If the internal energy of an ideal gas is u and volume is v then at constant temperature joule's law is :
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The internal energy of an ideal gas is independent of volume when considered as a function of volume and temperature. If we choose to consider internal energy as a function of volume and some other thermodynamic variable we will find that the dependence of the energy on volume will change because we are keeping a different variable constant as volume is varied.
So if we consider $U$ as a function of volume and entropy we get $$ \mathrm{d}U = \left(\frac{\partial U}{\partial S}\right)_V \mathrm{d}S + \left(\frac{\partial U}{\partial V}\right)_S \mathrm{d}V. $$ Now $P = \left(\frac{\partial U}{\partial V}\right)_S$ and is certainly not equal to 0.
The particular case of an ideal gas is unusual because it terns out that the internal energy is only a function of temperature. This means $\left(\frac{\partial U}{\partial X}\right)_T = 0$ for any variable $X$. If we choose our thermodynamic degrees of freedom to be variable other than $T$ however, say for concreteness $S$ and $V$ again, then we are treating $T$ as a function of $S$ and $V$ as well and so $U$ gains a dependence on $V$ and $S$ through $T$.
So if we consider $U$ as a function of volume and entropy we get $$ \mathrm{d}U = \left(\frac{\partial U}{\partial S}\right)_V \mathrm{d}S + \left(\frac{\partial U}{\partial V}\right)_S \mathrm{d}V. $$ Now $P = \left(\frac{\partial U}{\partial V}\right)_S$ and is certainly not equal to 0.
The particular case of an ideal gas is unusual because it terns out that the internal energy is only a function of temperature. This means $\left(\frac{\partial U}{\partial X}\right)_T = 0$ for any variable $X$. If we choose our thermodynamic degrees of freedom to be variable other than $T$ however, say for concreteness $S$ and $V$ again, then we are treating $T$ as a function of $S$ and $V$ as well and so $U$ gains a dependence on $V$ and $S$ through $T$.
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