Question

For a sample of perfect gas when its pressure is changed isothermally from $p_i$ to $p_f$, the entropy change is given by:

• A. $\Delta S=nRln\left(\frac{p_f}{p_i}\right)$
• B. $\Delta S=nRln\left(\frac{p_i}{p_f}\right)$
• C. $\Delta S=nRTln\left(\frac{p_f}{p_i}\right)$
• D. $\Delta S=RTln\left(\frac{p_i}{p_f}\right)$

Answer 1

Answer: (B)

$\Delta S=nRln\left(\frac{p_i}{p_f}\right)$

Solution:- (B) $\Delta S=nR\ln \left(\frac{P_i}{P_f}\right)$

For measurable change between two states, i.e., initial $\left(i\right)$ and final $\left(f\right)$-

$\Delta S={\int }_i^f\frac{dq}{T}$

As the temperature is constant,

$\therefore \Delta S=\frac{1}{T}{\int }_i^{f}d{q}_{rev.}$

As we know that,

$d{q}_{rev.}=-d{w}_{rev.}=nRT\left(\frac{dV}{V}\right)$

$\therefore \Delta S=\frac{1}{T}{\int }_i^{f}nRT\left(\frac{dV}{V}\right)$

$\Rightarrow \Delta S=\frac{nRT}{T}{\int }_i^f\left(\frac{dV}{V}\right)$

$\Rightarrow \Delta S=nR\ln \frac{V_f}{V_i}$

From Charle's law,

$P_f{V}_f=P_i{V}_i$

$\Rightarrow \frac{V_f}{V_i}=\frac{P_i}{P_f}$

$\therefore \Delta S=nR\ln \left(\frac{P_i}{P_f}\right)$