I Is this proof of cp - cv correct

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    Cv Proof
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The discussion focuses on the derivation of the relationship between specific heats, cp and cv, and the correct application of thermodynamic equations. It highlights the error in assuming that the change in volume (dV) is zero when calculating dU/dT. The correct starting point for the derivation is emphasized, using the equations for internal energy (dU) and enthalpy (dH). The importance of accurately applying the partial derivatives in thermodynamic relationships is stressed. Overall, the conversation aims to clarify the proper methodology for deriving cp - cv.
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cp=(dU/dT)P+P(dv/dT)P
cv=(dU/dT)V
cp-cv=(dU/dT)P+P(dv/dT)P- (dU/dT)V=(dU/dV)T(dV/dT)P+P(dv/dT)P- (dU/dV)T(dV/dT)V
since dV is zero (dU/dV)T(dV/dT)V is zero.
Hence
cp-cv=(dU/dV)T(dV/dT)P+P(dv/dT)P

I expanded both dU/dT and since one of them has no change in volume it is zero. is it acceptable? Did I multiply and divide both expressions by dV?
 
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This is done incorrectly. Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$and $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
 
Chestermiller said:
This is done incorrectly. Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$and $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
Thanks
Chestermiller said:
This is done incorrectly. Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$and $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
 
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