Entropy Thermodynamics: Calculate ∆S for 1 Mol of Diatomic Gas

AI Thread Summary
The discussion focuses on calculating the change in entropy (∆S) for 1 mol of a diatomic perfect gas heated from 100 ºC to 300 ºC at constant pressure. The specific heat at constant volume (Cv,m) is given as 3/2 R, leading to the calculation of the specific heat at constant pressure (Cpm) as 5/2 R. The formula used for entropy change is ∆S = Cp ln(T2/T1), resulting in a calculated value of 8.92 J/K. The calculation appears to be correct based on the provided values and equations. Overall, the approach to solving the problem is validated by the participants.
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Homework Statement


A sample consisting of 1 mol of a diatomic perfect gas with Cv,m = 3/2 R is heated from
100 ºC to 300 ºC at constant pressure. Calculate ∆S for the system.

Homework Equations


Cv,m = 3/2 R

The Attempt at a Solution


Cpm=Cvm +r because we want cp right isobaric
∆S= Cp ln T2/T1
= 5/2 . 8.314 . ln 573/373
=8.92 jk-1 correct or no
 
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Seems ok.
 
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