Entropy change in a reversible adiabatic expansion of ideal gas

Question

Below is P-V curves of 1 mole of ideal gase for temperature 700K, 500K and 300K each.

Now I would like to calculate entropy change of the system along the path from (1) to (3), which is adiabatic process.

If we assume that the given adiabatic process is reversible, then the entropy change must be zero, because in general, the entropy change consists of the entropy transfer by heat and the entropy generation by irreversibilities of the process.

On the other hand, in the below graph we can obtain the entropy change from (1) to (3) by adding the entropy change from (1) to (2) and the one from (2) to (3)

The entropy change from (1) to (2):

• is constant-pressure process hence $C_p \ln(T_f/T_i) = (5R/2) \ln(700/500)$

The entropy change from (2) to (3):

• is constant-volume process hence $C_v \ln(T_f / T_i) = (3R/2) \ln(300/700)$

Adding these two terms results in a non-zero entropy change from (1) to (3).

This contradicts zero entropy change from (1) to (3) obtained by assuming the adiabatic process has occurred reversibly: Does this mean that the given adiabatic process can not be reversibly carried out?

Answer 1

You are assuming, wrongly, I think, that (1) and (3) can be joined by an adiabatic curve. For an ideal gas, two points which can be gone between by a reversible adiabatic process are related by $$p_a\ V_a^\gamma=p_b\ V_b^\gamma$$ in which $\gamma=\frac{c_p}{c_V}$, so in this case$$\frac{p_3}{P_{1,2}}=\left(\frac{V_{1}}{V_{2,3}}\right)^\gamma$$

But we know by applying the ideal gas equation of state, $pV=nRT$, that$$\frac{p_3}{P_{1,2}}=\frac{300}{700}\ \ \ \text{and}\ \ \ \frac{V_1}{V_{2,3}}=\frac{500}{700}$$ We can show that, for any gas, $1.00<\gamma<1.67$

But $\left(\frac{500}{700}\right)^1=\frac{500}{700}\ \ \ \ \ $ and $\ \ \ \ \left(\frac{500}{700}\right)^{1.67}=0.57=\frac{400}{700}$

So it's never going to work!