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

Answer:

The vapor pressure is 170.6 mmHg.

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

$ln\dfrac{P_{2}}{P_{1}}=\dfrac{-\Delta H}{R}(\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}})$

Put the value into the relation

$ln\dfrac{P_{2}}{92.44}=\dfrac{-16690}{8.314}(\dfrac{1}{275.7}-\dfrac{1}{254.3})$

$ln\dfrac{P_{2}}{92.44}=0.61274$

$\dfrac{P_{2}}{92.44}=e^{0.61274}$

$P_{2}=1.84548\times92.44$

$P_{2}=170.6\ mmHg$

Hence, The vapor pressure is 170.6 mmHg.