Zemansky and Dittman's Heat and Thermodynamics (6th ed, p. 108) defines an ideal gas as one that satisfies the equations: $PV=nRT$ and $(\partial{U}/\partial{P})_T=0$.

However, Rossini and Frandsen's experiment (1932):"...found no pressure or temperature range in which the quantity $(\partial{U}/\partial{P})_T$ was equal to zero" (p. 108).

Why include the second equation in the definition of an ideal gas, if the low-pressure limit of a real gas disagrees with the predictions of the ideal gas model?

An earlier query does not clarify the matter: When do ideal gases stop behaving like ideal gases?


4 Answers 4


I too have found definitions based on pressure to be unsatisfactory.

The definition of an ideal gas varies somewhat from author to author.

You may prefer the definition used by both Landau/Lifshitz and Callen, which is simply:

The interactions between the molecules are negligible.


"...a gas in which the interaction between the particles (molecules) is so weak as to be negligible."


"...a gas composed of non-interacting molecules..."

That's it. Very simple. They use no other criteria for defining an ideal gas. As Landau emphasizes (see the link above), all real gases will approximate to an ideal gas when their density is sufficiently low.

Note as well that this definition has no criterion on:

  • being point masses
  • being monatomic
  • being spherical
  • disallowing rotational, vibrational, or electronic modes

Also to note: some authors state that low pressure or high temperature is a criterion for the applicability of ideal gas relations. But be careful with that: if low density is the core criterion, then, by the ideal gas law itself, you can see that the density is proportional to the ratio P/T.

So it's really the ratio of P/T that is the best criterion here, not P or T by themselves. If you have a low P and a low T, then the ideal gas rule may not be a good approximation...

  • In classical physics, the internal energy of an ideal gas is only a function of temperature. That makes sense because of the basic assumption that atoms of an ideal gas do not interact. They are point particles and hence no matter what, any volume reduction (or variation in pressure) does not affect these particles.

  • While it can be proven in Statistical Physics from the most basic Hamiltonian (a collection of non-interacting free particles), an experiment proposed by Joules hinted that $U$ is a function of only $T$.

  • For an adiabatic exchange of heat with a freely expanding piston, the first law of thermodynamics tells you that even in the case of heat transferred, the work done is nil and hence energy change is zero. Volume and pressure have varied, without resulting any change in temperature.

Why include the second equation in the definition of an ideal gas, if the low-pressure limit of a real gas disagrees with the predictions of the ideal gas model ?

  • Real gases are not point particles, and no real gases behave ideally. Ideal gas is a toy model to understand the behavior of real gases and furnishes a lot of nice results to begin with. In general, the internal energy is a function of two variables, since the dependence on third is given by the state equation.

  • We generally include an interaction potential $V(\vec{r_1},\vec{r_2},...,\vec{r_n})$ to model their real behavior.

  • Also, a quantity called compressibility factor $(Z)$ is defined to quantify the deviation of real gases from ideal behavior. Many gases at STP conditions have $Z$ value of 1, indicating minimal to negligible deviations.

Note: In QM, the ideal Fermi (or Boson) gas is characterised by number density, temperature and available energy states. High-temperature limit of quantum ideal gas yields us classical ideal gas.

  • $\begingroup$ An ideal gas cannot be a gas of non-interacting fermions or bosons? The internal energy of a fermion gas depends on its density too. I guess what you say is true only for classical physics. $\endgroup$
    – ProfRob
    Jan 24, 2020 at 20:37
  • $\begingroup$ @Rob Jeffries, yes you are correct. I restricted the definitions to classical physics since the OP mentioned about Zemansky's book. I'll add the above point. $\endgroup$
    – exp ikx
    Jan 25, 2020 at 3:29

I don’t have a copy of Rossini etc, but for an ideal gas the internal energy is a function of the temperature only. $\Delta U=C_{v}\Delta T$. It applies to any process not just a constant volume process, I imagine Zemansky has the derivation. As I recall it involves combining the first law with the ideal gas equation. Therefore the partial derivative at constant temperature is zero.

Perhaps the gas used in the experiment didn’t behave like an ideal gas. All real gases depart somewhat from ideal gas behavior.

Hope this helps


From the Tables in 1 the molar mass of air is $29g/mol$ and heat capacity is $C_v(T=300K)= 0.718 J/(gK)$, and if it were an ideal gas its internal energy would then be approximately $29 \times 0.718 \times 300 = 6.24kJ/mol$.

According to Zemansky-Dittman Figure 5-3, page 112, where the measurements of Rossini and Frandzen are plotted, if $p<17bar$ then the error relative to the ideal gas law, i.e., change in the internal energy at fixed temperature, is less than $100J/mol$, that is $<1.6\%$.

Zemansky states that "Their experiment has somewhat the same disadvantage as Joule's original experiment, in that the heat capacity of the gas is much smaller than that of the calorimeter and water bath. To keep the temperature of the gas constant within reasonable limits, the temperature of the water must be kept constant to within less than a thousandth of a degree. In Rossini and Frandsen's measurements, the final precision was estimated to be 2.5%." (As a recovering EE I find this (dis)agreement quite (un)impressive...)


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