Question

One mole of an ideal gas at $22.4\text{L}$ is expanded isothermally and reversibly at $300\text{K}$ to a volume of $224\text{L}$ at a constant pressure. Calculate the $\Delta G$ in $\text{kJ}$. (Given, $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$).

• A. $-5.74\text{kJ}$
• B. $+5.74\text{kJ}$
• C. $-3.74\text{kJ}$
• D. $+3.74\text{kJ}$

Answer 1

The correct option is A $-5.74\text{kJ}$

Given,
$n=1\text{mole}{V}_i=22.4\text{L}{V}_f=224\text{L}T=300\text{K}$
$W=-2.303\times nRT\log \frac{V_f}{V_i}=-2.303\times 1\times 8.314\times {10}^{-3}\times 300\times \log \frac{224}{22.4}=-2.303\times 2\times 0.3=-5.74\text{kJ}$
$\Delta G=$ work done in reversible process which is maximum work done $=-5.74\text{kJ}$.