Question

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

• A. $1.728kJmo{l}^{-1}{K}^{-1}$
• B. 0.0
• C. $-1.728kJmo{l}^{-1}{K}^{-1}$
• D. $0.75kJmo{l}^{-1}{K}^{-1}$

Answer 1

The correct option is C $-1.728kJmo{l}^{-1}{K}^{-1}$

$△G=△H-△\left(TS\right)$
$△G=△H-T△S$ (isothermal)
$△G=0-T△S=-T(\int \frac{d{q}_{rev}}{T})$
$△G=-\int d{q}_{rev}=-q_{rev}=W_{rev}$
as process is isothermal so $△U=0=q_{rev}+W_{rev}$
so $△G=-nRTln\left(\frac{V_f}{V_i}\right)$
$=-RTln2$
$=-8.314\times 300\times 0.693\times {10}^{-3}kJmo{l}^{-1}{K}^{-1}$
$=-1.728kJmo{l}^{-1}{K}^{-1}$