Question

The change in the entropy of a $1$ mole of an ideal gas which went through an isothermal process from an initial state $ ({{P}_{1}},{{V}_{1,}}T) $ to the final state $ ({{P}_{2}},{{V}_{2,}}T) $ is equal to:

• A. zero
• B. $R \ln T$
• C. $R \ln \frac{V_1}{V_2}$
• D. $ R \ln \frac{V_2}{V_1}$

Answer 1

Answer: (D)

Key Idea : In an isothermal process, there is no change in internal energy of gas,
ie., $\Delta U=0$.
The change in entropy of an ideal gas
$\Delta S=\frac{\Delta Q}{T} \ldots$(i)
In isothermal process, there is no change in internal energy of gas,
i.e., $\Delta U=0$
$\therefore \Delta U=\Delta Q-W$
$\Rightarrow 0 = \Delta Q-W $
$\Rightarrow \Delta Q = W$
i.e., $\Delta Q=$ work done by gas in isothermal process which went through from $\left(P_{1}, V_{1}, T\right)$ to $\left(P_{2}, V_{2}, T\right)$
or $\Delta Q=\mu R T \log _{e}\left(\frac{V_{2}}{V_{1}}\right) \ldots$(ii).
For 1 mole of an ideal gas, $\mu=1$
So, from Eqs. (i) and (ii), we get
or $\Delta S =R \log _{e}\left(\frac{V_{2}}{V_{1}}\right)$
$=R \ln \left(\frac{V_{2}}{V_{1}}\right)$