The normal boiling point of water at 760 , mm | vedclass

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(B) Using the Clausius-Clapeyron equation:$\log \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{2.303R} \left[ \frac{T_2 - T_1}{T_1 	imes T_2} ight]$Given:

Vedclass · Accepted Answer

$294$

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(B) Using the Clausius-Clapeyron equation:$\log \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{2.303R} \left[ \frac{T_2 - T_1}{T_1 	imes T_2} ight]$Given: $P_1 = 760 \, mm$,$T_1 = 373 \, K$,$P_2 = 23 \, mm$,$\Delta H_{vap} = 40656 \, J/mol$,$R = 8.314 \, J/mol \cdot K$.Substituting the values:$\log \frac{760}{23} = \frac{40656}{2.303 	imes 8.314} \left[ \frac{373 - T_2}{373 	imes T_2} ight]$$1.519 = 2123.6 \left[ \frac{373 - T_2}{373 	imes T_2} ight]$Solving for $T_2$,we get $T_2 \approx 294.4 \, K$.

The normal boiling point of water at $760 \, mm$ is $373 \, K$. The vapor pressure of water at $298 \, K$ is $23 \, mm$. If the enthalpy of vaporization is $40.656 \, kJ/mol$,then its boiling point at $23 \, mm$ pressure will be ........... $K$.

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