Question

The normal boiling point of water is $373K$. Vapour pressure of water at temperature $T$ is $19\text{mm Hg}$. If enthalpy of vaporisation is $40671.40\text{J/mol}$, then temperature $T$ would be:
(Use :$\log 2=0.3,R:8.3{\text{JK}}^{-1}{\text{mo1}}^{-1}$)

• A. $150.2\text{K}$
• B. $291.3\text{K}$
• C. $330.3\text{K}$
• D. $390\text{K}$

Answer 1

The correct option is B $291.3\text{K}$

Given $P_1=19\text{mm Hg}$,
$P_2=760\text{mm Hg}$
$\text{enthalpy of vaporisation,}\Delta H_{\text{vap}}=40671.40\text{J/mol}$
Applying Clausius-Clapeyron's equation,
$\log \frac{P_2}{P_1}=\frac{\Delta H_{\text{vap}}}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$\Rightarrow \log \frac{760}{19}=\frac{40671.40}{2.303\times 8.3}\left(\frac{373-T_1}{373T_1}\right)$
$\Rightarrow \log 40=\frac{40671.40}{2.303\times 8.3}\left(\frac{373-T_1}{373T_1}\right)$
$\Rightarrow 30.58=40671.40\left(\frac{373-T_1}{373T_1}\right)$
$\Rightarrow 373T_1=1330(373-T_1)$
$\Rightarrow 1703T_1=1330\times 373$
$\Rightarrow T_1=291.30\text{K}$