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Topic: cm vs cv (adiabatic equation)  (Read 863 times)

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Offline physikkk2

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cm vs cv (adiabatic equation)
« on: May 26, 2024, 11:40:58 AM »
Hello,

I have to solve the following problem: 2 moles of argon are expanded reversibly and adiabatically. Before expansion, the gas has a temperature of 30 ◦C (303,15 K) with a volume of 2 L. After expansion, the volume is 15 L. The
Molar heat capacity of argon is cm = 12.4 J/mol K. What is the temperature of the gas after
of expansion ? to solve that problem I use the adiabatic equation : T2= T1 * (V1/V2)^(nR/cv)   My question is if cm is the same as cv? When I use cm as cv, I get  20 K for T2 which seems wrong. Thanks in advance
« Last Edit: May 26, 2024, 11:57:41 AM by physikkk2 »

Offline physikkk2

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Re: cm vs cp (adiabatic equation)
« Reply #1 on: May 26, 2024, 11:56:46 AM »
alternative equation: T2=T1(v1/v2)^y-1 , cv is also needed for y

Offline mjc123

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Re: cm vs cv (adiabatic equation)
« Reply #2 on: May 26, 2024, 02:50:00 PM »
Looks like the problem is the factor of n in your expression of the exponent. You must be consistent in using either molar heat capacities or total heat capacities in both numerator and denominator. You can't use nR/cm.
The exponent is (Cp/Cv - 1), which in total heat capacities is nR/Cv, or in molar heat capacities is R/Cv,m.
I think just using γ-1 as the exponent is simpler and avoids this possible confusion.

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