Question

The vapor pressure of ethanol is 400 mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is vapor pressure of ethanol, in mmhg, at 34.9°c? (2 point clausius clapperon)

Answer 1

Answer:- The pressure of ethanol would be 109 mmHg.

Solution:- This problem is based on Clausius clapeyron equation--

$ln(\frac{P_1}{P_2})=(\frac{\Delta Hvap}{R})(\frac{1}{T_2}-\frac{1}{T_1})$

Given, $T_1$ = 63.5 + 273 = 336.5 K

$T_2$ = 34.9 + 273 = 307.9 K

$P_1$ = 400 mmHg

$P_2$ = ?

$\Delta Hvap$ = 39.3 kJ/mol = 39300 J/mol

R = 8.314 J/mol.K

Let's plug in the values in the equation and do the calculations.

$ln(\frac{400}{P_2})=(\frac{39300}{8.314})(\frac{1}{307.9}-\frac{1}{336.5})$

$ln(\frac{400}{P_2})$ = 1.30

On taking anti ln to both sides...

$\frac{400}{P_2}$ = $e^1^.^3^0$

$\frac{400}{P_2}$ = 3.67

$P_2$ = 400/3.67

$P_2$ = 109 mmHg