Physically, work is performed on an object when the object moves some distance s due to the application of a force F. Mathematically, it is the dot product of the force vector F and the distance vector s:
work 5F#s5 0F0 0s0 cos u (2.1) where u is the angle between the vectors. Work is a scalar, not vector, quantity.
Work has magnitude, but not direction. Figure 2.1 shows a force acting on an object. In Figure 2.1a, the object is not moving, so the amount of work is zero (despite the amount of force being exerted). In Figure 2.1b, an object has been moved, so work was done.
The First Law of Thermodynamics
2.1 Synopsis 2.2 Work and Heat
2.3 Internal Energy and the First Law of Thermodynamics 2.4 State Functions 2.5 Enthalpy
2.6 Changes in State Functions 2.7 Joule-Thomson Coefficients 2.8 More on Heat Capacities 2.9 Phase Changes 2.10 Chemical Changes 2.11 Changing Temperatures 2.12 Biochemical Reactions 2.13 Summary
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32 Chapter 2 | The First Law of Thermodynamics
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Work has units of joules, like energy. This is not without a reason: Work is a way to transfer energy. Energy is defined as the ability to do work, so it makes sense that energy and work are described using the same units.
The most common form of work studied by basic thermodynamics involves the changing volume of a system. Consider Figure 2.2a. A frictionless piston confines a sample of a gas in an initial volume Vi. The gas inside the chamber also has an initial pressure pi. Initially, what keeps the piston at a fixed position is the external pressure of the surroundings, pext.
If the piston moves out, Figure 2.2b, then the system is doing work on the surround- ings. That means that the system is losing energy in the form of work. The infinitesimal amount of work dw lost by the system to the surroundings for an infinitesimal change in volume dV while acting against a constant external pressure pext is defined as
dw5 2pext dV (2.2)
The negative sign indicates that the work done contributes to a decrease in the amount of energy of the system.* If the piston moves inward, Figure 2.2c, then the surroundings are doing work on the system, and the amount of energy in the system is increased. The infinitesimal amount of work done on the system is defined by equation 2.2, but because the volume change dV is in the opposite direction, the work now has a positive value. Notice that our focus is the system. The work is positive or negative with respect to the system, which is the part of the universe of interest to us.
If we add up all of the (infinite) infinitesimal changes that contribute to an over- all change, we get the total amount of work done on or by the system. Calculus uses the integral to add up infinitesimal changes. The total amount of work, w, for a change as represented in Figure 2.2 is therefore
Initial position F
s 5 0 Work 5 0
F
s . 0
Work . 0
Initial position
(a) (b)
Figure 2.1 When a force is exerted on an object, no work is done unless the object moves. (a) Because the wall does not move, no work is done. (b) Work is done because the force is acting through a distance.
Figure 2.2 (a) A frictionless piston with an enclosed gas is a simple example of how gases perform work on systems or surroundings. (b) The work is done on the surroundings. (c) The work is done on the system. The mathematical definition of work remains the same, however.
pext
Vi, pi
(a)
dw 5 2pext dV; and dV is positive work done by system
on surroundings (system loses energy)
(b)
dw 5 2pext dV; and dV is negative work done on system
by surroundings (system gains energy)
(c)
*It is easy to show that the two definitions of work are equivalent. Because pressure is force per unit area, equation 2.2 can be rewritten as work5forcearea3volume5force3distance, which is equation 2.1.
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2.2 | Work and Heat 33
w5 23pext dV (2.3)
Whether this integral can be simplified or not depends on the conditions of the process. If the external pressure remains constant throughout the process, then it can be removed to outside the integral and the expression becomes
w5 23pext dV5 2pext3dV5 2pext#V0VVfi
In this case, the limits on the integral are the initial volume, Vi, and the final volume, Vf, of the process. This is reflected in the last expression in the equation above. Evaluating the integral at its limits, we get
w5 2pext1Vf2Vi2
w5 2pext#DV (2.4)
If the external pressure is not constant throughout the process, then we will need some other way of evaluating the work in equation 2.3.
By using pressures in units of atm and volumes in units of L, we get a unit of work in L#atm. This is not a common work unit. The SI unit for work is the joule, J.
However, using the various values of R from the previous chapter, it can be shown that 1 L#atm = 101.32 J. This conversion factor is very useful to get work into its proper SI units. If volume were expressed in units of m3 and pressure in pascals, units of joules would be obtained directly because
Pa3m35 N
m23m35N3m5J
Figure 2.3 illustrates a condition that occasionally occurs with gases: the expansion of a gas into a larger volume, which is initially a vacuum. In such a case, because the gas is expanding against a pext of 0, by the definition of work in equation 2.4 the work done by the gas equals zero. Such a process is called a free expansion:
work 5 0 for free expansion (2.5)
ExamplE 2.1
Consider an ideal gas in a piston chamber, as in Figure 2.2, where the initial volume is 2.00 L and the initial pressure is 8.00 atm. Assume that the piston is moving up (that is, the system is expanding) to a final volume of 5.50 L against a constant external pressure of 1.75 atm. Also assume constant temperature for the process.
a. Calculate the work for the process.
b. Calculate the final pressure of the gas.
Solution
a. First, the change in volume is needed. We find it as follows:
DV 5 Vf2 Vi5 5.50 L 2 2.00 L 5 3.50 L
To calculate the work against the constant external pressure, we use equation 2.4:
w 52pext DV 52(1.75 atm)(3.50 L) 526.13 L # atm
A change is always “final value minus initial value.”
Note how the final units are “liter atmospheres.”
p 5 0 p fi 0
p fi 0 p fi 0
Figure 2.3 No work is performed if a sample of gas expands into a vacuum.
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34 Chapter 2 | The First Law of Thermodynamics
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ExamplE 2.2
From the conditions and the given definitions of the system, determine whether there is work done by the system, work done on the system, or no work done.
a. A balloon expands as a small piece of dry ice (solid CO2) inside the balloon sublimes (balloon 5 system).
b. The space shuttle’s cargo bay doors are opened to space, releasing a little bit of residual atmosphere (cargo bay 5 system).
c. Gaseous CHF2Cl, a refrigerant, is compressed in the compressor of an air conditioner, to liquefy it (CHF2Cl 5 system).
d. A can of spray paint is discharged (can 5 system).
e. Same as part d, but consider the spray to be the system.
Solution
a. Because the balloon is increasing in volume, it is undoubtedly doing work:
Work is done by the system.
b. When the shuttle’s cargo bay doors are opened in space, the bay is being opened to vacuum (although not a perfect one), so we are considering relatively free expansion. Therefore, no work is done.
c. When CHF2Cl is compressed, its volume is decreased, so work is being done on the system.
d, e. When a can of spray paint is discharged, the can itself usually does not change in volume. Therefore, if the can itself is defined as the system, the amount of work it does is zero. However, work is done by the spray itself as it expands against the atmosphere. This last example shows how important it is to define the system as specifically as possible.
If it were possible, we could change the volume of the gas inside the piston chamber in infinitesimally small steps, allowing the system to react to each infinitesimal change before making the next change. At each step, the system comes to equilibrium with its surroundings so that the entire process is one of a continuous equilibrium state. (In reality, that would require an infinite number of steps for any finite change in volume.
Sufficiently slow changes are a good approximation.) Such a process is called reversible.
Processes that are not performed this way (or are not approximated this way) are called If we want to convert units to the SI units of joules, we use the appropriate con- version factor:
26.13 L#atm3 101.32 J
1 L#atm 5 2621 J
That is, 621 joules have been lost by the system during the expansion.
b. Because of the assumption of an ideal gas, we can use Boyle’s law to calculate the final gas pressure. We get
(2.00 L)(8.00 atm) 5 (5.50 L)(pf) pf5 2.91 atm ExamplE 2.1 (continued)
Recall that Boyle’s law is piVi5 pfVf.
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2.2 | Work and Heat 35 irreversible. Many thermodynamic ideas are based on systems that undergo reversible
processes. Volume changes aren’t the only processes that can be reversible. Thermal changes, mechanical changes (that is, moving a piece of matter), and other changes can be modeled as reversible or irreversible.
Gaseous systems are useful examples for thermodynamics because we can use various gas laws to help us calculate the amount of pressure-volume work when a system changes volume. This is especially so for reversible changes, because most reversible changes occur by letting the external pressure equal the internal pressure:
pext5pint for reversible change
The following substitution can then be made for reversible changes:
wrev5 23pint dV (2.6)
So now we can determine the work for a process in terms of the internal pressure.
The ideal gas law can be used to substitute for the internal pressure, because if the system is composed of an ideal gas, the ideal gas law must hold. We can get
wrev5 23nRT V dV
when we substitute for pressure. Although n and R are constants, the temperature T is a variable and may change. However, if the temperature does remain constant for the process, the term isothermal is used to describe the process, and the tem- perature “variable” can be taken outside of the integral sign. Volume remains inside the integral because it is the variable being integrated. We get
wrev5 2nRT31 V dV
This integral has a standard form; we can evaluate it. The equation becomes wrev5 2nRT (ln V0VVfi)
The “ln” refers to the natural logarithm, not the base 10 logarithm (which is repre- sented by “log”). Evaluating the integral at its limits,
wrev5 2nRT (ln Vf2ln Vi) which, using the properties of logarithms, is
wrev5 2nRT ln Vf
Vi (2.7)
for an isothermal, reversible change in the conditions of an ideal gas. Using Boyle’s law for gas, we can substitute the expression piypf for the volumes in the logarithm and also see that
wrev5 2nRT ln pi
pf (2.8)
for an ideal gas undergoing an isothermal process.
ExamplE 2.3
Gas in a piston chamber kept in a constant-temperature bath at 25.0°C expands from 25.0 mL to 75.0 mL very, very slowly, as illustrated in Figure 2.4. If there is 0.00100 mole of ideal gas in the chamber, calculate the work done by the system.
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36 Chapter 2 | The First Law of Thermodynamics
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Solution
Because the system is kept in a constant-temperature bath, the change is an isothermal one. Also, because the change is very, very slow, we can presume that the change is reversible. Therefore, we can use equation 2.7. We find
wrev5 2(0.00100 mol)a8.314 J
mol#Kb(298.15 K)aln 75.0 mL 25.0 mLb wrev5 22.72 J
That is, 2.72 J is lost by the system.
ExamplE 2.3 (continued)
Note that we use the value of R that has joules as part of its units. This gives the answer directly in joule units.
Other equations of state can be used to calculate work, starting with the definition in equation 2.6. The next example illustrates this.
ExamplE 2.4
Repeat the previous example using 0.00100 mol of SO2 and using the van der Waals equation of state. The van der Waals parameters for SO2 are a 5 6.714 atm # L2ymol2 and b 5 0.05636 Lymol.
Solution
First, we need to derive an expression for work caused by a van der Waals gas.
The original definition for work from equation 2.6 is w5 23pext dV
If the expansion is reversible, then pext 5 pint. The pressure for a van der Waals gas can be expressed by rearranging the van der Waals equation of state:
ap1an2
V2b(V2nb) 5nRT p5 nRT
V2nb2 an2 V2 This is the van der Waals equation of state.
Here we have rearranged to solve for p.
n 5 0.001 mol Constant- temperature bath
Slowly 75.0 mL
25.0°C
25.0 mL
Figure 2.4 A piston chamber in a constant-temperature bath, undergoing a reversible change in volume. See Example 2.3.
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2.2 | Work and Heat 37
Heat, symbolized by the letter q, is more difficult to define than work. Heat is a mea- sure of thermal energy transfer that can be determined by the change in the temperature of an object. That is, heat is a way of following a change in energy of a system. Because heat is a change in energy, we use the same units for heat as we do for energy: joules.
Even historically, heat was a difficult concept. It was thought that heat was a property of a system that could be isolated and bottled as a substance in its own right. This sub- stance was even given a name: “caloric.” However, around 1780 Benjamin Thompson, later Count Rumford, kept track of the production of heat during the boring of cannon barrels and concluded that the amount of heat was related to the amount of work done in the process. In the 1840s, careful experiments by the English physicist James Prescott Joule (Figure 2.5) verified this. A brewer at the time, Joule used an apparatus like the one shown in Figure 2.6 to perform the work of mixing a quantity of water using a weight on a pulley. By making careful measurements of the temperature of the water and of the work being performed by the falling weight (using equation 2.1), Joule was able to support the idea that work and heat were manifestations of the same thing. (In fact, the phrase “me- chanical equivalent of heat” is still used occasionally and emphasizes their relationship.) The SI unit of energy and work and heat, the joule, is named in Joule’s honor.
The older unit of energy and heat and work, the calorie, is defined as the amount of heat needed to raise the temperature of exactly 1 mL of water by 1°C from 15°C to 16°C. The relationship between the calorie and the joule is
1 calorie 5 4.184 joules
Although joules are the accepted SI unit, the unit of calorie is still used often, especially in the United States.
Substituting this expression for the pressure:
w5 23
Vf
Vi
a nRT
V2nb2 an2 V2b dV We can actually integrate this. First, we separate it into two parts:
w5 23
Vf
Vi
a nRT
V2nbb dV13
Vf
Vi
aan2 V2b dV
Now we can integrate. The first term integrates to the natural logarithm, and the second term integrates into a function of 1/V:
w5 2nRT ln(V2nb)0VVfi2an2a1 Vb 0VVfi Substituting the limits and simplifying, this becomes
w5 2nRT ln (Vf2nb)
(Vi2nb) 2an2a l Vf2 l
Vib
We have all of the data we need to perform the calculation. To be consistent with units, the volumes need to be converted to L: Vi5 0.025 L and Vf5 0.075 L. Substi- tuting and solving:
w5 22.73 J2(0.01814) J w5 22.71 J
ExamplE 2.4 (continued)
A minus sign is reintroduced in the second term because of the integration.
This is very close to the answer from the previous example, but it shows that we can use nonideal equations of state to calculate work.
Figure 2.5 James Prescott Joule (1818–1889), English physicist. His work established the inter-conversion of heat and work as forms of energy, and laid the foun- dation for the first law of thermodynamics.
© Bettmann/Corbis
The two minus signs cancel, so the second term is positive.
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38 Chapter 2 | The First Law of Thermodynamics
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Heat can go into a system, so that the temperature of the system increases, or it can come out of a system, in which case the temperature of the system decreases.
For any change where heat goes into a system, q is positive. On the other hand, if heat comes out of a system, q is negative. The sign on q therefore tells one the direction of the heat transfer.
The same change in temperature requires a different amount of heat for different materials. For example, a system composed of 10 cm3 of iron metal gets hotter with less heat than does 10 cm3 of water. In fact, the amount of heat necessary to change the temperature is proportional to the magnitude of the temperature change, DT, and the mass of the system m:
q~m #DT
In order to convert a proportionality to an equality, a proportionality constant is needed. For the above expression, the proportionality constant is represented by the letter c (sometimes s) and is called the specific heat (or specific heat capacity):
q5m # c #DT (2.9)
The specific heat is an intensive characteristic of the material composing the system. Materials with a low specific heat, like many metals, need little heat for a relatively large change in temperature. Table 2.1 lists specific heat of select materials.
Units for specific heat are (energy)y(mass#temperature) or (energy)y(moles# temperature), so although the SI units for specific heat are Jyg # K or Jymol # K, it is not unusual to see specific heat having units of calymol # °C or some other set of units. Notice that, because equation 2.9 involves the change in temperature, it does not matter if the temperature has units of kelvins or degrees Celsius.
Table 2.1 Specific heat capacities of various materials
Material c (Jyg # K)
Al 0.900
Al2O3 1.275
C2H5OH, ethanol 2.42 C6H6, benzene (vapors) 1.05 C6H14, n-hexane 1.65
Cu 0.385
Fe 0.452
Fe2O3 0.651
H2 (g) 14.304
H2O (s) 2.06
H2O (,), 25°C 4.184 H2O (g), 25°C 1.864 H2O, steam, 100°C 2.04
He 5.193
Hg 0.138
N2 1.040
NaCl 0.864
O2 (g) 0.918
Figure 2.6 Joule used this apparatus to measure what was once called the “mechanical equivalent of heat.”
Science and Society/SuperStock
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2.2 | Work and Heat 39 Heat capacity C is an extensive property that includes the amount of material in
the system, so equation 2.9 would be written as q5C # DT Units on C are typically J/K.
ExamplE 2.5
a. Assuming that 400. J of energy is put into 7.50 g of iron, what will be the change in temperature? Use c 5 0.450 Jyg # K.
b. If the initial temperature of the iron is 65.0°C, what is the final temperature?
Solution
a. Using equation 2.9:
1400. J 5 (7.50 g)(0.450 Jyg # K)DT Solving for DT:
DT 5 1118 K