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!

Easy Math Editor

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Comments

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 - 8 years, 3 months ago

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

Sachin Kukreja - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$