Question

A certain substance has a heat of vaporization of 56.92 kJ/mol. At what Kelvin temperature will the vapor pressure be 4.00 times higher than it was at 313 K?

Answer #1

**Solution :-**

Using the Clausius clyeperon equation we can find the final temperature as follows

Ln[P2/P1] = Delta H vap /R [(1/T1)-(1/T2)]

Ln [4.00] = 56920 J per mol / 8.314 J per mol K [(1/313)-(1/T2)]

1.386 = 6846.283 * [0.003195 – (1/T2)]

1.386 / 6846.283 = 0.003195 – (1/T2)

0.000202 = 0.003195 – (1/T2)

0.000202 – 0.003195 = -1/T2

-0.00299 = -1/T2

-1/-0.00299 = T2

334 K = T2

So the final temperature is 334 K where the vapor pressure will be 4.00 times than it is at 313 K

**So the answer is 334 K**

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