Gas is isothermally and reversibly condensed

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SUMMARY

The discussion focuses on the isothermal and reversible condensation of a gas, specifically with 2 moles at a temperature of 337K and a heat of vaporization of 35.3 kJ/mol. The calculations for work (w), heat (q), internal energy (U), and enthalpy (H) reveal discrepancies, particularly with U being calculated as -76.2 kJ, while the reference book states -65 kJ. The relationship ΔH = q is confirmed for reversible processes at constant pressure, supported by thermodynamic equations involving enthalpy and internal energy.

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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
 

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