thermodynamic derivation of heat capacityby tarletontexan Tags: capacity, derivation, heat, thermodynamic 

#1
Dec1111, 10:36 PM

P: 32

1. The problem statement, all variables and given/known data
c_{p}=c_{v}+TV?^2/? 2. Relevant equations c_{p}=T/N([itex]\partialS[/itex]/[itex]\partialT[/itex])_{p} 3. The attempt at a solution I have the equation, just not sure how to apply it? Any help would be appreciated 



#2
Dec1111, 11:07 PM

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AM 



#3
Dec1111, 11:55 PM

P: 32

yes, I know that there are several maxwell relations to get to the solution I just dont know how to apply them.




#4
Dec1211, 09:00 AM

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thermodynamic derivation of heat capacityTdS = dU + PdV C_{P} = (∂Q/∂T)_{P} = T(∂S/∂T)_{P} = (∂U/∂T)_{P} + P(∂V/∂T)_{P} AM 


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