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Not all of the heat input to a gas is absorbed if it is over a finite temperature difference and thus not reversible.

If we have say, a isochoric irreversible process, so no work can be done either, where does the heat that isn’t absorbed go?

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  • $\begingroup$ You need to be more specific.Take your isochoric example; it could be a gas in a box with one side in contact with a heat reservoir and the all other sides insulated. In that case, the gas will eventually reach the temperature of the reservoir. Or, you could have one side in contact with a heat reservoir and all other sides free to release heat to the surroundings. In that case, a temperature gradient will develop in the gas, and the heat that isn't absorbed will be released to the surroundings. To answer your question, heat that isn't absorbed is simply passed through. $\endgroup$ – Drew Apr 17 '20 at 19:08
  • $\begingroup$ @Drew Okay. What about in the case where all the other sides are isolated, where would the heat go then? It can’t be absorbed right otherwise it would be the same result as a reversible process. $\endgroup$ – Alex Gower Apr 17 '20 at 19:59
  • $\begingroup$ If all sides are insulated, then all of the heat gets absorbed by the gas. You can do this reversibly by heating the gas by a tiny infinitesimal amount of heat, or you can do it irreversibly by heating the gas with a finite amount of heat. In either case, all of the heat gets absorbed by the gas. $\endgroup$ – Drew Apr 17 '20 at 20:09
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Not all of the heat input to a gas is absorbed if it is over a finite temperature difference and thus not reversible.

That's not true. If you add some amount of heat $Q$ to a gas and don't allow it to perform any work, then the internal energy of the gas will increase by $Q$ whether the process is reversible or not.

The difference between a reversible process and an irreversible process is that in the latter case, the entropy of the universe increases. If the heat is transferred over a finite temperature difference, the entropy of the gas changes by

$$\Delta S_{gas} = \int \frac{\delta Q}{T_{gas}} = \int_{t_0}^{t_f} \frac{\dot Q}{T_{gas}(t)} dt$$

where $\dot Q$ is the rate at which heat is transferred to the gas and $T_{gas}$ is the temperature of the gas at time $t$, while the entropy of the environment changes by $$\Delta S_{env} = \int \frac{-\delta Q}{T_{env}} = \int_{t_0}^{t_f} \frac{-\dot Q}{T_{env}(t)} dt$$

The net change in entropy of the universe is therefore

$$\Delta S = \Delta S_{gas} + \Delta S_{env} = \int_{t_0}^{t_f} \dot Q \left(\frac{1}{T_{gas}} - \frac{1}{T_{env}}\right) dt$$

If the environment is at a higher temperature than the gas, then $\dot Q>0$ and $\frac{1}{T_{gas}}-\frac{1}{T_{env}} > 0$ the whole time, which means the integral is strictly positive and the entropy of the universe increases.


It can’t be absorbed right otherwise it would be the same result as a reversible process.

If you transfer some amount of heat $Q$ to a gas which is not permitted to do any work, then the increase in entropy of the gas is the same whether the process is reversible or not. It is the entropy change of the system plus its environment which depends on the irreversibility of the transfer.

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  • $\begingroup$ This answer does make intuitive sense. What confuses me is that the variable dS represents the entropy increase of just the gas system and we are told $$dS = \frac{dQ_{rev}}{T} >= \frac{dQ}{T} $$. Now for the irreversible isochoric case with a finite temperature difference, you’re saying these dS for the gas should be the same right (with the dS only different for the surroundings)? Does that mean the equality case can occur for this inequality even for irreversible processes? Or is it due to something being different when considering infinitesimal changes of the finite temperature difference? $\endgroup$ – Alex Gower Apr 17 '20 at 20:25
  • $\begingroup$ Does this answer your question? physics.stackexchange.com/questions/297386/… $\endgroup$ – Vivek Apr 17 '20 at 20:36
  • $\begingroup$ Maybe my original question, but how does it answer this bit? $\endgroup$ – Alex Gower Apr 17 '20 at 20:42
  • $\begingroup$ @AlexGower Yes, the equality can be satisfied even for irreversible processes. It does not generally hold for irreversible processes, but it can (as it does in this example). $\endgroup$ – J. Murray Apr 17 '20 at 20:57

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