Question

The molar entropy content of 1 mole of oxygen $\left(O_2\right)$ gas at $300\text{K}$ and $1\text{atm}$ is $250{\text{J mol}}^{-1}{\text{K}}^{-1}$. Calculate $\Delta G$ when 1 mole of oxygen is expanded reversibly and isothermally from $300\text{K}$, $1\text{atm}$ to double its volume (Take $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1},\text{log}e=2.303$)

• A. $1.728{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
• B. $0$
• C. $-1.728{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
• D. $0.75{\text{kJ mol}}^{-1}{\text{K}}^{-1}$

Answer 1

The correct option is A $1.728{\text{kJ mol}}^{-1}{\text{K}}^{-1}$

$\Delta G=\Delta H-\Delta \left(TS\right)$
$=\Delta H-T\Delta S$ (isothermal)
$=0-T\Delta S=-T\left(\int \frac{d{q}_{\text{rev}}}{T}\right)$
$=-\int d{q}_{\text{rev}}=-q_{\text{rev}}=W_{\text{rev}}$
as process is isothermal so $\Delta E=0=q_{\text{rev}}+W_{\text{rev}}$
so $\Delta G=-nRT\text{ln}\left(\frac{V_f}{V_i}\right)$
$=-RT\text{ln}2$
$=-8.314\times 300\times 0.693\times {10}^{-3}{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
$=1.728{\text{kJ mol}}^{-1}{\text{K}}^{-1}$