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
[tex]ln\dfrac{P_{2}}{P_{1}}=\dfrac{-\Delta H}{R}(\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}})[/tex]
Put the value into the relation
[tex]ln\dfrac{P_{2}}{92.44}=\dfrac{-16690}{8.314}(\dfrac{1}{275.7}-\dfrac{1}{254.3})[/tex]
[tex]ln\dfrac{P_{2}}{92.44}=0.61274[/tex]
[tex]\dfrac{P_{2}}{92.44}=e^{0.61274}[/tex]
[tex]P_{2}=1.84548\times92.44[/tex]
[tex]P_{2}=170.6\ mmHg[/tex]
Hence, The vapor pressure is 170.6 mmHg.
