Undergrad Is this proof of cp - cv correct

[planck999]

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?

 

[Chestermiller]

This is done incorrectly. Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$and $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

 

[planck999]

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

This is done incorrectly. Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$and $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$