Background (in case it clarifies my question): I'm a 10th-grade school student. We've just learnt about the ideal gas model, the ideal gas law, and the formulae relating pressure, average kinetic energy of a gas particle, volume, and temperature. The question itself: so, the ideal gas law was presented to us as follows: $$ pV = \nu RT $$ where $p$ is pressure, $V$ is volume, $\nu$ is the amount of substance, $R$ is the universal gas constant, and $T$ is absolute temperature. If I expand $\nu$ as $\frac{N}{N_{A}}$ ($N$ is the amount of particles, $N_A$ is Avogadro's constant) and $R$ as $N_{A}k$ ($k$ is Boltzmann's constant) I get $$pV = NkT$$ One of the formulae our teacher presented us was $\overline{E} = \frac{3}{2}kT$ where $\overline{E}$ is the average kinetic energy of a particle. So, $kT = \frac{2}{3} \overline{E}$, which I can sustitute above: $$pV = \frac{2}{3}N\overline{E}$$ I thought (but I am not sure) that it is reasonable enough to assume that $N\overline{E}=E$, the overall energy the ideal gas has. So, pressure times volume is two-thirds of all energy. My question is: where is the rest? Where is another one-third of energy?
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$\begingroup$ Looked through my notebook, realised there is a much shorter way to arrive at my question, but still $\endgroup$PeteInThePit– PeteInThePit2025-12-03 22:37:41 +00:00Commented yesterday
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3$\begingroup$ Your question originates from an uncontrolled assumption: that $pV$ should be equivalent to the internal energy. Why? It is not that way, and there is no missing one-third of energy. $\endgroup$GiorgioP-DoomsdayClockIsAt-89– GiorgioP-DoomsdayClockIsAt-892025-12-04 08:07:46 +00:00Commented yesterday
3 Answers
You're wondering about the equation $$pV = \frac{2}{3} E.$$ I think your confusion might come from reading the equation as "two-thirds of the total energy is manifesting in the pressure times volume term". As was already stated by Giorgio's comment, this stems from the faulty assumption that $pV$ is itself a kind of energy. It's not.
Here's how to read it properly: $pV = \frac{2}{3} E$ is what we call an "equation of state". It relates 3 global properties of a thermodynamic system: Its pressure, its volume and its total internal energy. So if you know two of these properties, (e.g. pressure and volume), you can compute a third quantity from the equation of state. Such as the total energy $$E = \frac{3}{2} pV.$$
Similary, if you knew the energy and the volume, you could calculate the pressure as
$$p = \frac{2}{3}\frac{E}{V}$$
or if you know the energy and pressure, you could calculate the volume as
$$V = \frac{2}{3}\frac{E}{p}$$
This is how you'll usually write equations of state: expressing one system variable as a function of the others. Even though these are all mathematically equivalent, how you write the equation matters for interpretation. The way you wrote it suggests that it is about energy-budgets, which is not the case.
Let me know if this clears things up. It's a good question and it shows that you're thinking deeply about what the equations are saying.
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5$\begingroup$ +1 To add a simple analogy where you don't need thermodynamics. Just consider one Newtonian free particle of mass $m$, speed $v$ and energy $E$. Because all energy is kinetic, we know that $E = 1/2 m v^2$, or in other words $m v^2 = 2E$. So OP's question is kinda like saying "Mass times speed squared is 200% of the total energy. Where did half of that disappear?" In this formulation it's obvious that the question doesn't really make sense. $\endgroup$JiK– JiK2025-12-04 10:17:17 +00:00Commented yesterday
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$\begingroup$ And just to clarify: By "doesn't really make sense", I didn't mean that OP's question is bad, but that it stems from a confusion. I agree that it's great that OP asks such questions! $\endgroup$JiK– JiK2025-12-04 10:19:57 +00:00Commented yesterday
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$\begingroup$ Oh, yes, that makes sense! Thank you and JiK too! Come to think of it, I don't why have I thought that energy is "missing". $\endgroup$PeteInThePit– PeteInThePit2025-12-05 12:36:01 +00:00Commented 6 hours ago
The pressure $p$ is the average force (per unit of time and per unit of surface area) exerted on the walls of the gas container by the gas particles, bouncing against them. If you do the calculation, it is essentially $1$-dimensional motion, perpendicular to the wall, that contributes. By contrast, the average kinetic energy takes into account that the motion of the gas particles is $3$-dimensional.
If our world had a spatial dimension other than $3$, the relation between pressure and kinetic energy would be similar but with a different pre-factor.
While a couple of helpful and relevant answers have already been provided, an alternative discussion may provide some useful perspective. The questioner shows that pressure has units of energy per unit volume (the two quantities differ by a dimensionless factor). This is equivalent to the more common expression in terms of force per unit area. However, this does not mean that pressure and energy per unit volume represent the same physical quantity or that one is a component of the other.
Pressure can be defined as the amount of work which the gas does on its surroundings when it expands per unit increase in volume. Therefore, $$W = p\Delta V $$ where $W$ is the work done and $\Delta V$ is a small change in volume. The first law of thermodynamics states that the change in internal energy of the gas $\Delta E$ in the absence of heat transfer from/to the surroundings will be $$\Delta E = - W = - p \Delta V .$$ Note that for this expansion $$ \frac{\Delta E}{\Delta V} \ne \frac{E}{V}$$ as $p$ varies with $V$ (meaning that $E$ is not directly proportional to $V$).
It is also worth noting that the formula $\bar{E} = \frac{3}{2}kT$ only applies to an ideal gas for which the only motion is translation in the $x$, $y$, and $z$ directions (typically a monatomic gas). For molecules which contain more than one atom vibration and rotation may be possible. The additional energy associated with $n_{vib}$ vibrational modes and $n_{rot}$ rotational modes means that $\bar{E} = \frac{3 + n_{vib} + n_{rot}}{2}kT$ and hence $$ pV = \frac{2}{3 + n_{vib} + n_{rot}} E .$$
