Why does an ideal gas satisfy ##(\partial U/\partial P)_T=0##?

[zenterix]

TL;DR

Why does an ideal gas satisfy $\left (\frac{\partial U}{\partial P}\right )_T = 0$?

The book I am reading says that by definition, the ideal gas satisfies the equations

$$PV=nRT\tag{1}$$

$$\left (\frac{\partial U}{\partial P}\right )_T = 0\tag{2}$$

where does (2) come from? In other words, what justifies this equation in the definition above?

 

[Orodruin]

The internal energy of an ideal (monoatomic) gas is $3RnT/2$. Differentiating with respect to $P$ with $T$ constant is clearly zero.

 

[zenterix]

The internal energy of an ideal (monoatomic) gas is $3RnT/2$. Differentiating with respect to $P$ with $T$ constant is clearly zero.

The thing is, I believe that equation comes from the kinetic theory of the ideal gas right.

The chapter of the book that I am on is a few chapters before talking about that theory. The only reason I know about that equation is from looking ahead.

I am wondering about some other justification not based on that theory.

 

[Chestermiller]

See my answer in this thread in another forum: https://chemistry.stackexchange.com...rtia/177559?noredirect=1#comment377712_177559

 

[Demystifier]

Ideal gas is made of particles which do not interact with each other, or more realistically, of particles for which the interaction energy is negligible. By compressing ideal gas while keeping temperature constant you increase the pressure $P$, but since the interaction energy is negligible, $U$ does not change. This answers the OP's question.