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ioda
2 years ago
8

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 p

ressure at 275.7 K.
Physics
1 answer:
adoni [48]2 years ago
5 0

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.

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