Question

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

Answer 1

`331\text{ K}`

Step-by-step explanation:

Here, we want to get the vapor pressure

We can find this by using Clausius-Clapeyron equation

Mathematically, we have it that:

`ln((p_2)/(p_1))\text{ = -}(\Delta H_(vap))/(R)((1)/(T_1)-(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((760)/(145))\text{ = -}(40700)/(8.314)((1)/(373)-(1)/(T_2)_) \\ \\ -0.0003384\text{ =0.002681 - }(1)/(T_2) \\ (1)/(T_2)\text{ = 0.002681 +0.0003384} \\ \\ (1)/(T_2)\text{ = 0.00301936429} \\ \\ T_2\text{ = }(1)/(0.00301936429) \\ \\ T_2\text{ = 331.20 K} \end{gathered}`