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I'm confused about the internal energy of an ideal gas.

From the gas laws, $PV=nRT$, and both terms have a dimension of energy. In SI units it's Joules.

So, for air $(80/20 \;\mathrm{N}_2, \mathrm{O}_2)$ $n$ is about 29.

$R$ is 8.134 Joules/mole/°K

So one mole of air (29 grams) at stp (273°K, 101,300 Pa) occupies about 0.65 cu m. $V=nRT/P$: $= 28*8.314*273/101,300 = .65 \; m^3$

So $PV = nRT = 101300*.65 = 65,800 \; \mathrm{Joules}$.

On the other hand, Kinetic theory says the internal energy of an ideal diatomic gas (both O2 and N2 are diatomic) is $5/2nRT$ , which is 2½ times the PV value, i.e. 165,000 Joules.

Both numbers are energy in relation to 1 mole of gas at stp, so what are they in simple terms?

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3 Answers 3

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Dimensional analysis may provide proportionality relations but does not say anything about the numerical factors.

It turns out that for a classical (i.e., non-relativistic) ideal gas, there is a relation between $PV$ and internal energy of the form $$ PV=\alpha U $$ where $\alpha= 2/3$ for a monoatomic gas, $2/5$ for a diatomic gas, and in general, depends on the number of the internal degrees of freedom of polyatomic molecules.

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  • $\begingroup$ Thanks, but not an answer because it fails to address the question. $\endgroup$
    – alan.raceQs
    Commented Oct 8, 2022 at 21:07
  • $\begingroup$ @alan.raceQs if you read carefully my answer, you'll find that the correct relation between $PV$ and $U$ in the case of a diatomic molecule is $PV=\frac25 U$, I assume that you know that the equation of state is $PV=nRT$ for every perfect gas. Then the answer to all your question is there. Please explain why my answer does not address your questions if you need clarification. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Oct 9, 2022 at 4:47
  • $\begingroup$ Because it doesn't explain how to interpret, i.e. explain the meaning of, the energy/work/heat value returned from PV (or nRT) as @Chemomechanics did in their answer below. $\endgroup$
    – alan.raceQs
    Commented Oct 9, 2022 at 20:16
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There are many combinations of parameters with units of energy; that doesn't mean they all equal the internal energy!

$PV$ is the work required to insert a volume $V$ of some substance into an atmosphere at pressure $P$. This is distinct from the ideal gas internal energy $U=nc_VT$, where $c_V$ is the molar heat capacity ($\frac{5}{2}R$ for a diatomic gas).

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  • $\begingroup$ OK, I think I understand conceptually. So would it be correct to say that when heat is applied to a fixed volume of ideal gas some of the heat energy goes to increasing the linear momentum of the molecules which results in a rise in temperature and pressure and some of the heat energy goes to increasing the rotational momentum of the molecules which does not increase the temperature or pressure? And that ratio is reflected in the 5/2 ratio that we see for the internal energy from Kinetic theory. That makes sense, thanks. $\endgroup$
    – alan.raceQs
    Commented Oct 8, 2022 at 6:47
  • $\begingroup$ Heat capacity of a diatomic gas depends on temperature so $U \ne C_V T$. $\endgroup$
    – Andrew Steane
    Commented Oct 8, 2022 at 22:14
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    $\begingroup$ @AndrewSteane I don't consider that nuance relevant to the context of this question, which is why the internal energy of the ideal gas isn't $PV=nRT$. $U=C_VT$ and $PV=nRT$ are arguably perfectly acceptable equations of state in this context. $\endgroup$
    – Chemomechanics
    Commented Oct 8, 2022 at 22:28
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Unfortunately neither of those is the correct formula for internal energy.

The product $PV$ cannot be the internal energy, because it makes no account for the temperature $T$ of the gas. Now for an ideal gas, the internal energy $u$ not only depends upon the temperature, it depends only upon the temperature.

That formula is:

$$ u(T) = C_V T $$

where $C_V$ is the molar specific heat (at constant volume). But you have to be very careful that the specific heat is by mole or by mass, as needed, and that the reference condition for $u=0$ is consistent. In practice, the usual way to determine specific heat and other properties of fluids is simply reference tables, such as NIST Webbook, or any thermodynamics textbook, [viz] 1

Your second formula the specific heat:

$$C_V = \frac 5 2 R$$

based on the fact that a diatomic gas like $\mathrm N_2$ or $\mathrm O_2$ has 5 degrees of freedom in which to store energy.

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    $\begingroup$ Once the equation of state is known, $PV$ does make an account for the temperature. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Oct 8, 2022 at 5:39
  • $\begingroup$ Thanks but that is not an answer since it doesn't address the question which asked for an answer in simple terms, and the equations are wrong. Cv is NOT 5/2RT. That would make Cv a function of T which it clearly is not. Cv is constant (within wide temperature ranges) $\endgroup$
    – alan.raceQs
    Commented Oct 8, 2022 at 21:18
  • $\begingroup$ Correct, that was a typo. Should be $Cv=\frac 5 2 R$, my mistake. This page explains it in more detail than I can go into here, but hopefully still "simple" hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/shegas.html $\endgroup$
    – RC_23
    Commented Oct 8, 2022 at 21:52
  • $\begingroup$ careful: heat capacity of a diatomic gas depends on temperature so $U \ne C_V T$. $\endgroup$
    – Andrew Steane
    Commented Oct 8, 2022 at 22:13
  • $\begingroup$ Are you referring to the variation in Cv over extreme ranges : H2 3/2 R below ~100K , 5/2R below ~ 600K rising to 7/2R approaching 10,000 K or something else? $\endgroup$
    – alan.raceQs
    Commented Oct 8, 2022 at 22:37

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