0
$\begingroup$

During a reversible process: $$du=dq-dw=c_vdT-Pdv=\left(\frac{\partial u}{\partial T}\right)_vdT+\left(\frac{\partial u}{\partial v}\right)_Tdv$$

From the above, I get $\left(\frac{\partial u}{\partial v}\right)_T=-P$, but for an ideal gas $\left(\frac{\partial u}{\partial v}\right)_T= 0$ because internal energy is only a function of temperature, so combine the above

$$P=0$$

Which is wrong, so what is the mistake I made during the above derivation?

$\endgroup$
1
  • 2
    $\begingroup$ The problem is that, for an ideal gas that is expanding, dq is not equal to $c_vdT$. For an ideal gas, $dU=c_vdT$ always, irrespective of the pressure and volume variations. $\endgroup$ Mar 31, 2022 at 10:06

1 Answer 1

2
$\begingroup$

You are wrong, $\left(\frac{\partial U}{\partial V}\right)_S = -p$.

You can show that $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V -p$ and everything checks out.

$\endgroup$
2
  • $\begingroup$ I agree with your derivation, but what is the problem of defining $\left(\frac{\partial u}{\partial v}\right)_T=-P$ from the expression $du=c_vdT-Pdv$ ? I can't find what's wrong here $\endgroup$
    – P'bD_KU7B2
    Mar 31, 2022 at 7:06
  • 2
    $\begingroup$ The problem is that $\delta q \neq c_v dT$. Actually, $\delta q = T ds = c_v dT + A dv$, where $A$ is some function of state. $\endgroup$ Mar 31, 2022 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.