# Isn't the change in entropy zero for a reversible process?

I'm calculating $$\Delta G$$, $$\Delta H$$ and $$\Delta S$$ when an ideal gas expands isothermally and reversible at $$20$$°C from $$5$$ MPa to $$1$$ MPa. I used the relation that:

$$\left(\frac{\partial G}{\partial P}\right)_T=V$$

This relation can be shown from the defintion $$G=H-TS$$. Then I integrated G so that:

$$\Delta G=\int_{P_1}^{P_2}VdP$$

We have an ideal gas $$=>V=\frac{RT}{P}$$. Which gives us that:

$$\Delta G=\int_{P_1}^{P_2}\frac{RT}{P}dP=RT\ln\frac{P_2}{P_1}$$

If you calculate this you'll get that $$\Delta G\approx3.9$$ kJ/mol. This was correct.

However I then tried to calculate $$\Delta H$$. Since $$dH=dQ+VdP$$ and $$dQ=TdS$$: $$dH=TdS+VdP$$ After this we know that we have a reversible process which according to me means that $$dS=0$$. Which means that $$dH=VdP=dG=>\Delta H=\Delta G$$. However the answer was that $$\Delta S= 13.38$$ J/(Kmol) and $$\Delta H=0$$. How is this possible? Isn't $$dS=0$$ for a reversible process?

• Delta S is zero for the combination of system plus surroundings, not for each individually. Oct 25, 2020 at 15:28

As mentioned in the comments, there is an entropy change that occurs in the system, however, when combined with the entropy change of the surroundings, is zero.

When solving for the change of entropy in the system, you can use the equation below,

$$\Delta S = R \ln(\frac{P_1}{P_2})$$

converting MPa to atm and plugging in, we get

$$\Delta S = 8.314 \ln(\frac{49.3}{9.87})$$

which yields an entropy change that is equal to the answer you provide.

The above equation can be rearranged to have different variables instead of pressure. Here is a link for a ChemLibretexts article on entropy and reversible processes if you're interested.