The normal boiling point of water is 373 K ( | vedclass

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(C) To find the boiling point at a different pressure,we use the Clausius-Clapeyron equation: $\ln(\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R} (\frac{

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$298$

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(C) To find the boiling point at a different pressure,we use the Clausius-Clapeyron equation: $\ln(\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.Given: $P_1 = 760 \ mm$,$T_1 = 373 \ K$,$P_2 = 23 \ mm$,$\Delta H_{vap} = 40.656 \ kJ/mol = 40656 \ J/mol$,$R = 8.314 \ J/mol \cdot K$.Substituting the values: $\ln(\frac{23}{760}) = \frac{40656}{8.314} (\frac{1}{373} - \frac{1}{T_2})$.$-3.498 = 4890.06 (0.00268 - \frac{1}{T_2})$.$-0.000715 = 0.00268 - \frac{1}{T_2}$.$\frac{1}{T_2} = 0.00268 + 0.000715 = 0.003395$.$T_2 = \frac{1}{0.003395} \approx 294.5 \ K$.Rounding to the nearest provided option,the answer is $298 \ K$.

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

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