0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ From where are you obtaining $\Delta S_{sys}=nC_v\ln\left( \frac{T_f}{T_i} \right)$? $\endgroup$ Jul 6, 2022 at 4:41
  • $\begingroup$ @Chemomechanics From $\Delta S_{sys}=\int \frac{dq_{sys}}{T} $ over a reversible isochoric path $\endgroup$
    – Boson
    Jul 6, 2022 at 5:01
  • $\begingroup$ Then you cannot use that $dq_\text{sys}$ to calculate the entropy change in the environment. It is not the actual heat transfer but an imaginary value that allowed the system progression to be modeled as reversible. $\endgroup$ Jul 6, 2022 at 15:13

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ in a related answer I read that we take temperature of the system as equal to the temperature of the environment(where the temperature is taken at the boundary). Now, if $T_{sys}$ is varying then how is $T_{env}$ constant? $\endgroup$
    – Boson
    Jul 6, 2022 at 10:49
  • $\begingroup$ @Boson I explained this to you in my answer to your post physics.stackexchange.com/questions/716000/… I don't know what else I can add $\endgroup$
    – Bob D
    Jul 6, 2022 at 13:38
  • $\begingroup$ Can we start a chat? I need some clarification on a number of minute points which is difficult in this setting. $\endgroup$
    – Boson
    Jul 6, 2022 at 13:39
  • $\begingroup$ @Boson Sure. You set up the Chat room. $\endgroup$
    – Bob D
    Jul 6, 2022 at 13:50
  • $\begingroup$ please wait a minute I'm new I need to figure out how to do it $\endgroup$
    – Boson
    Jul 6, 2022 at 13:55
1
$\begingroup$

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 $\Delta S_{env} = -\int{dq_{sys} \over T}$.

$\endgroup$
2
1
$\begingroup$

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 $$\Delta S=\int{\frac{dq_{syst,rev}}{T_{rev}}}-\int{\frac{dq_{syst,irrev}}{T_I(q_{syst,irrev})}}$$

$\endgroup$
11
  • $\begingroup$ Just to clarify, you are saying $T_{sys}=T_{env}$ (where the temperature is taken at the boundary) in both reversible and irreversible but the value of temperature itself is different? $\endgroup$
    – Boson
    Jul 6, 2022 at 10:51
  • $\begingroup$ Also, $dq_{sys, rev}=dq_{sys,irrev}$ in the entropy change of universe calculation, right? $\endgroup$
    – Boson
    Jul 6, 2022 at 11:03
  • $\begingroup$ To. get the changes in entropy of both system and surroundings for an irreversible process, you first separate them and then subject each to a separate reversible path. The reversible path for the system does not match the reversible path for the surroundings. Only the initial and final states have to match that of the irreversible path. $\endgroup$ Jul 6, 2022 at 12:56
  • $\begingroup$ Regarding you second question, no. dq for the system for the reversible path does not necessarily have to match dq for the system for the irreversible path. For example, for an adibatic irreversible process, dq is zero but dq between the same two end states for a reversible path is not. $\endgroup$ Jul 6, 2022 at 13:01
  • $\begingroup$ Can we continue this discussion in chat? $\endgroup$
    – Boson
    Jul 6, 2022 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.