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The expression for entropy change given by, $$\Delta S = - nRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right),$$ holds good for
Answer & Solution
Correct Answer:
Option
B
We know $$S$$ as a function of $$T$$ and $$P$$ as :
$$dS = \frac{{{C_P}dT}}{T} - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}dP$$
For isothermal change $$dT = 0$$ for ideal gas $$\frac{{\partial V}}{{\partial {T_P}}} = \frac{{nR}}{P}$$
$${\text{So, }}dS = - nRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$$
For isothermal volume change of an ideal gas for both reversible and irreversible process.
$$dS = \frac{{{C_P}dT}}{T} - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}dP$$
For isothermal change $$dT = 0$$ for ideal gas $$\frac{{\partial V}}{{\partial {T_P}}} = \frac{{nR}}{P}$$
$${\text{So, }}dS = - nRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$$
For isothermal volume change of an ideal gas for both reversible and irreversible process.
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