# How can both of these equations for pressure be correct?

Consider the Gibbs equation:

$$du=Tds-pdv$$

Identifying partial derivatives, one obtains:

$$-p=\left( \frac{\partial u}{\partial v} \right)_T$$

But you can also show that:

$$p=T\left( \frac{\partial s}{\partial v}\right)_T -\left( \frac{\partial u}{\partial v} \right)_T$$

In fact for an ideal gas, the latter partial derivative is $$0$$ and therefore it is the first term the one that determines its pressure. But how come both of these equations are true, at the same time?

Consider the Gibbs equation: $$du=Tds-pdv$$ Identifying partial derivatives, one obtains: $$-p=\left( \frac{\partial u}{\partial v} \right)_T$$

No.

$$-p=\left( \frac{\partial u}{\partial v} \right)_s$$

But you can also show that: $$p=T\left( \frac{\partial s}{\partial v}\right)_T -\left( \frac{\partial u}{\partial v} \right)_T$$ But how come both of these equations are true, at the same time?

Because you are using the wrong expression for $$p$$. You should use: $$-p=\left( \frac{\partial u}{\partial v} \right)_s$$

You can then consider: $$T=\left( \frac{\partial u}{\partial s} \right)_v = T(s,v)$$ to see that we can write $$s = s(T,v)$$. Then you can compute $$\left( \frac{\partial u}{\partial v} \right)_T$$ by considering the derivative of $$u(s(T,v),v)$$ with respect to $$T$$ at constant $$v$$.

• Term for pressure is only valid for adiabatic process or isolated process. What about isobaric or isothermal process. Nov 22, 2022 at 5:02
• Neil. What are you talking about. Look at this equation: $du=Tds-pdv$. I am literally just using the definition of the partial derivatives. Literally, by inspection: $T=\left(\frac{\partial u}{\partial s}\right)_v$ and $-p=\left(\frac{\partial u}{\partial v}\right)_s$.
– hft
Nov 22, 2022 at 5:55
• Your definition is based on internal energy as a function of entropy and volume, but question is about at constant temperature. Look at your temperature, which is constant that means isothermal. Now how volume kept constant and change in internal energy while entropy is changing at constant gibbs energy, constant volume and at constant temperature. Is this possible. Mathematics is different but look at situation. Nov 22, 2022 at 6:32
• @NeilLibertine If you are interested in an isothermal situation, then the equation for $dU$ is not useful. Instead, you need to use one of the free energies: Helmholtz, $dF=-S\,dT-p\,dV$, or Gibbs, $dG=-S\,dT+V\,dp$. From the equation for $F$, you can infer that, for example $p=-\left(\frac{\partial F}{\partial V}\right)_{T}$, since considering an isothermal process means exactly that $dT=0$; hence, for such a process, $dF=-p\,dV$ only.
– Buzz
Nov 22, 2022 at 8:10
• @NeilLibertine I'm afraid I don't understand what your most recent comment is about, at all.
– Buzz
Nov 22, 2022 at 8:20

Yes, both are true. Let consider this equation first, $$p=T\left( \frac{\partial S}{\partial V}\right)_T -\left( \frac{\partial U}{\partial V} \right)_T$$ For an ideal gas at constant temperature $$\frac{\partial U}{\partial V}_T$$ is not zero, but this is derived from constant gibbs energy thus it becomes zero.

Now, $$\ T\frac{\partial S}{\partial V}_T=-\frac{\partial U}{\partial V}$$. That is why both are correct. Although first term in it is, $$\frac{-Nk}{V}=\frac{\partial S}{\partial V}$$.

Reason: From first law,$$pdV=TdS-dU$$$$p=T\left(\frac{\partial S}{\partial V}\right)_T-\left(\frac{\partial U}{\partial V}\right)_T$$$$\text{Also,}\ \left (\frac{\partial U}{\partial V}\right)_T=kT\left(\frac{\partial N}{\partial V}\right)_T$$Now if, $$dG=0=\mu dN$$, then $$\frac{\partial N}{\partial V}=0$$

You can check it also, https://physics.stackexchange.com/a/736889/344834

• Downvotes for correct answer, this seems people are not doing science but practicing religion. Nov 29, 2022 at 11:49
• The first of the two pressure equations presented by the OP (i.e. the second equation) is not true. Your answer states "yes both are true". I believe you're trying to say that the first and third equations in the OP's question are true, but that's not what was confusing the OP, so your answer doesn't address the question.
– Rick
Jan 9 at 14:11
• @Rick I am sorry but if you omit $p=\left(\frac{\partial U}{\partial V}\right)_T$ or equals to zero, you get no radiation or radiation pressure. Jan 10 at 4:36