# Pressure becomes zero for ideal gas?

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?

• 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. Mar 31, 2022 at 10:06

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.
• 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 Mar 31, 2022 at 7:06
• 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. Mar 31, 2022 at 7:22