Question

1 mole of an ideal gas at ${25}^{0}C$ is subjected to expand reversibly ten times of its initial volume. The change in entropy due to expansion is:

• A. $19.15{JK}^{-1}{mole}^{-1}$
• B. $16.15{JK}^{-1}{mole}^{-1}$
• C. $22.15{JK}^{-1}{mole}^{-1}$
• D. none

Answer 1

The correct option is A $19.15{JK}^{-1}{mole}^{-1}$

We have the state relation -

$TdS=dU+PdV$

$R=8.314Jmo{l}^{-1}K,\frac{V_2}{V_1}=10$

Since expansion is isothermal, there will be no change in temperature and consequently no change in internal energy (dU = 0).

It reduces to -

$dS=\frac{PdV}{T}$
PdVT

From ideal gas relation, PV = nRT,

$dS=\frac{nRdV}{V}$

$S_2-S_1=nRln\left(\frac{V_2}{V_1}\right)$

n = 1 (1 mole of gas), $R=8.314Jmo{l}^{-1}K,\frac{V_2}{V_1}=10$

$S_2-S_1=8.314\times ln\left(10\right)$

$⟹S_2-S_1=19.15K^{-1}mol{e}^{-1}$