Question

A substance has a heat of vaporization of 16.69 kJ/mole. At 254.3 K it has a vaporpressure of 92.44 mm Hg. Calculate its vapor pressure at 275.7 K.

Answer 1

The vapor pressure is 170.6 mmHg.

Step-by-step explanation:

Given that,

Heat of vaporization = 16.69 kJ/mole

Temperature = 254.3

Pressure = 92.44 mmHg

Temperature = 275.7 K

We need to calculate the vapor pressure

Using relation pressure and temperature

`ln(P_(2))/(P_(1))=(-\Delta H)/(R)((1)/(T_(2))-(1)/(T_(1)))`

Put the value into the relation

`ln(P_(2))/(92.44)=(-16690)/(8.314)((1)/(275.7)-(1)/(254.3))`

`ln(P_(2))/(92.44)=0.61274`

`(P_(2))/(92.44)=e^(0.61274)`

`P_(2)=1.84548*92.44`

`P_(2)=170.6\ mmHg`

Hence, The vapor pressure is 170.6 mmHg.