# Is isothermal expansion of a gas a reversible process?

I have a homework question that asks me to calculate the change in the entropy during the isothermal expansion using the Sackur-Tetrode. Then, show that it is a reversible process.

I have finished the first part and got $\Delta S = Nk\ln(\frac{V_f}{V_i})$. According to the definition of the book, I am using, a process is reversible if the change in entropy is 0. If a process increases entropy, then it is irreversible. So, base on the equation that I got for the first part, the only way for isothermal expansion to be reversible, the change in volume must be 0. However, it is an expansion, so therefore $V_f > V_i$, which then implies that this process is irreversible. Can someone point out what is wrong with this reasoning?

## 3 Answers

Reversibility is when $dS=\frac{\delta Q}T$ so entropy can change but equality sign should be satisfied.

So you have already got $\Delta S = Nk\ln(\frac{V_f}{V_i})$

Note for an isothermal process, $\delta Q=dW$. For ideal gas, $PV=NkT$ or $P=\frac{NkT}{V}$

So, the work done during the expansion is $dW=\int_{V_i}^{V_f}PdV$

replacing P with ideal gas law, $dW=\int_{V_i}^{V_f}\frac{NkT}{V}dV=NkT\ln \frac{V_f}{V_i}$.

You can then show $\frac{\delta Q}T=Nk\ln \frac{V_f}{V_i}=dS$ for ideal gas isothermal expansion process and conclude it is a reversible process.

• I arrived at this conclusion shortly after I post the question. Thank you. – Kane Billiot Feb 26 '17 at 22:03

A process is reversible if the change in entropy of the system and its surroundings total zero. The entropy of the system can change and the entropy of its surroundings can change in a reversible process as long as their sum doesn't change. In your case, if the gas was in contact with a constant temperature bath during the expansion (to keep the gas temperature constant), the change in entropy of the bath would have been minus the change in entropy of the gas.

The slow compression/expansion of a gas does not always cause an change in S. In wave function terms, with a slow enough compression/expansion, you change the energy of the system, but the multiplicity does not change, each molecule or particle undergoes an energy change , but the number of energy levels does not change, even though the energy in each level does.

A particle in the $n$th level will stay in the $n$th level, even though it has undergone $\Delta E$.

So S is constant, as the multiplicity does not change.

A free expansion will create new entropy however.