Question

A certain substance has a heat of vaporization of 61.65 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 301 K?

Answer 1

Answer :- T2 = 317.13 K

You must use the Clausius-Clapeyron equation.

Therefore,

ln( P1 / P2 ) = ( ΔHvap / R )( 1 / T2 – 1 / T1 ),

where,

P1 - the vapor pressure measured at T1
P2 - the vapor pressure measured at P2
ΔHvap - the enthalpy of vaporization
R - the gas constant - expressed in Joules per mol K

You have everything you need to solve for T2. Since the pressure measured at this new temperature will be 3.50 times bigger than P1, you can write it as P2=3.50P1 and use it in this form in the equation.

So, plug all in and you'll get

ln(P1 / 3.50P1 ) = [ ( 61650 J/mol ) / ( 8.314 J/mol k ) ] * ( 1 / T2−1 / 301 K )

ln( 1 / 3.50 ) = ( 7415.20 )( 1 / T2 ) – ( 7415.20 )( 1 / 301 )

−1.2528 = 7415.20 T2 – 24.6352

23.3824 = 7415.20 T2

T2 = 7415.20 / 23.3824

T2 = 317.13K