- #1

Robert Davidson

- 32

- 4

- TL;DR Summary
- Consider a reversible ideal gas cycle consisting of: 1. An isochoric heat addition, 2. An isothermal expansion to the initial pressure, and 3. An isobaric compression to the initial volume. What, if any, is the difference in net work done by the gas if the isochoric heat addition was irreversible instead of reversible?

Consider a reversible ideal gas cycle consisting of: 1. An isochoric heat addition, 2. An isothermal expansion to the initial pressure, and 3. An isobaric compression to the initial volume. What, if any, is the difference in net work done by the gas in the cycle if the isochoric heat addition was irreversible instead of reversible?

Since no work is done in the isochoric process,

$$\Delta U_{1-2}=Q=nC_{v}(T_{2}-T_{1})$$

for both the reversible and irreversible process. The change in entropy for both a reversible and irreversible process is, using a reversible path involving an infinite series of thermal reservoirs,

$$\Delta S_{sys}=nC_{v}ln\frac{T_2}{T_1}$$

If we assume the irreversible process involves heat addition from a single thermal reservoir at temperature T2, then I believe the entropy generated would be

$$S_{gen}= nC_{v}ln\frac{T_2}{T_1}-\frac{Q}{T_2}$$

In order to complete the cycle, I would think that this entropy generated would need to be transferred to the surroundings in the form of additional heat and that the only opportunity to do this is during the reversible isobaric compression. Then it would seem that the negative work done in the isobaric process should be greater than for the reversible cycle for an overall positive net work of less. However, that can't be the case if the isobaric process connects the same two volumes for both the reversible cycle and irreversible cycle.

Can someone find the error in my reasoning? Thanks in advance.

Since no work is done in the isochoric process,

$$\Delta U_{1-2}=Q=nC_{v}(T_{2}-T_{1})$$

for both the reversible and irreversible process. The change in entropy for both a reversible and irreversible process is, using a reversible path involving an infinite series of thermal reservoirs,

$$\Delta S_{sys}=nC_{v}ln\frac{T_2}{T_1}$$

If we assume the irreversible process involves heat addition from a single thermal reservoir at temperature T2, then I believe the entropy generated would be

$$S_{gen}= nC_{v}ln\frac{T_2}{T_1}-\frac{Q}{T_2}$$

In order to complete the cycle, I would think that this entropy generated would need to be transferred to the surroundings in the form of additional heat and that the only opportunity to do this is during the reversible isobaric compression. Then it would seem that the negative work done in the isobaric process should be greater than for the reversible cycle for an overall positive net work of less. However, that can't be the case if the isobaric process connects the same two volumes for both the reversible cycle and irreversible cycle.

Can someone find the error in my reasoning? Thanks in advance.