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

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SUMMARY

The discussion focuses on calculating the change in entropy (∆S) for 1 mol of a diatomic perfect gas, specifically using the heat capacity at constant volume (Cv,m = 3/2 R). The calculation involves determining the heat capacity at constant pressure (Cp) using the relation Cp = Cv + R, resulting in Cp = 5/2 R. The final entropy change is computed as ∆S = Cp ln(T2/T1), yielding a value of 8.92 J/K for the system when heated from 100 ºC to 300 ºC.

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  • Understanding of thermodynamic principles, particularly entropy.
  • Familiarity with the ideal gas law and properties of diatomic gases.
  • Knowledge of heat capacities (Cv and Cp) and their relationships.
  • Basic logarithmic functions and their application in thermodynamics.
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  • Study the derivation of heat capacities for different types of gases.
  • Learn about the implications of entropy changes in thermodynamic processes.
  • Explore the concept of isobaric processes and their calculations.
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Diamond101
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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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