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I want to know that if Cp= Cv. If it is how can we derive it?

Note by Sachin Kukreja 3 years, 7 months ago

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How dH = dU + Pdv??? The correct relation is dH = dU + Pdv + Vdp – Saurabh Dubey · 3 years, 7 months ago

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but by 1st law of thermodynamics I can prove 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! – Sachin Kukreja · 3 years, 7 months ago

NO!!! infact Cp=Cv+R R is gas constant. – Saurabh Dubey · 3 years, 7 months ago

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TopNewestHow dH = dU + Pdv???

The correct relation is

dH = dU + Pdv + Vdp – Saurabh Dubey · 3 years, 7 months ago

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but by 1st law of thermodynamics I can prove 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

CpdT

V = 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! – Sachin Kukreja · 3 years, 7 months ago

Log in to reply

NO!!! infact Cp=Cv+R R is gas constant. – Saurabh Dubey · 3 years, 7 months ago

Log in to reply