[BreakingBad20]

A fluorocarbon liquid has a standard enthalpy of vaporization of 32 kJ/mol. Calculate q (heat), w (work), ΔH (enthalpy change) and ΔU (internal energy change) when 2,50 mol are vaporized at 250 K and 750 torr.

dH_vap = 32kJ/mol
q = 32kJ/mol x 2.5 mol
q = 80kJ

pV = nRT
(750 x 133)V = (2.5)(8.31451)(250)
V= .0521 J/Pa

w = p_ex x dV
w = (750 x 133)Pa x .0521 J/Pa
w = 5196.975 J

dU = w + q
dU = 5196.975 + 80000
dU = 85196.975 J

dH = dU + pV
dH = 85196.975 + 5196.975
dH = 90393.95 J

Would you agree with the following calculations or am I going wrong somewhere in regards to understanding the theory?
All feedback is appreciated because I have my exam on the 14th of May and between work and other exams I am struggling to grasp the full understanding!