Question

the heat of vaporization of water is 40.7 kj/mol. at what temperature is the vapor pressure 145 torr334339331324320

Answer 1

Answer:

$331\text{ K}$

Explanation:

Here, we want to get the vapor pressure

We can find this by using Clausius-Clapeyron equation

Mathematically, we have it that:

$ln(\frac{p_2}{p_1})\text{ = -}\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where:

P2 equals final pressure which is temperature at boiling point which is 760 torr

P1 is 145 torr

Heat of vaporization is 40,700 Kj/mol

R is molar gas constant which is 8.314

T1 is initial temperature which is 373 K (the boiling point of water)

T2 is final temperature which is what we want to calculate

Substituting the values, we have it that:

$\begin{gathered} ln(\frac{760}{145})\text{ = -}\frac{40700}{8.314}(\frac{1}{373}-\frac{1}{T_2}_) \\ \\ -0.0003384\text{ =0.002681 - }\frac{1}{T_2} \\ \frac{1}{T_2}\text{ = 0.002681 +0.0003384} \\ \\ \frac{1}{T_2}\text{ = 0.00301936429} \\ \\ T_2\text{ = }\frac{1}{0.00301936429} \\ \\ T_2\text{ = 331.20 K} \end{gathered}$