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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?

Qmechanic
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agaminon
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2 Answers2

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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 $v$.

hft
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  • Term for pressure is only valid for adiabatic process or isolated process. What about isobaric or isothermal process. – Neil Libertine Nov 22 '22 at 05:02
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    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 '22 at 05:55
  • 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. – Neil Libertine Nov 22 '22 at 06:32
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    @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 '22 at 08:10
  • @Buzz We are discussing the question. A process not necessarily means a cycle. So helmholtz free energy to keep again system to its initial position, but for further process that again needs energy. This is way to extract more energy because entropy makes energy unavailable. $T\frac{\partial s}{\partial v}$ is the answer as asked in question. – Neil Libertine Nov 22 '22 at 08:19
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    @NeilLibertine I'm afraid I don't understand what your most recent comment is about, at all. – Buzz Nov 22 '22 at 08:20
  • @Buzz Why you said for isothermal this is not, while this is answered by me below. Why you use helmholtz free energy while it is to put system by adiabitically for further process. Read the question, read my answer and then comment also. What @ hft answered was one way, because all these internal energy, free energy depends upon many variables, so why pick any two. – Neil Libertine Nov 22 '22 at 08:58
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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

Neil Libertine
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  • Downvotes for correct answer, this seems people are not doing science but practicing religion. – Neil Libertine Nov 29 '22 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 09 '23 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. – Neil Libertine Jan 10 '23 at 04:36