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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Dharsh on August 05, 2019, 11:17:35 AM

Title: First law of thermodynamics (numerical problem)
Post by: Dharsh on August 05, 2019, 11:17:35 AM
Hey guys!

I tried solving this problem from Atkins physical chemistry 8ed, exercise 2.6a (pg 71)

The problem goes like this:
A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100°C. The standard enthalpy of vaporisation of water at 100°C is 40.656kJmol-1. Find w,q,ΔU and ΔH for this process.

Since it's given that the process is isothermal, I took ΔU to be zero and proceeded but the answer turned out to be something else(ΔU=-37.555kJmol-1 )I want to know why my assumption of taking ΔU=0 for an isothermal process is wrong here.
Title: Re: First law of thermodynamics (numerical problem)
Post by: Corribus on August 05, 2019, 04:59:54 PM
A short answer is: ΔU = 0 for an isothermal process is true only for an ideal gas, in which all the heat transferred to/from the system is converted perfectly to work. This is not true for a real gas, and especially not true for a phase change, in which much of the heat transferred to/from the system is used in association with the breaking or forming of intermolecular forces. This is embodied in the "heat of vaporization".