PROBLEMS
2.8 PERFECT GASES AND THE FIRST LAW Perfect Gases
Chapter 2
The First Law of Thermodynamics 58
2.8 PERFECT GASES AND THE FIRST LAW
Section 2.8 Perfect Gases and the First Law 59
EXAMPLE 2.4
Calculation of q, w, and USuppose 0.100 mol of a perfect gas having CV,m 1.50R independent of tem-perature undergoes the reversible cyclic process 1 → 2 → 3 → 4 → 1 shown in Fig. 2.10, where either P or V is held constant in each step. Calculate q, w, and
U for each step and for the complete cycle.
Since we know how P varies in each step and since the steps are reversible, we can readily find w for each step by integrating dwrev P dV. Since either V or P is constant in each step, we can integrate dqV CVdT and dqP CPdT [Eqs. (2.51) and (2.52)] to find the heat in each step. The first law U q w then allows calculation of U.
To evaluate integrals like 兰21CVdT, we will need to know the temperatures of states 1, 2, 3, and 4. We therefore begin by using PV nRT to find these tem-peratures. For example, T1 P1V1/nR 122 K. Similarly, T2 366 K, T3 732 K, T4 244 K.
Step 1 → 2 is at constant volume, no work is done, and w1→2 0. Step 2 → 3 is at constant pressure and
where two values of R were used to convert to joules. Similarly, w3→4 0 and w4→1 101 J. The work w for the complete cycle is the sum of the works for the four steps, so w 304 J 0 101 J 0 203 J.
Step 1 → 2 is at constant volume, and
Step 2 → 3 is at constant pressure, and q2→3 兰32CPdT. Equation (2.72) gives CP,m CV,m R 2.50R, and we find q2→3 761 J. Similarly, q3→4
608 J and q4→1 253 J. The total heat for the cycle is q 304 J We have U1→2 q1→2 w1→2 304 J 0 304 J. Similarly, we find
U2→3 457 J, U3→4 608 J, U4→1 152 J. For the complete cycle,
U 304 J 457 J 608 J 152 J 0, which can also be found from q w as 203 J 203 J 0. An alternative procedure is to use the perfect-gas equation dU CVdT to find U for each step.
For this cyclic process, we found
are consistent with the fact that U is a state function but q and w are not.
Exercise
Use the perfect-gas equation dU CVdT to find U for each step in the cycle of Fig. 2.10. (Answer: 304 J, 456 J, 609 J, 152 J.)
Exercise
Verify that w for the reversible cyclic process in this example equals minus the area enclosed by the lines in Fig. 2.10.
1 2 1
2
1 2 1
2
761 J 60812 J 25312 J 203 J.
1 2 1
2
10.100 mol21.5038.314 J>1mol K2 4 1366 K 122 K2 304 J q1S2
冮
12CVdT nCV,m冮
12dT n11.50R2 1T2 T123000 cm3 atm 18.314 J2>182.06 cm3 atm2 304 J
w2S3
冮
23P dV P1V3 V22 13.00 atm2 12000 cm3 1000 cm32Figure 2.10
A reversible cyclic process.
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Chapter 2
The First Law of Thermodynamics 60
Reversible Isothermal Process in a Perfect Gas
Consider the special case of a reversible isothermal (constant-T) process in a perfect gas. (Throughout this section, the system is assumed closed.) For a fixed amount of a perfect gas, U depends only on T [Eq. (2.67)]. Therefore U 0 for an isothermal change of state in a perfect gas. This also follows from dU CVdT for a perfect gas.
The first law U q w becomes 0 q w and q w. Integration of dwrev
P dV and use of PV nRT give
(2.74) where Boyle’s law was used. If the process is an expansion (V2 V1), then w (the work done on the gas) is negative and q (the heat added to the gas) is positive. All the added heat appears as work done by the gas, maintaining U as constant for the perfect gas.
It is best not to memorize an equation like (2.74), since it can be quickly derived from dw P dV.
To carry out a reversible isothermal volume change in a gas, we imagine the gas to be in a cylinder fitted with a frictionless piston. We place the cylinder in a very large constant-temperature bath (Fig. 2.11) and change the external pressure on the piston at an infinitesimal rate. If we increase the pressure, the gas is slowly compressed. The work done on it will transfer energy to the gas and will tend to increase its temperature at an infinitesimal rate. This infinitesimal temperature increase will cause heat to flow out of the gas to the surrounding bath, thereby maintaining the gas at an essentially constant temperature. If we decrease the pressure, the gas slowly expands, thereby doing work on its surroundings, and the resulting infinitesimal drop in gas temperature will cause heat to flow into the gas from the bath, maintaining constant temperature in the gas.
EXAMPLE 2.5
Calculation of q, w, and UA cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P 1.00 atm and is in a large constant-temperature bath at 400 K. The pressure is re-versibly increased to 5.00 atm. Find w, q, and U for this process.
It is an excellent approximation to consider the helium as a perfect gas.
Since T is constant, U is zero [Eq. (2.68)]. Equation (2.74) gives
Also, q w 1.61 104J. Of course, w (the work done on the gas) is pos-itive for the compression. The heat q is negative because heat must flow from the gas to the surrounding constant-temperature bath to maintain the gas at 400 K as it is compressed.
Exercise
0.100 mol of a perfect gas with CV,m 1.50R expands reversibly and isother-mally at 300 K from 1.00 to 3.00 L. Find q, w, and U for this process. (Answer:
274 J, 274 J, 0.)
Reversible Constant-P (or Constant-V ) Process in a Perfect Gas The calculations of q, w, and U for these processes were shown in Example 2.4.
w19980 J2 11.6092 1.61 104 J
w13.00 mol2 18.314 J mol1 K12 1400 K2 ln 15.00>1.002 19980 J2 ln 5.00 w q nRT ln V1
V2 nRT ln P2
P1 rev. isothermal proc., perf. gas w
冮
12PdV冮
12 nRTV dV nRT冮
12V1 dV nRT1ln V2 ln V12System
Bath
Figure 2.11
Setup for an isothermal volume change.
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Reversible Adiabatic Process in a Perfect Gas
For an adiabatic process, dq 0. For a reversible process in a system with only P-V work, dw P dV. For a perfect gas, dU CVdT [Eq. (2.68)]. Therefore, for a re-versible adiabatic process in a perfect gas, the first law dU dq dw becomes
where PV nRT and CV,m CV/n were used. To integrate this equation, we separate the variables, putting all functions of T on one side and all functions of V on the other side. We get (CV,m/T ) dT (R/V)dV. Integration gives
(2.75) For a perfect gas, CV,mis a function of T [Eq. (2.69)]. If the temperature change in the process is small, CV,mwill not change greatly and can be taken as approximately constant.
Another case where CV,mis nearly constant is for monatomic gases, where CV,mis essen-tially independent of T over a very wide temperature range (Sec. 2.11 and Fig. 2.15).
The approximation that CV,m is constant gives 兰21 (CV,m/T ) dT CV,m 兰21 T1dT CV,mln (T2/T1), and Eq. (2.75) becomes CV,mln (T2/T1) R ln (V1/V2) or
where k ln x ln xk[Eq. (1.70)] was used. If ln a ln b, then a b. Therefore (2.76) Since CVis always positive [Eq. (2.56)], Eq. (2.76) says that, when V2 V1, we will have T2 T1. A perfect gas is cooled by a reversible adiabatic expansion. In ex-panding adiabatically, the gas does work on its surroundings, and since q is zero, U must decrease; therefore T decreases. A near-reversible, near-adiabatic expansion is one method used in refrigeration.
An alternative equation is obtained by using P1V1/T1 P2V2/T2. Equation (2.76) becomes
The exponent is 1 R/CV,m (CV,m R)/CV,m CP,m/CV,m, since CP,m CV,m R for a perfect gas [Eq. (2.72)]. Defining the heat-capacity ratio g (gamma) as
we have
(2.77) For an adiabatic process, U q w w. For a perfect gas, dU CVdT. With the approximation of constant CV, we have
(2.78) To carry out a reversible adiabatic process in a gas, the surrounding constant-temperature bath in Fig. 2.11 is replaced by adiabatic walls, and the external pressure is slowly changed.
We might compare a reversible isothermal expansion of a perfect gas with a reversible adiabatic expansion of the gas. Let the gas start from the same initial P1and V1 and go to the same V2. For the isothermal process, T2 T1. For the adiabatic expansion, we showed that T2 T1. Hence the final pressure P2 for the adiabatic expansion must be less than P2for the isothermal expansion (Fig. 2.12).
¢U CV1T2 T12 w perf. gas, ad. proc., CV const.
P1V1g P2Vg2 perf. gas, rev. ad. proc., CV const.
g⬅ CP>CV
P2V2>P1V1 1V1>V22R>CV, m and P1V11R>CV, m P2V21R>CV, m T2
T1
aV1
V2bR>CV, m perf. gas, rev. adiabatic proc., CV const.
ln 1T2>T12 ln 1V1>V22R>CV, m
冮
12CTV,m dT冮
12RV dV R1ln V2 ln V12 R ln VV12CV,mdT 1RT>V2dV CV dT P dV 1nRT>V2dV
Section 2.8 Perfect Gases and the First Law 61
Figure 2.12
Ideal-gas reversible isothermal and adiabatic expansions that start from the same state.
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Chapter 2
The First Law of Thermodynamics 62
Summary
A perfect gas obeys PV nRT, has (U/V)T 0 (H/P)T, has U, H, CV, and CP depending on T only, has CP CV nR, and has dU CVdT and dH CPdT. These equations are valid only for a perfect gas. A common error students make is to use one of these equations where it does not apply.