Doubt in entropy change of environment in an irreversible isochoric process

Question

I have read the following:-

1. The entropy change of the environment is calculated by $\Delta S_{env}=-\int \frac{dq_{sys}}{T} $ for that process. Unlike for a system, we do not assume a reversible path connecting the endpoints and then integrate.

2. For an isochoric process $dq_{sys}=dU_{sys}=nC_vdT$

3. Now, we plug this into the integral and get $\Delta S_{env}=-nC_v\ln( \frac{T_f}{T_i} )$

4. $\Delta S_{universe}=\Delta S_{sys}+\Delta S_{env}$

5. Now, $\Delta S_{sys}=nC_v\ln( \frac{T_f}{T_i} )$ so $\Delta S_{universe}=0$

However, this is wrong because $\Delta S_{universe}\neq0$ for an irreversible process

Where is the mistake in this proof?

Answer 1

As you say the entropy increase of the system (in this case an ideal gas) is calculated assuming a reversible process regardless of the actual process, since entropy is a state function. In such a reversible isochoric heat addition process the environment consists of an infinite series of thermal reservoirs ranging in temperature between the initial and final temperatures. The increase in entropy of the system is then equal to the decrease in entropy of the environment for a total entropy change of zero.

However, for an irreversible heat transfer process the environment consists of a single thermal reservoir at temperature $T_f$. The change in entropy of the environment is then

$$\Delta S_{env}=-\frac{Q}{T_f}=-\frac{nC_{v}(T_{f}-T_i)}{T_f}$$

This decrease in entropy of the environment less for the irreversible process than the reversible process, for a total entropy change greater than zero.

Example: let $T_f$ = 300 K and $T_i$ = 100 K

$$\Delta S_{sys}= nC_{v}\ln\frac{300}{100}=+1.1 nC_v$$

$$\Delta S_{env}= -\frac{nC_{v}(300-100)}{300}=-0.66 nC_v$$

$$\Delta S_{tot}=+0.44 nC_v$$

Hope this helps.

Answer 2

If the process is reversible, you can have $\Delta S_{env}=-\int{dq_{sys} \over T}$. However, this does not work if the process is irreversible.

Recall that entropy measures the disorder of a state. The state with higher disorder is more likely to occur. If the process is reversible, the entropy of the initial state and that of the final state is the same. That is to say, there is no probabilistic preference of which state should occur or which state the system and the environment tend to evolve into.

However, if the process is irreversible, $\Delta S_{universe} > 0$, and $\Delta S_{env} > -\int{dq_{sys} \over T}$. You no longer have $\Delta S_{env} = -\int{dq_{sys} \over T}$.

Answer 3

To elaborate on what @Bob D said, in the actual irreversible process, the temperature at the interface between the system and surroundings $T_I$ is not the same as that for the reversible path suffered by the system. In conventional thermodynamics, we typically treat the surroundings as ideal (i.e., reversible), such that $$\Delta S_{surr}=-\int{\frac{dq_{syst,irrev}}{T_I(q_{syst,irrev})}}$$where $T_I(q_{syst,irrev})$ is the temperature at the interface between the system and surroundings in the actual irreversible process that has occurred up to the point in the process where the cumulative amount of heat $q_{syst}$ has been transferred to the surroundings. So the overall change in entropy of the universe is then $$\Delta S=\int{\frac{dq_{syst,rev}}{T_{rev}}}-\int{\frac{dq_{syst,irrev}}{T_I(q_{syst,irrev})}}$$