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) $\Delta S=nR\ln (\dfrac{{{p}_{f}}}{{{p}_{i}}})$
(B) $\Delta S=nR\ln (\dfrac{{{p}_{i}}}{{{p}_{f}}})$
(C) $\Delta S=nRT\ln (\dfrac{{{p}_{f}}}{{{p}_{i}}})$
(D) $\Delta S=RT\ln (\dfrac{{{p}_{f}}}{{{p}_{i}}})$

Answer 1

Hint: Entropy is a measure of disorders and a state function. The entropy change is observed only in its initial and final states. The symbol of entropy is S and units are J/K. If a container of ideal gas an entropy value, pressure, volume and temperature will be easy to determine, but the entropy of the system will calculate as entropy change of system.

Complete step by step answer:
Entropy change will observe when there is a transfer of heat. The change in entropy is the ratio of heat to temperature at which transfer took place. If the heat transfer Q takes place at a single temperature, the change in entropy simply as,
$\Delta S=\dfrac{Q}{T}$ , this process is isothermal.
According to the second law of thermodynamics,
The entropy of a closed system is constant for reversible processes and increases for the irreversible process. Entropy never closed for a closed system.
$\Delta S\ge 0$

Given perfect gas, measurable initial and final states are ‘i’ and ‘f’.
Entropy change, \[\Delta S=\int\limits_{i}^{f}{\dfrac{dQ}{T}}\]
As the temperature is constant,
$\Delta S=\dfrac{1}{T}\int\limits_{i}^{f}{d{{Q}_{rev}}}$

For a reversible process,
$d{{Q}_{rec}}=-d{{W}_{rev}}=nRT(\dfrac{dV}{V})$
$\therefore \Delta S=\frac{1}{T}\int\limits_{i}^{f}{nRT(\dfrac{dV}{V})}$
$\Rightarrow \Delta S=\dfrac{nRT}{T}\int\limits_{i}^{f}{(\dfrac{dV}{V})}$
$\Rightarrow \Delta S=nR\ln \dfrac{{{V}_{f}}}{{{V}_{i}}}$ --- (1)

From Charle’s law,
${{P}_{i}}{{V}_{i}}={{P}_{f}}{{V}_{f}}$
$\Rightarrow \dfrac{{{V}_{f}}}{{{V}_{i}}}=\dfrac{{{P}_{f}}}{{{P}_{i}}}$ --- (2)
Therefore, from equation (1) and (2),
 $\Delta S=nR\ln (\dfrac{{{P}_{f}}}{{{P}_{i}}})$

Hence, for a sample of perfect gas when its pressure is changed isothermally from ${{p}_{i}}$ to ${{p}_{f}}$ , the entropy change, $\Delta S=nR\ln (\dfrac{{{P}_{f}}}{{{P}_{i}}})$
So, the correct answer is “Option A”.

Note: If an ideal gas changes from ${{P}_{1}},{{V}_{1}},{{T}_{1}}$ to ${{P}_{2}},{{V}_{2}},{{T}_{2}}$ , the change in entropy determines with two reversible paths. The first path is reversible at constant pressure and the second path will be reversible with the isothermal process.