, a less organized process, is driven by temperature differences. Work, a quite organized process, involves a macroscopic force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfer into a system, such as when the Sun warms the air in a bicycle tire, can increase its temperature, and so can work done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat transfer or by doing work. This uncertainty is an important point. Heat transfer and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy[latex]\boldsymbol{U}[/latex]of a system. Internal energy is a form of energy completely different from either heat or work.

Internal Energy U

We can think about the internal energy of a system in two different but consistent ways. The first is the atomic and molecular view, which examines the system on the atomic and molecular scale. The internal energy [latex]\boldsymbol{U}[/latex]of a system is the sum of the kinetic and potential energies of its atoms and molecules. Recall that kinetic plus potential energy is called mechanical energy. Thus internal energy is the sum of atomic and molecular mechanical energy. Because it is impossible to keep track of all individual atoms and molecules, we must deal with averages and distributions. A second way to view the internal energy of a system is in terms of its macroscopic characteristics, which are very similar to atomic and molecular average values.

Macroscopically, we define the change in internal energy[latex]\boldsymbol{\Delta{U}}[/latex]to be that given by the first law of thermodynamics:

Many detailed experiments have verified that[latex]\boldsymbol{\Delta{U}=Q-W},[/latex]where[latex]\boldsymbol{\Delta{U}}[/latex]is the change in total kinetic and potential energy of all atoms and molecules in a system. It has also been determined experimentally that the internal energy[latex]\boldsymbol{U}[/latex]of a system depends only on the state of the system and not how it reached that state . More specifically,[latex]\boldsymbol{U}[/latex]is found to be a function of a few macroscopic quantities (pressure, volume, and temperature, for example), independent of past history such as whether there has been heat transfer or work done. This independence means that if we know the state of a system, we can calculate changes in its internal energy[latex]\boldsymbol{U}[/latex]from a few macroscopic variables.

MAKING CONNECTIONS: MACROSCOPIC AND MICROSCOPIC

In thermodynamics, we often use the macroscopic picture when making calculations of how a system behaves, while the atomic and molecular picture gives underlying explanations in terms of averages and distributions. We shall see this again in later sections of this chapter. For example, in the topic of entropy, calculations will be made using the atomic and molecular view.

To get a better idea of how to think about the internal energy of a system, let us examine a system going from State 1 to State 2. The system has internal energy[latex]\boldsymbol{U_1}[/latex]in State 1, and it has internal energy[latex]\boldsymbol{U_2}[/latex]in State 2, no matter how it got to either state. So the change in internal energy[latex]\boldsymbol{\Delta{U}=U_2-U_1}[/latex]is independent of what caused the change. In other words,[latex]\boldsymbol{\Delta{U}}[/latex] is independent of path . By path, we mean the method of getting from the starting point to the ending point. Why is this independence important? Note that[latex]\boldsymbol{\Delta{U}=Q-W}.[/latex]Both[latex]\boldsymbol{Q}[/latex]and[latex]\boldsymbol{W}[/latex] depend on path , but[latex]\boldsymbol{\Delta{U}}[/latex]does not. This path independence means that internal energy[latex]\boldsymbol{U}[/latex]is easier to consider than either heat transfer or work done.

Example 1: Calculating Change in Internal Energy: The Same Change in U is Produced by Two Different Processes

(a) Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system?

(b) What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system? (See Figure 3 ).

In part (a), we must first find the net heat transfer and net work done from the given information. Then the first law of thermodynamics[latex]\boldsymbol{(\Delta{U}=Q-W)}[/latex]can be used to find the change in internal energy. In part (b), the net heat transfer and work done are given, so the equation can be used directly.

Solution for (a)

The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or

Similarly, the total work is the work done by the system minus the work done on the system, or

Thus the change in internal energy is given by the first law of thermodynamics:

We can also find the change in internal energy for each of the two steps. First, consider 40.00 J of heat transfer in and 10.00 J of work out, or

Now consider 25.00 J of heat transfer out and 4.00 J of work in, or

The total change is the sum of these two steps, or

Discussion on (a)

No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.

Solution for (b)

Here the net heat transfer and total work are given directly to be[latex]\boldsymbol{Q=-150.00\textbf{ J}}[/latex]and[latex]\boldsymbol{W=-159.00\textbf{ J}},[/latex]so that

Discussion on (b)

A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to[latex]\boldsymbol{\Delta{U}}[/latex]and not to the individual[latex]\boldsymbol{Q}\text{s}[/latex]or[latex]\boldsymbol{W}\text{s}[/latex]involved. The system ends up in the same state in both (a) and (b). Parts (a) and (b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path.

Human Metabolism and the First Law of Thermodynamics

Human metabolism is the conversion of food into heat transfer, work, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. We now take another look at these topics via the first law of thermodynamics. Considering the body as the system of interest, we can use the first law to examine heat transfer, doing work, and internal energy in activities ranging from sleep to heavy exercise. What are some of the major characteristics of heat transfer, doing work, and energy in the body? For one, body temperature is normally kept constant by heat transfer to the surroundings. This means[latex]\boldsymbol{Q}[/latex]is negative. Another fact is that the body usually does work on the outside world. This means[latex]\boldsymbol{W}[/latex]is positive. In such situations, then, the body loses internal energy, since[latex]\boldsymbol{\Delta{U}=Q-W}[/latex]is negative.

Now consider the effects of eating. Eating increases the internal energy of the body by adding chemical potential energy (this is an unromantic view of a good steak). The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Food energy is reported in a special unit, known as the Calorie. This energy is measured by burning food in a calorimeter, which is how the units are determined.

In chemistry and biochemistry, one calorie (spelled with a lowercase c) is defined as the energy (or heat transfer) required to raise the temperature of one gram of pure water by one degree Celsius. Nutritionists and weight-watchers tend to use the dietary calorie, which is frequently called a Calorie (spelled with a capital C). One food Calorie is the energy needed to raise the temperature of one kilogram of water by one degree Celsius. This means that one dietary Calorie is equal to one kilocalorie for the chemist, and one must be careful to avoid confusion between the two.

Again, consider the internal energy the body has lost. There are three places this internal energy can go—to heat transfer, to doing work, and to stored fat (a tiny fraction also goes to cell repair and growth). Heat transfer and doing work take internal energy out of the body, and food puts it back. If you eat just the right amount of food, then your average internal energy remains constant. Whatever you lose to heat transfer and doing work is replaced by food, so that, in the long run,[latex]\boldsymbol{\Delta{U}=0}.[/latex]If you overeat repeatedly, then[latex]\boldsymbol{\Delta{U}}[/latex]is always positive, and your body stores this extra internal energy as fat. The reverse is true if you eat too little. If[latex]\boldsymbol{\Delta{U}}[/latex]is negative for a few days, then the body metabolizes its own fat to maintain body temperature and do work that takes energy from the body. This process is how dieting produces weight loss.

Life is not always this simple, as any dieter knows. The body stores fat or metabolizes it only if energy intake changes for a period of several days. Once you have been on a major diet, the next one is less successful because your body alters the way it responds to low energy intake. Your basal metabolic rate (BMR) is the rate at which food is converted into heat transfer and work done while the body is at complete rest. The body adjusts its basal metabolic rate to partially compensate for over-eating or under-eating. The body will decrease the metabolic rate rather than eliminate its own fat to replace lost food intake. You will chill more easily and feel less energetic as a result of the lower metabolic rate, and you will not lose weight as fast as before. Exercise helps to lose weight, because it produces both heat transfer from your body and work, and raises your metabolic rate even when you are at rest. Weight loss is also aided by the quite low efficiency of the body in converting internal energy to work, so that the loss of internal energy resulting from doing work is much greater than the work done. It should be noted, however, that living systems are not in thermal equilibrium.

The body provides us with an excellent indication that many thermodynamic processes are irreversible . An irreversible process can go in one direction but not the reverse, under a given set of conditions. For example, although body fat can be converted to do work and produce heat transfer, work done on the body and heat transfer into it cannot be converted to body fat. Otherwise, we could skip lunch by sunning ourselves or by walking down stairs. Another example of an irreversible thermodynamic process is photosynthesis. This process is the intake of one form of energy—light—by plants and its conversion to chemical potential energy. Both applications of the first law of thermodynamics are illustrated in Figure 4 . One great advantage of conservation laws such as the first law of thermodynamics is that they accurately describe the beginning and ending points of complex processes, such as metabolism and photosynthesis, without regard to the complications in between. Table 1 presents a summary of terms relevant to the first law of thermodynamics.

Section Summary

• The first law of thermodynamics is given as[latex]\boldsymbol{\Delta{U}=Q-W},[/latex]where[latex]\boldsymbol{\Delta{U}}[/latex]is the change in internal energy of a system,[latex]\boldsymbol{Q}[/latex]is the net heat transfer (the sum of all heat transfer into and out of the system), and[latex]\boldsymbol{W}[/latex]is the net work done (the sum of all work done on or by the system).
• Both[latex]\boldsymbol{Q}[/latex]and[latex]\boldsymbol{W}[/latex]are energy in transit; only[latex]\boldsymbol{\Delta{U}}[/latex] represents an independent quantity capable of being stored.
• The internal energy[latex]\boldsymbol{U}[/latex]of a system depends only on the state of the system and not how it reached that state.
• Metabolism of living organisms, and photosynthesis of plants, are specialized types of heat transfer, doing work, and internal energy of systems.

Conceptual Questions

1: Describe the photo of the tea kettle at the beginning of this section in terms of heat transfer, work done, and internal energy. How is heat being transferred? What is the work done and what is doing it? How does the kettle maintain its internal energy?

2: The first law of thermodynamics and the conservation of energy, as discussed in Chapter 7.6 Conservation of Energy , are clearly related. How do they differ in the types of energy considered?

3: Heat transfer[latex]\boldsymbol{Q}[/latex]and work done[latex]\boldsymbol{W}[/latex]are always energy in transit, whereas internal energy[latex]\boldsymbol{U}[/latex]is energy stored in a system. Give an example of each type of energy, and state specifically how it is either in transit or resides in a system.

4: How do heat transfer and internal energy differ? In particular, which can be stored as such in a system and which cannot?

5: If you run down some stairs and stop, what happens to your kinetic energy and your initial gravitational potential energy?

6: Give an explanation of how food energy (calories) can be viewed as molecular potential energy (consistent with the atomic and molecular definition of internal energy).

7: Identify the type of energy transferred to your body in each of the following as either internal energy, heat transfer, or doing work: (a) basking in sunlight; (b) eating food; (c) riding an elevator to a higher floor.

Problems & Exercises

1: What is the change in internal energy of a car if you put 12.0 gal of gasoline into its tank? The energy content of gasoline is[latex]\boldsymbol{1.3\times10^8\textbf{ J/gal}}.[/latex]All other factors, such as the car’s temperature, are constant.

2: How much heat transfer occurs from a system, if its internal energy decreased by 150 J while it was doing 30.0 J of work?

3: A system does[latex]\boldsymbol{1.80\times10^8\textbf{ J}}[/latex]of work while[latex]\boldsymbol{7.50\times10^8\textbf{ J}}[/latex]of heat transfer occurs to the environment. What is the change in internal energy of the system assuming no other changes (such as in temperature or by the addition of fuel)?

4: What is the change in internal energy of a system which does[latex]\boldsymbol{4.50\times10^5\textbf{ J}}[/latex]of work while[latex]\boldsymbol{3.00\times10^6\textbf{ J}}[/latex]of heat transfer occurs into the system, and[latex]\boldsymbol{8.00\times10^6\textbf{ J}}[/latex]of heat transfer occurs to the environment?

5: Suppose a woman does 500 J of work and 9500 J of heat transfer occurs into the environment in the process. (a) What is the decrease in her internal energy, assuming no change in temperature or consumption of food? (That is, there is no other energy transfer.) (b) What is her efficiency?

6: (a) How much food energy will a man metabolize in the process of doing 35.0 kJ of work with an efficiency of 5.00%? (b) How much heat transfer occurs to the environment to keep his temperature constant? Explicitly show how you follow the steps in the Problem-Solving Strategy for thermodynamics found in Chapter 15.5 Problem-Solving Strategies for Thermodynamics .

7: (a) What is the average metabolic rate in watts of a man who metabolizes 10,500 kJ of food energy in one day? (b) What is the maximum amount of work in joules he can do without breaking down fat, assuming a maximum efficiency of 20.0%? (c) Compare his work output with the daily output of a 187-W (0.250-horsepower) motor.

8: (a) How long will the energy in a 1470-kJ (350-kcal) cup of yogurt last in a woman doing work at the rate of 150 W with an efficiency of 20.0% (such as in leisurely climbing stairs)? (b) Does the time found in part (a) imply that it is easy to consume more food energy than you can reasonably expect to work off with exercise?

9: (a) A woman climbing the Washington Monument metabolizes[latex]\boldsymbol{6.00\times10^2\textbf{ kJ}}[/latex]of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

[latex]\boldsymbol{1.6\times10^9\textbf{ J}}[/latex]

[latex]\boldsymbol{-9.30\times10^8\textbf{ J}}[/latex]

(a)[latex]\boldsymbol{-1.0\times10^4\textbf{ J}},[/latex]or[latex]\boldsymbol{-2.39\textbf{ kcal}}[/latex]

(b)[latex]\boldsymbol{2.10\times10^6\textbf{ J}}[/latex]

(c) Work done by the motor is[latex]\boldsymbol{1.61\times10^7\textbf{ J}}[/latex];thus the motor produces 7.67 times the work done by the man

(b) This amount of heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc.

College Physics: OpenStax Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Table of content

Full table of contents

The first law of thermodynamics states that the change in internal energy of the system is equal to the net heat transfer into the system minus the net work done by the system. This equation is a generalized form of energy conservation and can be applied to any thermodynamic process.

The following strategies can be used to solve any problem involving the first law of thermodynamics.

• The thermodynamic system should be identified.
• The initial and final states of the thermodynamic process should be listed.
• The known and unknown quantities for the given problem are identified. It should be ensured that the units for all the quantities are consistent. For example, if pressure is in Pascals and volume is in cubic meters, the work should be expressed in Joules.
• The change in internal energy is path independent. As a result, it can be estimated for all possible paths if the internal energy for the initial and final states is known.
• Heat, work, and change in internal energy for any process are related via the equation for the first law of thermodynamics.
• Conventionally, Q is positive when heat is added to the system and negative when the heat is removed from it. Also, W is positive when work is done by the system and negative when work is done on the system. This means that the polarity of work and heat should be considered in the first law of thermodynamics.
• The known quantities can then be plugged into the equation for the first law of thermodynamics to evaluate the unknown quantity.

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Chapter 20: The First Law of Thermodynamics

Back to chapter, first law of thermodynamics: problem-solving, previous video 20.6: first law of thermodynamics, next video 20.8: cyclic processes and isolated systems.

The first law of thermodynamics states that the heat added to a system is utilized in performing the work and increasing the system's internal energy.

Consider an example where 100 grams of water at one atmospheric pressure converts into steam at 100 degrees Celsius. What is the change in the internal energy for this thermodynamic process?

This thermodynamic system changes from the initial liquid to the final gaseous state at constant temperature and pressure when water converts to steam.

The heat transfer into the system is the product of mass and the latent heat of the vaporization of water.

The work done is equal to the product of pressure and volume change. Using the value of mass and density of the steam and water, the volume change and so the work done is estimated.

Here the unknown quantity is the change in the internal energy, while the known quantities are heat supplied and work done by the system. So, applying the first law of thermodynamics, a change in internal energy can be estimated.

The first law of thermodynamics states that the change in internal energy of the system is equal to the net heat transfer into the system minus the net work done by the system. This equation is a generalized form of energy conservation and can be applied to any thermodynamic process.

The following strategies can be used to solve any problem involving the first law of thermodynamics.

• The thermodynamic system should be identified.
• The initial and final states of the thermodynamic process should be listed.
• The known and unknown quantities for the given problem are identified. It should be ensured that the units for all the quantities are consistent. For example, if pressure is in Pascals and volume is in cubic meters, the work should be expressed in Joules.
• The change in internal energy is path independent. As a result, it can be estimated for all possible paths if the internal energy for the initial and final states is known.
• Heat, work, and change in internal energy for any process are related via the equation for the first law of thermodynamics.
• Conventionally, Q is positive when heat is added to the system and negative when the heat is removed from it. Also, W is positive when work is done by the system and negative when work is done on the system. This means that the polarity of work and heat should be considered in the first law of thermodynamics.
• The known quantities can then be plugged into the equation for the first law of thermodynamics to evaluate the unknown quantity.

Suggested Reading

• Young, H. D, and Freedman, R.A. (2012). University Physics with Modern Physics . San Francisco, CA: Pearson. pp 632.
• OpenStax. (2019). University Physics Vol. 2 . [Web version]. pp 116 Retrieved from  https://openstax.org/books/college-physics/pages/15-1-the-first-law-of-thermodynamics

15.2 The First Law of Thermodynamics and Some Simple Processes

Learning objectives.

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

• Describe the processes of a simple heat engine.
• Explain the differences among the simple thermodynamic processes—isobaric, isochoric, isothermal, and adiabatic.
• Calculate total work done in a cyclical thermodynamic process.

One of the most important things we can do with heat transfer is to use it to do work for us. Such a device is called a heat engine . Car engines and steam turbines that generate electricity are examples of heat engines. Figure 15.7 shows schematically how the first law of thermodynamics applies to the typical heat engine.

The illustrations above show one of the ways in which heat transfer does work. Fuel combustion produces heat transfer to a gas in a cylinder, increasing the pressure of the gas and thereby the force it exerts on a movable piston. The gas does work on the outside world, as this force moves the piston through some distance. Heat transfer to the gas cylinder results in work being done. To repeat this process, the piston needs to be returned to its starting point. Heat transfer now occurs from the gas to the surroundings so that its pressure decreases, and a force is exerted by the surroundings to push the piston back through some distance. Variations of this process are employed daily in hundreds of millions of heat engines. We will examine heat engines in detail in the next section. In this section, we consider some of the simpler underlying processes on which heat engines are based.

PV Diagrams and their Relationship to Work Done on or by a Gas

A process by which a gas does work on a piston at constant pressure is called an isobaric process . Since the pressure is constant, the force exerted is constant and the work done is given as

See the symbols as shown in Figure 15.9 . Now F = PA F = PA , and so

Because the volume of a cylinder is its cross-sectional area A A times its length d d , we see that Ad = Δ V Ad = Δ V , the change in volume; thus,

Note that if Δ V Δ V is positive, then W W is positive, meaning that work is done by the gas on the outside world.

(Note that the pressure involved in this work that we’ve called P P is the pressure of the gas inside the tank. If we call the pressure outside the tank P ext P ext , an expanding gas would be working against the external pressure; the work done would therefore be W = − P ext Δ V W = − P ext Δ V (isobaric process). Many texts use this definition of work, and not the definition based on internal pressure, as the basis of the First Law of Thermodynamics. This definition reverses the sign conventions for work, and results in a statement of the first law that becomes Δ E int = Q + W Δ E int = Q + W .)

It is not surprising that W = P Δ V W = P Δ V , since we have already noted in our treatment of fluids that pressure is a type of potential energy per unit volume and that pressure in fact has units of energy divided by volume. We also noted in our discussion of the ideal gas law that PV PV has units of energy. In this case, some of the energy associated with pressure becomes work.

Figure 15.10 shows a graph of pressure versus volume (that is, a PV PV diagram for an isobaric process. You can see in the figure that the work done is the area under the graph. This property of PV PV diagrams is very useful and broadly applicable: the work done on or by a system in going from one state to another equals the area under the curve on a PV PV diagram .

We can see where this leads by considering Figure 15.11 (a), which shows a more general process in which both pressure and volume change. The area under the curve is closely approximated by dividing it into strips, each having an average constant pressure P i ( ave ) P i ( ave ) . The work done is W i = P i ( ave ) Δ V i W i = P i ( ave ) Δ V i for each strip, and the total work done is the sum of the W i W i . Thus the total work done is the total area under the curve. If the path is reversed, as in Figure 15.11 (b), then work is done on the system. The area under the curve in that case is negative, because Δ V Δ V is negative.

PV PV diagrams clearly illustrate that the work done depends on the path taken and not just the endpoints . This path dependence is seen in Figure 15.12 (a), where more work is done in going from A to C by the path via point B than by the path via point D. The vertical paths, where volume is constant, are called isochoric processes. Since volume is constant, Δ V = 0 Δ V = 0 , and no work is done in an isochoric process. Now, if the system follows the cyclical path ABCDA, as in Figure 15.12 (b), then the total work done is the area inside the loop. The negative area below path CD subtracts, leaving only the area inside the rectangle. In fact, the work done in any cyclical process (one that returns to its starting point) is the area inside the loop it forms on a PV PV diagram, as Figure 15.12 (c) illustrates for a general cyclical process. Note that the loop must be traversed in the clockwise direction for work to be positive—that is, for there to be a net work output.

Example 15.2

Total work done in a cyclical process equals the area inside the closed loop on a pv diagram.

Calculate the total work done in the cyclical process ABCDA shown in Figure 15.12 (b) by the following two methods to verify that work equals the area inside the closed loop on the PV PV diagram. (Take the data in the figure to be precise to three significant figures.) (a) Calculate the work done along each segment of the path and add these values to get the total work. (b) Calculate the area inside the rectangle ABCDA.

To find the work along any path on a PV PV diagram, you use the fact that work is pressure times change in volume, or W = P Δ V W = P Δ V . So in part (a), this value is calculated for each leg of the path around the closed loop.

Solution for (a)

The work along path AB is

Since the path BC is isochoric, Δ V BC = 0 Δ V BC = 0 , and so W BC = 0 W BC = 0 . The work along path CD is negative, since Δ V CD Δ V CD is negative (the volume decreases). The work is

Again, since the path DA is isochoric, Δ V DA = 0 Δ V DA = 0 , and so W DA = 0 W DA = 0 . Now the total work is

Solution for (b)

The area inside the rectangle is its height times its width, or

The result, as anticipated, is that the area inside the closed loop equals the work done. The area is often easier to calculate than is the work done along each path. It is also convenient to visualize the area inside different curves on PV PV diagrams in order to see which processes might produce the most work. Recall that work can be done to the system, or by the system, depending on the sign of W W . A positive W W is work that is done by the system on the outside environment; a negative W W represents work done by the environment on the system.

Figure 15.13 (a) shows two other important processes on a PV PV diagram. For comparison, both are shown starting from the same point A. The upper curve ending at point B is an isothermal process—that is, one in which temperature is kept constant. If the gas behaves like an ideal gas, as is often the case, and if no phase change occurs, then PV = nRT PV = nRT . Since T T is constant, PV PV is a constant for an isothermal process. We ordinarily expect the temperature of a gas to decrease as it expands, and so we correctly suspect that heat transfer must occur from the surroundings to the gas to keep the temperature constant during an isothermal expansion. To show this more rigorously for the special case of a monatomic ideal gas, we note that the average kinetic energy of an atom in such a gas is given by

The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy E int E int is

where N N is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, Δ E int = 0 Δ E int = 0 . If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because Δ E int = Q − W = 0 Δ E int = Q − W = 0 here, Q = W Q = W . We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions.

Note that diatomic gases, such as those in air, have different formulas for average kinetic energy of an atom and total internal energy.

Also shown in Figure 15.13 (a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, Q = 0 Q = 0 . Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic expansion process, since work is done at the expense of internal energy:

(You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because Q = 0,  Δ E int = – W Q = 0,  Δ E int = – W for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), Figure 15.13 (b), there would be a net work output.

Reversible Processes

Both isothermal and adiabatic processes such as shown in Figure 15.13 are reversible in principle. A reversible process is one in which both the system and its environment can return to exactly the states they were in by following the reverse path. The reverse isothermal and adiabatic paths are BA and CA, respectively. Real macroscopic processes are never exactly reversible. In the previous examples, our system is a gas (like that in Figure 15.9 ), and its environment is the piston, cylinder, and the rest of the universe. If there are any energy-dissipating mechanisms, such as friction or turbulence, then heat transfer to the environment occurs for either direction of the piston. So, for example, if the path BA is followed and there is friction, then the gas will be returned to its original state but the environment will not—it will have been heated in both directions. Reversibility requires the direction of heat transfer to reverse for the reverse path. Since dissipative mechanisms cannot be completely eliminated, real processes cannot be reversible.

There must be reasons that real macroscopic processes cannot be reversible. We can imagine them going in reverse. For example, heat transfer occurs spontaneously from hot to cold and never spontaneously the reverse. Yet it would not violate the first law of thermodynamics for this to happen. In fact, all spontaneous processes, such as bubbles bursting, never go in reverse. There is a second thermodynamic law that forbids them from going in reverse. When we study this law, we will learn something about nature and also find that such a law limits the efficiency of heat engines. We will find that heat engines with the greatest possible theoretical efficiency would have to use reversible processes, and even they cannot convert all heat transfer into doing work. Table 15.2 summarizes the simpler thermodynamic processes and their definitions.

PhET Explorations

States of matter.

Watch different types of molecules form a solid, liquid, or gas. Add or remove heat and watch the phase change. Change the temperature or volume of a container and see a pressure-temperature diagram respond in real time. Relate the interaction potential to the forces between molecules.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

• Authors: Paul Peter Urone, Roger Hinrichs
• Publisher/website: OpenStax
• Book title: College Physics 2e
• Publication date: Jul 13, 2022
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
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First Law Of Thermodynamics

The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. This means that heat energy cannot be created or destroyed. It can, however, be transferred from one location to another and converted to and from other forms of energy.

Introducing State Variables

Thermodynamic state variables are the macroscopic quantities that determine a system’s thermodynamic equilibrium state. A system not in equilibrium cannot be described by state variables. State variables can further be classified as intensive or extensive variables. Intensive variables are independent of the dimensions of the system like pressure and temperature, while extensive variables depend on dimensions of the system like volume, mass, internal energy etc.

Explaining the first law of thermodynamics

The first law of thermodynamics relates to heat, internal energy, and work.

The first law of thermodynamics, also known as the law of conservation of energy, states that energy can neither be created nor destroyed, but it can be changed from one form to another.

It can be represented mathematically as

• ΔQ is the heat given or lost
• ΔU is the change in internal energy
• W is the work done

We can also represent the above equation as follows,

So we can infer from the above equation that the quantity (ΔQ – W) is independent of the path taken to change the state. Further, we can say that internal energy increases when the heat is given to the system and vice versa.

Sign Conventions

The table below shows the appropriate sign conventions for all three quantities under different conditions:

Recommended Video

First Law of Thermodynamics Solved Examples

1. Calculate the change in the system’s internal energy if 3000 J of heat is added to a system and a work of 2500 J is done.

Solution: The following sign conventions are followed in the numerical: Solution: The following sign conventions are followed in the numerical:

• Q is positive as heat is added to the system
• W is positive if work is done on the system

Hence, the change in internal energy is given as: \(\begin{array}{l}\Delta U=3000-2500\end{array} \) \(\begin{array}{l}\Delta U=500\end{array} \) The internal energy of the system is 500 J.

2. What is the change in the internal energy of the system if 2000 J of heat leaves the system and 2500 J of work is done on the system? Solution: The change in the internal energy of the system can be identified using the formula:

Substituting the values in the following equation, we get

ΔU = -2000-(-3000)

ΔU = -2000+3000

ΔU = 1000 Joule

Internal energy increases by 4500 Joules.

Suggested Reading

Frequently Asked Questions – FAQs

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12.2 First law of Thermodynamics: Thermal Energy and Work

Learning objectives.

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

• Describe how pressure, volume, and temperature relate to one another and to work, based on the ideal gas law
• Describe pressure–volume work
• Describe the first law of thermodynamics verbally and mathematically
• Solve problems involving the first law of thermodynamics

Pressure, Volume, Temperature, and the Ideal Gas Law

Before covering the first law of thermodynamics, it is first important to understand the relationship between pressure , volume, and temperature. Pressure, P , is defined as

where F is a force applied to an area, A , that is perpendicular to the force.

Depending on the area over which it is exerted, a given force can have a significantly different effect, as shown in Figure 12.3 .

The SI unit for pressure is the pascal , where 1 Pa = 1 N/m 2 . 1 Pa = 1 N/m 2 .

Pressure is defined for all states of matter but is particularly important when discussing fluids (such as air). You have probably heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids.

The relationship between the pressure, volume, and temperature for an ideal gas is given by the ideal gas law . A gas is considered ideal at low pressure and fairly high temperature, and forces between its component particles can be ignored. The ideal gas law states that

where P is the pressure of a gas, V is the volume it occupies, N is the number of particles (atoms or molecules) in the gas, and T is its absolute temperature. The constant k is called the Boltzmann constant and has the value k = 1.38 × 10 −23 J/K , k = 1.38 × 10 −23 J/K , For the purposes of this chapter, we will not go into calculations using the ideal gas law. Instead, it is important for us to notice from the equation that the following are true for a given mass of gas:

• When volume is constant, pressure is directly proportional to temperature.
• When temperature is constant, pressure is inversely proportional to volume.
• When pressure is constant, volume is directly proportional to temperature.

This last point describes thermal expansion —the change in size or volume of a given mass with temperature. What is the underlying cause of thermal expansion? An increase in temperature means that there’s an increase in the kinetic energy of the individual atoms. Gases are especially affected by thermal expansion, although liquids expand to a lesser extent with similar increases in temperature, and even solids have minor expansions at higher temperatures. This is why railroad tracks and bridges have expansion joints that allow them to freely expand and contract with temperature changes.

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 a 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 you continue to pump air into tire (which now has a nearly constant volume), the pressure increases with increasing temperature (see Figure 12.4 ).

Pressure–Volume Work

Pressure–volume work is the work that is done by the compression or expansion of a fluid. Whenever there is a change in volume and external pressure remains constant, pressure–volume work is taking place. During a compression, a decrease in volume increases the internal pressure of a system as work is done on the system. During an expansion ( Figure 12.5 ), an increase in volume decreases the internal pressure of a system as the system does work.

Recall that the formula for work is W = F d . W = F d . We can rearrange the definition of pressure, P = F A , P = F A , to get an expression for force in terms of pressure.

Substituting this expression for force into the definition of work, we get

Because area multiplied by displacement is the change in volume, W = P Δ V W = P Δ V , the mathematical expression for pressure–volume work is

Just as we say that work is force acting over a distance, for fluids, we can say that work is the pressure acting through the change in volume. For pressure–volume work, pressure is analogous to force, and volume is analogous to distance in the traditional definition of work.

Watch Physics

Work from expansion.

This video describes work from expansion (or pressure–volume work). Sal combines the equations W = P Δ V W = P Δ V and Δ U = Q − W Δ U = Q − W to get Δ U = Q − P Δ V Δ U = Q − P Δ V .

If the volume of a system increases while pressure remains constant, is the value of work done by the system W positive or negative? Will this increase or decrease the internal energy of the system?

• Positive; internal energy will decrease
• Positive; internal energy will increase
• Negative; internal energy will decrease
• Negative; internal energy will increase

The First Law of Thermodynamics

Heat ( Q ) and work ( W ) are the two ways to add or remove energy from a system. The processes are very different. Heat is driven by temperature differences, while work involves a force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air’s temperature. Similarly, work can be done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat or work. Heat and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy, U , of a system.

Internal energy is the sum of the kinetic and potential energies of a system’s atoms and molecules. It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P , V , and T ), not on how the energy enters or leaves the system.

In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics . The first law of thermodynamics applies the conservation of energy principle to systems where heat and work are the methods of transferring energy into and out of the systems. It can also be used to describe how energy transferred by heat is converted and transferred again by work.

Tips For Success

Recall that the principle of conservation of energy states that energy cannot be created or destroyed, but it can be altered from one form to another.

The first law of thermodynamics states that the change in internal energy of a closed system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is

Here, Δ U Δ U is the change in internal energy , U , of the system. As shown in Figure 12.6 , Q is the net heat transferred into the system —that is, Q is the sum of all heat transfers into and out of the system. W is the net work done by the system —that is, W is the sum of all work done on or by the system. By convention, if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system by heat, and positive W takes energy from the system by work. Note that if heat transfers more energy into the system than that which is done by work, the difference is stored as internal energy.

It follows also that negative Q indicates that energy is transferred away from the system by heat and so decreases the system’s internal energy, whereas negative W is work done on the system, which increases the internal energy.

First Law of Thermodynamics/Internal Energy

This video explains the first law of thermodynamics, conservation of energy, and internal energy. It goes over an example of energy transforming between kinetic energy, potential energy, and heat transfer due to air resistance.

• Both will decrease. Energy is transferred to the air by heat due to air resistance.
• Both will increase. Energy is transferred from the air to the ball due to air resistance.
• Final velocity will increase, but final kinetic energy will decrease. Energy is transferred by heat to the air from the ball through air resistance.
• Final velocity will decrease, but final kinetic energy will increase. Energy is transferred by heat from the air to the ball through air resistance.

More on Internal Energy

This video goes into further detail, explaining internal energy and how to use the equation Δ U = Q − W . Δ U = Q − W . Note that Sal uses the equation Δ U = Q + W Δ U = Q + W , where W is the work done on the system, whereas we use W to represent work done by the system.

Links To Physics

Biology: biological thermodynamics.

We often think about thermodynamics as being useful for inventing or testing machinery, such as engines or steam turbines. However, thermodynamics also applies to living systems, such as our own bodies. This forms the basis of the biological thermodynamics ( Figure 12.7 ).

Life itself depends on the biological transfer of energy. Through photosynthesis, plants absorb solar energy from the sun and use this energy to convert carbon dioxide and water into glucose and oxygen. Photosynthesis takes in one form of energy—light—and converts it into another form—chemical potential energy (glucose and other carbohydrates).

Human metabolism is the conversion of food into energy given off by heat, work done by the body’s cells, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. Eating increases the internal energy of the body by adding chemical potential energy; this is an unromantic view of a good burrito.

The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Exercise helps you lose weight, because it provides energy transfer from your body by both heat and work and raises your metabolic rate even when you are at rest.

Biological thermodynamics also involves the study of transductions between cells and living organisms. Transduction is a process where genetic material—DNA—is transferred from one cell to another. This often occurs during a viral infection (e.g., influenza) and is how the virus spreads, namely, by transferring its genetic material to an increasing number of previously healthy cells. Once enough cells become infected, you begin to feel the effects of the virus (flu symptoms—muscle weakness, coughing, and congestion).

Energy is transferred along with the genetic material and so obeys the first law of thermodynamics. Energy is transferred—not created or destroyed—in the process. When work is done on a cell or heat transfers energy to a cell, the cell’s internal energy increases. When a cell does work or loses heat, its internal energy decreases. If the amount of work done by a cell is the same as the amount of energy transferred in by heat, or the amount of work performed on a cell matches the amount of energy transferred out by heat, there will be no net change in internal energy.

Based on what you know about heat transfer and the first law of thermodynamics, do you need to eat more or less to maintain a constant weight in colder weather? Explain why.

• more; as more energy is lost by the body in colder weather, the need to eat increases so as to maintain a constant weight
• more; eating more food means accumulating more fat, which will insulate the body from colder weather and will reduce the energy loss
• less; as less energy is lost by the body in colder weather, the need to eat decreases so as to maintain a constant weight
• less; eating less food means accumulating less fat, so less energy will be required to burn the fat, and, as a result, weight will remain constant

Solving Problems Involving the First Law of Thermodynamics

Worked example, calculating change in internal energy.

Suppose 40.00 J of energy is transferred by heat to a system, while the system does 10.00 J of work. Later, heat transfers 25.00 J out of the system, while 4.00 J is done by work on the system. What is the net change in the system’s internal energy?

You must first calculate the net heat and net work. Then, using the first law of thermodynamics, Δ U = Q − W , Δ U = Q − W , find the change in internal energy.

The net heat is the transfer into the system by heat minus the transfer out of the system by heat, or

The total work is the work done by the system minus the work done on the system, or

The change in internal energy is given by the first law of thermodynamics.

A different way to solve this problem is to find the change in internal energy for each of the two steps separately and then add the two changes to get the total change in internal energy. This approach would look as follows:

For 40.00 J of heat in and 10.00 J of work out, the change in internal energy is

For 25.00 J of heat out and 4.00 J of work in, the change in internal energy is

The total change is

No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.

Calculating Change in Internal Energy: The Same Change in U is Produced by Two Different Processes

What is the change in the internal energy of a system when a total of 150.00 J is transferred by heat from the system and 159.00 J is done by work on the system?

The net heat and work are already given, so simply use these values in the equation Δ U = Q − W . Δ U = Q − W .

Here, the net heat and total work are given directly as Q = − 150 .00 J and  W = − 159.00 J, Q = − 150 .00 J and  W = − 159.00 J, so that

A very different process in this second worked example produces the same 9.00 J change in internal energy as in the first worked example. Note that the change in the system in both parts is related to Δ U Δ U and not to the individual Q ’s or W ’s involved. The system ends up in the same state in both problems. Note that, as usual, in Figure 12.8 above, W out W out is work done by the system, and W in W in is work done on the system.

Practice Problems

Check your understanding.

• Pressure is force divided by length.
• Pressure is force divided by area.
• Pressure is force divided by volume.
• Pressure is force divided by mass.

What is the SI unit for pressure?

• pascal, or N/m 3
• pascal, or N/m 2
• It is the work that is done by the compression or expansion of a fluid.
• It is the work that is done by a force on an object to produce a certain displacement.
• It is the work that is done by the surface molecules of a fluid.
• It is the work that is done by the high-energy molecules of a fluid.

When is pressure-volume work said to be done ON a system?

• When there is an increase in both volume and internal pressure.
• When there is a decrease in both volume and internal pressure.
• When there is a decrease in volume and an increase in internal pressure.
• When there is an increase in volume and a decrease in internal pressure.
• Transferring energy by heat is the only way to add energy to or remove energy from a system.
• Doing compression work is the only way to add energy to or remove energy from a system.
• Doing expansion work is the only way to add energy to or remove energy from a system.
• Transferring energy by heat or by doing work are the ways to add energy to or remove energy from a system.
• It is the sum of the kinetic energies of a system’s atoms and molecules.
• It is the sum of the potential energies of a system’s atoms and molecules.
• It is the sum of the kinetic and potential energies of a system’s atoms and molecules.
• It is the difference between the magnitudes of the kinetic and potential energies of a system’s atoms and molecules.

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Related Items

First Law of Thermodynamics

The first law of thermodynamics simply states that energy is conserved and cannot be destroyed. Energy can only be converted from one form to another. This means that energy can only be converted from potential to kinetic.

• Zeroth Law of Thermodynamics

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