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

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The ideal gas is defined by the equations PV=nRT and (∂U/∂P)_T = 0, with the latter indicating that internal energy (U) does not depend on pressure (P) at constant temperature (T). This is justified by the assumption that ideal gas particles do not interact, meaning changes in pressure do not affect internal energy. The internal energy of an ideal monoatomic gas is expressed as 3RnT/2, and differentiating it with respect to pressure at constant temperature yields zero. The discussion highlights that the lack of interaction energy in ideal gases supports this relationship. Thus, the equation (∂U/∂P)_T = 0 is valid due to the negligible interaction energy among gas particles.
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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?
 
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The internal energy of an ideal (monoatomic) gas is ##3RnT/2##. Differentiating with respect to ##P## with ##T## constant is clearly zero.
 
Orodruin said:
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.
 
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.
 
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