Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 351.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.

Answer 1

Answer : The value of $\Delta H_{vap}$ is 47.627 kJ/mol

Explanation :

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $23^oC$ = 92 torr

$P_2$ = vapor pressure at temperature $45^oC$ = 351 torr

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $23^oC=[23+2730]K=296K$

$T_2$ = final temperature = $45^oC=[45+2730]K=318K$

Putting values in above equation, we get:

$\ln(\frac{351torr}{92torr})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{296}-\frac{1}{318}]\\\\\Delta H_{vap}=47627.347J/mol=47.627kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 47.627 kJ/mol

Answer 2

There is a specific formula to use for these type of problems. 

ln (P2/ P1)= Δvap/ R x (1/T1 - 1/T2)

R= 8.314
P1= 92.0 torr
T1= 23 C + 273= 296 K
P2= 351.0 torr
T2= 45.0 C + 273= 318 K

plug the values and solve for the unknown

ln( 351.0/ 92.0)= Δvap/ 8.314 x (1/296 - 1/318)

Δvap= 47630.6 joules