Question

The normal boiling point of a liquid $X$ is $400\text{K}$. $△H_{\text{vap}}$ at normal boiling point is $40\text{kJ/mol}$.

• A. $△S_{\text{vaporization}}<100\text{J/ mol.K at}400\text{K}\text{and}2\text{atm}$
• B. $△S_{\text{vaporization}}<10\text{J/mol.K}\text{at}400\text{K}\text{and}1\text{atm}$
• C. $△G_{\text{vaporization}}<0\text{at}410\text{K}\text{and}1\text{atm}$
• D. $△U=43.32\text{kJ/ mol.K at}400\text{K}\text{and}1\text{atm}$

Answer 1

Answer: (A), (C)

$△G_{\text{vaporization}}<0\text{at}410\text{K}\text{and}1\text{atm}$

$\text{At}{S}_{\text{vap}}=\frac{△H}{\text{Boiling Point}}=\frac{40\times 1000}{400}=100\text{J/mol. K}$

(A) If the external pressure is $2\text{atm}$, then $\text{B.P}>400\text{K}$
Since boiling point will increase $△S_{vap}<100$

(B) At $400\text{K}$, and $1\text{atm}$, $\Delta S_{\text{vaporization}}$ is $100\text{J/mol. K}$

(C) $△G=△H-T△S$
$△G=40\times 1000-410\times 100=-1000\text{J}$
$△G<0$

(D) $△H=△U+△n_{g}RT$
$40=△U+1\times 8.314\times {10}^{-3}\times 400$
$△U=36.67\text{kJ/mol}$