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

  • #1
zenterix
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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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  • #2
The internal energy of an ideal (monoatomic) gas is ##3RnT/2##. Differentiating with respect to ##P## with ##T## constant is clearly zero.
 
  • #3
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.
 
  • #5
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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FAQ: Why does an ideal gas satisfy ##(\partial U/\partial P)_T=0##?

What is the significance of the partial derivative ##(\partial U/\partial P)_T=0## for an ideal gas?

The partial derivative ##(\partial U/\partial P)_T=0## signifies that the internal energy (U) of an ideal gas is independent of pressure (P) when the temperature (T) is held constant. This is a fundamental property of ideal gases, indicating that changes in pressure do not affect the internal energy as long as the temperature remains unchanged.

Why does the internal energy of an ideal gas depend only on temperature?

For an ideal gas, the internal energy is solely a function of temperature because the gas molecules do not interact with each other except through elastic collisions. The kinetic theory of gases states that the internal energy is proportional to the kinetic energy of the molecules, which depends only on temperature. Therefore, changes in pressure or volume do not affect the internal energy.

How does the ideal gas law relate to the partial derivative ##(\partial U/\partial P)_T=0##?

The ideal gas law, PV=nRT, describes the relationship between pressure, volume, and temperature for an ideal gas. Since the internal energy of an ideal gas depends only on temperature and not on volume or pressure, this leads to the result ##(\partial U/\partial P)_T=0##. The internal energy remains constant with changes in pressure at a fixed temperature.

Can the partial derivative ##(\partial U/\partial P)_T=0## be derived from the first law of thermodynamics?

Yes, the partial derivative ##(\partial U/\partial P)_T=0## can be derived from the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system. For an ideal gas, since the internal energy is a function only of temperature, any change in pressure at constant temperature does not contribute to a change in internal energy, leading to ##(\partial U/\partial P)_T=0##.

Does the partial derivative ##(\partial U/\partial P)_T=0## apply to real gases?

No, the partial derivative ##(\partial U/\partial P)_T=0## applies specifically to ideal gases. Real gases exhibit intermolecular forces and deviations from ideal behavior, especially at high pressures and low temperatures. For real gases, the internal energy can depend on both temperature and pressure, and thus ##(\partial U/\partial P)_T## may not be zero.

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