Consider the following processes: The temperature of two identical gases are increased from the same initial temperature to the same final temperature. Reversible processes are used. For gas A the process is carried out at constant volume while for gas B it is carried out at constant pressure. The change in entropy:
A) is the same for A and B
B) is greater for A
C) is greater for B
D) is greater for A only if the initial temperature is low
E) is greater for A only if the initial temperature is high

Question

Quizplus · Accepted Answer

For a reversible process, the entropy change is given by: 
ΔS = nC ln(T2/T1)
where n is the number of moles of gas and C is the molar heat capacity. Since the two gases are identical, n and C are the same for both. The only difference between the two processes is the temperature change. For gas A, since the process is carried out at constant volume, the temperature increase will lead to an increase in pressure (by the ideal gas law). This will result in a smaller temperature increase compared to gas B, which is at constant pressure. Therefore, T2/T1 will be larger for gas B than for gas A, leading to a larger entropy change for gas B.

Consider the Following Processes: the Temperature of Two Identical Gases