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?
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
Get Answers For Free
Most questions answered within 1 hours.