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Topic: First law of thermodynamics (numerical problem)  (Read 167 times)

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Offline Dharsh

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First law of thermodynamics (numerical problem)
« 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.

Offline Corribus

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Re: First law of thermodynamics (numerical problem)
« Reply #1 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".
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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