Question

He normal boiling point of water is 373 k. vapour pressure of water at temperature t is 19 mm hg. if enthalpy of vaporization is 40.67 kj/mol, then temperature t would be:

Answer 1

Answer:- Value of t is 291 K.

Solution:- Normal boiling point is the temperature at which the pressure equals to atmospheric pressure which is 1 atm or 760 mmHg.

We are asked to calculate the temperature at which the vapor pressure is 19 mmHg.

This type of problems are solved by using Clausius Clapeyron equation:

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

$P_1$ = 760 mmHg

$T_1$ = 373 K

$P_2$ = 19 mmHg

$T_2$ = t

$\Delta H_v_a_p=40.67\frac{kj}{mol}$

R is universal gas constant and it's value in terms of kj is $0.008134\frac{kj}{mol.K}$ .

Let's plug in the values in the equation and solve it:

$ln(\frac{19}{760})=\frac{40.67}{0.008314}(\frac{1}{373}-\frac{1}{t})$

$-3.69=4891.75(0.00268-\frac{1}{t})$

$\frac{-3.69}{4891.75}=(0.00268-\frac{1}{t})$

$-0.000754=(0.00268-\frac{1}{t})$

Let's keep the similar terms on same side:

$\frac{1}{t}=0.00268+0.000754$

$\frac{1}{t}$ = 0.003434

$t=\frac{1}{0.003434}$

t = 291 K

So, the value of t is 291 K.