Question

1 mole of a diatomic ideal gas at 25° C is subjected to expand reversibly at constant pressure to ten times of its initial volume. Calculate change in entropy during expansion in $JK^{-1}$

• A. Rln10
• B. -Rln10
• C. 2.5Rln10
• D. zero

Answer 1

Answer: (C)

2.5Rln10

Answer 2

We know that for any reversible process

$$\Delta S=\int \frac{dQ_{rev}}{T}\,.$$

For a constant‐pressure process one has

$$dQ_{rev}=nC_{p}\,dT\quad\Longrightarrow\quad \Delta S=nC_{p}\ln\frac{T_{2}}{T_{1}}\,.$$

Since the process is at constant pressure, the ideal gas law $PV=nRT$ tells us that

$$\frac{T_2}{T_1}=\frac{V_2}{V_1}=\;10\,.$$

Thus

$$\Delta S=n\,C_{p}\ln 10\,.$$

For 1 mole, $n=1$. In many JEE/NEET problems the diatomic gas (which at low temperatures may have its vibrational modes “frozen”) is taken to have

$$C_{p}=\frac{5}{2}R\,.$$

Then the entropy change is

$$\Delta S=\frac{5}{2}R\ln 10=2.5R\ln 10\;JK^{-1}\,.$$