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EXAMPLE 2.4 Calculation of q, w, and U Section 2.8
Perfect Gases and the First Law
Suppose 0.100 mol of a perfect gas having C V,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 dw rev PdV. Since either
V or P is constant in each step, we can integrate dq C dT and dq C dT
V
P
V
P
[Eqs. (2.51) and (2.52)] to find the heat in each step. The first law U q w
then allows calculation of U.
2
To evaluate integrals like C dT, we will need to know the temperatures
V
1
of states 1, 2, 3, and 4. We therefore begin by using PV nRT to find these tem- Figure 2.10
peratures. For example, T P V /nR 122 K. Similarly, T 366 K, T A reversible cyclic process.
3
1
1 1
2
732 K, T 244 K.
4
Step 1 → 2 is at constant volume, no work is done, and w 1→2 0. Step
2 → 3 is at constant pressure and
3
3
3
w P dV P1V V 2 13.00 atm212000 cm 1000 cm 2
2S3 3 2
2
3
3
3000 cm atm 18.314 J2>182.06 cm atm2 304 J
where two values of R were used to convert to joules. Similarly, w 0 and
3→4
w 4→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
1S2 2 2
q C dT nC V,m dT n11.50R21T T 2
V
1
2
1 1
10.100 mol21.5038.314 J>1mol K241366 K 122 K2 304 J
3
Step 2 → 3 is at constant pressure, and q 2→3 C dT. Equation (2.72) gives
2
P
C P,m C V,m R 2.50R, and we find q 2→3 761 J. Similarly, q 3→4
1
1
608 J and q 4→1 253 J. The total heat for the cycle is q 304 J
2
2
1
1
761 J 608 J 253 J 203 J.
2
2
We have U 1→2 q 1→2 w 1→2 304 J 0 304 J. Similarly, we find
1
1
U 457 J, U 608 J, U 152 J. For the complete cycle,
2→3 3→4 2 4→1 2
1
U 304 J 457 J 608 J 152 1 J 0, which can also be found from
2 2
q w as 203 J 203 J 0. An alternative procedure is to use the perfect-gas
equation dU C dT to find U for each step.
V
For this cyclic process, we found U 0, q
0, and w
0. These results
are consistent with the fact that U is a state function but q and w are not.
Exercise
Use the perfect-gas equation dU C dT to find U for each step in the cycle
V
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.