This may get you started. What is the boiling point for CH3OH? I found 64.7 C.
For q, you are moving gaseous CH3OH from 122 C to 64.7, then you condense it, then you move the liquid from 64.7 C to 34.0 C. Here are the three steps.
q1 = moles x Cp(g) x (T2-T1) = 0.956 x 44.1 x (122.0 - 64.7) = ?for gas
q2 = dHvap x mols = ? for condensation
q3 = moles x Cp(l) x (64.7 - 34.0) = ? for liquid
q total = q1 + q2 + q3
Work. That is -p(delta V) or = p(V2-V1)
Use PV = nRT to solve V of the gas. That is V1. The problem tells you to ignore V2(the liquid volume).
any one know how i can solve this question?
What are the values of q, w, ΔU, ΔH, ΔS, ΔSsurr, and ΔSuniv for the following constant pressure process for a system containing 0.956 moles of CH3OH ?
CH3OH(g, 122.0 ºC, 1.00 atm) ⟶ CH3OH(l, 34.0 ºC, 1.00 atm)
Assume that CH3OH(g) behaves as an ideal gas and the volume of CH3OH(l) is much less than that of CH3OH(g). Also, assume that the temperature of the surroundings is 34.0 ºC.
Data:
Molar heat capacity for CH3OH(g), Cp,m = 44.1 J K−1 mol−1
Molar heat capacity for CH3OH(l), Cp,m = 81.1 J K−1 mol−1
Enthalpy of vaporization, ΔvapH = 35.2 kJ mol−1 at 64.7 ºC and 1.00 atm
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