A certain substance has a heat of vaporization of 57.85kJ/mol. At what kelvin te

Question

A certain substance has a heat of vaporization of 57.85kJ/mol. At what kelvin temperature will the vapor pressure be 4.00 times higher than it was at 307 K?
I need help solving and working through this problem. Please help. A certain substance has a heat of vaporization of 57.85kJ/mol. At what kelvin temperature will the vapor pressure be 4.00 times higher than it was at 307 K?
I need help solving and working through this problem. Please help.
I need help solving and working through this problem. Please help.

Explanation / Answer

Solution :-

Given data

Heat of vaporization ?Hvap = 57.85 kJ/ mol

Initial temperature T1 = 307 K

Final temperature T2 = ?

Initial vapor pressure = 1 atm

Final vapor pressure = 4 atm ( since vapor pressure incease by 4 times so initial is assumed 1 and then final is 4 atm)

We need to use the Clausius-Clapeyron Equation

Formula is as follows

ln (P1/P2) = (?Hvap /R)* [(1/T2)-(1/T1)]

here R= gas constant =8.314 J/ mol K

now convert ?Hvap from kJ to Joules since R has unit of Joules

(57.85 kJ per mol * 1000 J) / 1 kJ = 57850 J/mol

lets put the values in the formula

ln(1/4) = (57850 J per mol / 8.314 J per mol .K) * [(1/T2)-(1/307 K)]

-1.386 = 6958.143 * [(1/T2)-(0.003257)]

-1.386 / 6958.143 = (1/T2)

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