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. $\Delta S=nR\ln \left(\frac{P_f}{P_i}\right)$
• B. $\Delta S=nR\ln \left(\frac{P_i}{P_f}\right)$
• C. $\Delta S-nRT\ln \left(\frac{P_f}{P_i}\right)$
• D. $\Delta S=RT\ln \left(\frac{P_i}{P_f}\right)$

Answer 1

The correct option is B $\Delta S=nR\ln \left(\frac{P_i}{P_f}\right)$

For measurable change between two states, i.e., initial (i) and final (f)-
$\Delta S={\int }_i^f\frac{dq}{T}$

As the temperature is constant,

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

As we know that,

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

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

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

$\Rightarrow \Delta S=nRIn\frac{V_f}{V_i}$

From Charle's law,

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

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

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