Gas is isothermally and reversibly condensed

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Homework Statement


gas is isothermally and reversibly condensed
n=2(mol)
T=337K
H(vap)=35.3kJ/mol
find w, q, U, H


Homework Equations





The Attempt at a Solution


H=-H(vap)*2=-70.6kJ/mol
H=U+n(g)RT
w=-n(g)RT n(g)=-2 w=5.6kJ
So U=H-w U Should be -70.6-5.6=-76.2, but the book gives -65, Where am I wrong?
And why is H=q in this process?
 
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In the condensation from a gas to a liquid, what is the sign of ΔV? What then, should be the sign on w? Is work being done on the gas by the surroundings or is work being done on the surroundings by the gas?

Also, to answer your second question, ΔH = q for any reversible process done at constant pressure.
proof: From the definition of enthalpy, we know that
ΔH = ΔU + Δ(PV)
Next, we combine this expression with the first law of thermodynamics, ΔU = q + w, to get:
ΔH = q + w + Δ(PV)
At constant pressure, w = -integral(PdV) = -PΔV, and Δ(PV) = PΔV. Plugging these two expressions into the expression above gives the desired result:
ΔH = q