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Cp=Cv

By 1st law of thermodynamics I proved that Cp = Cv.

dH=dU + PdV

but, PV=RT (for 1 mole of gas)

P=RT/V & dH=CpdT & dU=CvdT

therefore, CpdT=CvdT + RT/V*dV

CpdTV = CvdTV + RTdV

& R=Cp-Cv

CpVdT=CvVdt + (Cp-Cv)TdV

CpVdT= CvVdT + CpTdV - CvTdV

Cp(VdT- TdV) = Cv(VdT- TdV)

therefore, Cp=Cv

Is there anything wrong I did there? Please tell me!

Note by Sachin Kukreja
4 years, 8 months ago

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the problem is in the equation for dH,

actually dH=dU + PdV +VdP for a general process AND PV=RT, so PdV + VdP= RdT.

Substituting this in equation and using dH=CpdT, dU=CvdT, we get

CpdT= CvdT + RdT, giving Cp-Cv=R

Nidhin Kurian - 4 years, 8 months ago

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I did not know that. Thank you. :-)

Sachin Kukreja - 4 years, 8 months ago

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