Derivations in adiabatic process for ideal gas with C_V and C_P

  • Context: Chemistry 
  • Thread starter Thread starter zenterix
  • Start date Start date
  • Tags Tags
    Thermodynamics first law
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
zenterix
Messages
774
Reaction score
84
Homework Statement
I am quite confused by the calculations involving an adiabatic process for an ideal gas.
Relevant Equations
Below I show the calculations.
Consider an ideal gas undergoing an adiabatic process.

The first law says that

$$dU=\delta Q+\delta w=\delta w=-PdV$$

since ##\delta Q=0## for an adiabatic process.

##U## is a function of any two of ##P,V##, and ##T##.

Consider ##U_1=U_1(T,V)## and ##U_2=U_2(T,P)##.

For an ideal gas we have

$$dU_1=\left (\frac{\partial U_1}{\partial T}\right )_VdT=C_VdT=-PdV=\frac{nRT}{V}dV\tag{1}$$

$$dU_2=\left (\frac{\partial U_2}{\partial T}\right )_PdT=C_PdT=-PdV=-\frac{nRT}{V}dV\tag{2}$$

Are these equations both correct?
For (1) we have

$$C_VdT=-\frac{nRT}{V}dV$$

and after integrating we reach

$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$

where ##k## is a constant and ##\gamma=1+\frac{R}{C_V}##.

Note the implicit assumption that ##C_V## is constant.

Can we do the same thing for (2) to reach
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$

where ##k## is a constant and ##\gamma=1+\frac{R}{C_P}##?

Something is fishy here since ##C_V\neq C_P##.
 
Physics news on Phys.org
I think the mistake is that I assumed that

$$C_P=\left (\frac{\partial U}{\partial T}\right )_P\tag{Incorrect}$$

but actually

$$C_P=\left (\frac{\partial H}{\partial T}\right )_P$$

The calculations for (2) in the OP would be

$$\left (\frac{\partial U_2}{\partial T}\right )_PdT=-\frac{nRT}{V}dV$$

$$\frac{1}{T}\left (\frac{\partial U_2}{\partial T}\right )_PdT=-\frac{nR}{V}dV$$