One useful application of thermodynamics is in the calculation of work and heat effects for many different processes by applying the fi rst law. This information allows engineers to use energy more effi ciently, saving costs and resources. Since heat and work are path dependent, the specifi c process must be defi ned in order to perform the necessary cal- culations. In this section, we go through two such examples of these types of calculations using an ideal gas undergoing reversible processes. We will look at nonideal gases in Chapter 5. The intent is to gain some experience with applying the fi rst law to get values for work and heat as well as to develop expressions that are useful in understanding the Carnot cycle (Section 2.9).
Reversible, Isothermal Expansion (Compression)
Consider a reversible, isothermal expansion of an ideal gas. A schematic of a piston–
cylinder assembly undergoing such a process is shown in Figure 2.15. The gas is kept at constant temperature by keeping it in contact with a thermal reservoir. A thermal reservoir contains enough mass so that its temperature does not noticeably change dur- ing the process. Can you predict the signs of DU, Q, and W ?
Since the internal energy of an ideal gas is only a function of temperature, DU50
For a reversible process, we can integrate over the system pressure (see Section 2.3):
W5 23PdV (2.36)
Applying the ideal gas relationship:
V5 nRT P
Figure 2.15 An ideal gas in a piston–cylinder assembly undergoing a reversible, isothermal expan- sion. See if you can predict the signs of DU, Q, and W for this process in the table.
Ideal gas W Q ΔU
Ideal gas
Initial state (1) Constant T
reservoir
Constant T reservoir
Final state (2) Process
Positive (+) Negative (−) Zero (0)
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2.7 Reversible Processes in Closed Systems ◄ 93 the differential in volume can be transformed into a differential in pressure (remember- ing, for this case, that T is constant):
dV5 2nRT
P2 dP (2.37)
Substituting Equation (2.37) into Equation (2.36) and integrating gives:
W53
2
1
nRT
P dP5nRT ln P2
P1
(2.38) Now applying the fi rst law, we get:
Q5 DU2W5 2nRT ln P2
P1
(2.39) Since P2,P1, the sign for W is negative and for Q is positive. Did you get the sign right in the table in Figure 2.15? How do Equations (2.38) and (2.39) change if the gas under- goes a compression process instead of an expansion?
Adiabatic Expansion (Compression) with Constant Heat Capacity
Consider when the same ideal gas undergoes an adiabatic, reversible expansion (as opposed to isothermal). We will assume that the heat capacity of this gas does not change with temperature, that is, constant heat capacity. This process is illustrated in Figure 2.16.
Again, can you predict the signs of DU, Q, and W ?
Neglecting macroscopic kinetic and potential energy, the fi rst law for a closed sys- tem in differential form is obtained from Equation (2.14):
dU5 dQ1 dW (2.40) where the heat transfer was set to zero, since this process is adiabatic. From Equation (2.24) we get:
dU5ncvdT (2.41)
and for a reversible process:
dW5 2PdV (2.42)
Substituting Equations (2.41) and (2.42) into Equation (2.40) yields:
ncvdT5 2PdV (2.43)
We can use the ideal gas law to relate the measured properties T, V, and P:
d1nRT2 5d1PV25PdV1VdP (2.44)
where we applied the product rule. Solving Equation (2.44) for dT and then plugging back into Equation (2.43) and rearranging gives:
cvVdP5 21cv1R2PdV5 2cPPdV (2.45) 0
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94 ► Chapter 2. The First Law of Thermodynamics
Separating variables in Equation (2.45):
2cP
cv
dV V 5dP
P (2.46)
Now we integrate Equation (2.46) from the initial state 1 to the fi nal state 2, 2k ln¢V2
V1≤ 5ln¢P2
P1≤ (2.47)
where k5cP/cv. Applying mathematical relationships of the natural logarithm, we can rewrite the left-hand side of Equation (2.47) as:
2k ln¢V2
V1≤ 5ln¢V2
V1≤2k5ln¢V1
V2≤
k
so,
ln1P1V1k2 5ln1P2V2k2 or,
PVk5const (2.48)
Now integrating for work:
W5 23PdV5 23const V2kdV5 const k21B 1
V2k21 2 1 V1k21R 5 1
k213P2V22P1V145 nR
k213T22T14 From the fi rst law:
DU5W5 1
k213P2V22P1V14 5 nR
k213T22T14
Ideal gas
Cv ≠ Cv(T) Well-
insulated
Well- insulated Ideal gas
W Q ΔU
Initial state (1) Final state (2) Process
Positive (+) Negative (−) Zero (0)
Figure 2.16 An ideal gas in a piston–cylinder assembly undergoing a reversible, adiabatic expan- sion. In this example, cv is constant. See if you can predict the signs of DU, Q, and W for this process in the table.
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►Summary
A summary of the two cases presented in this section is shown in Table 2.2. In both cases, expansion of a piston provides useful energy to the surroundings in the form of work.
However, each case represents a limit. In the isothermal process, all the energy delivered as work is provided by the surroundings in the form of heat. On the other hand, for the adiabatic case, the energy for work is provided by the internal energy of the gas in the system. An intermediate case exists where there is some heat adsorbed from the sur- roundings as well as some “cooling” of the gas in the system.
A process is defi ned as polytropic if it follows the relation:
PVg5const (2.49)
Both the processes in this section can be considered polytropic. The isothermal expan- sion of an ideal gas follows Equation (2.49) with g 51 while the reversible, adiabatic expansion of an ideal gas with constant heat capacity has g 5k5cP/cv. Can you think of another example of a polytropic process?
TABLE 2.2 Summary of Expressions for Change in Internal Energy, Heat, and Work for an Ideal Gas Undergoing a Reversible Process
Isothermal Adiabatic, cv2cv1T2
DU 0 nR
k213T22T14 Q 2nRT ln P2
P1 0
W nRT ln P2
P1
nR
k213T22T14