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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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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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