I Is this proof of cp - cv correct

  • I
  • Thread starter Thread starter planck999
  • Start date Start date
  • Tags Tags
    Cv Proof
AI Thread Summary
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.
planck999
Messages
23
Reaction score
7
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?
 
Science news on Phys.org
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 $$dH=dU+d(PV)=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
 
Thread 'Thermo Hydrodynamic Effect'
Vídeo: The footage was filmed in real time. The rotor takes advantage of the thermal agitation of the water. The agitation is uniform, so the resultant is zero. When the aluminum cylinders containing frozen water are immersed in the water, about 30% of their surface is in contact with the water, and the rest is thermally insulated by styrofoam. This creates an imbalance in the agitation: the cold side of the water "shrinks," so that the hot side pushes the cylinders toward the cold...
Back
Top