I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?

  • $\begingroup$ Pressure is force per area as you know. The force is proportional to the number of molecules striking the wall which is in turn proportional to area. If you double the area you get double the force. If you divide that force by the area you'll get something that doesn't depend on the system scale. $\endgroup$ Dec 20, 2018 at 0:06
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    $\begingroup$ Pressure is also energy per unit volume, which I think is easier to think about with ideal gasses. If you cut the volume in half, you get half the (total) energy too because you have half the atoms. The ratio [pressure] remains constant however. $\endgroup$ Dec 20, 2018 at 19:54
  • $\begingroup$ @JonathanStott I think yours is ultimately the real reason. Beware of course that the pressure is not the energy density (though they have the same dimension) but the proof works the same (see my answer). $\endgroup$
    – lcv
    Dec 21, 2018 at 4:33

6 Answers 6


If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?

You are right that if we only halved the number of particles we would have a smaller pressure. But you have also halved the volume of the container. The fewer number of particles hits the walls more frequently due to the smaller volume. In other words, the number of particles goes down, but the number of collisions per particle goes up. The two effects cancel out, leading to the same pressure as before you put in the partition.


But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?

The pressure is not dependent on the number of molecules alone. You can simply examine the ideal gas law: $PV=nRT$. If the temperature is constant then reducing both $n$ and $V$ by half leaves pressure unchanged.

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    $\begingroup$ Physical answer vs. mathematical answer. Let's see who wins :p $\endgroup$ Dec 19, 2018 at 2:59
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    $\begingroup$ Usually physical does, but the mathematical one was so simple in this case. $\endgroup$
    – Dale
    Dec 19, 2018 at 3:11
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    $\begingroup$ Isn't this sort of a circular answer? The formula was (assumingly) derived from how this works in nature, so answering "why does it work like this?" with "because of this formula" seems like a circular argument. $\endgroup$ Dec 19, 2018 at 21:48
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    $\begingroup$ By that logic it would seem that formulas have no place in answering any physics question. I emphatically disagree. Essentially I showed how the property of intensiveness is obtained from the applicable formula. Nothing circular there. $\endgroup$
    – Dale
    Dec 20, 2018 at 17:46
  • $\begingroup$ @Dale intensiveness of pressure is not a property unique to ideal gases though (or maybe it is?) $\endgroup$
    – jacob1729
    Dec 20, 2018 at 19:54

Yet another way to think of it: if we use instead of $V$ and $n$ the molar density $\rho_n = \frac{n}{V}$, we get

$$P = \rho_n RT$$

or at molecular level, the molecular density (also number density) $\rho_N = \frac{N}{V}$ giving

$$P = \rho_N k_B T$$

which shows that the pressure is an intensive property, since the volume ($V$) does not appear. This $\rho_n$ is itself an intensive property for the same reason ordinary mass density is an intensive property.

Thinking about this more physically, since pressure is force over area, and the force is proportional to the number of molecules hitting it which in turn is proportional to how many happen to be in proximity, then we can think about it like this: the amount of molecules that each tiny piece of surface area "sees" remains the same in each case despite that we have cut off another half of the box, and thus it feels the same force. Think about a box of (reasonably small) ordinary macroscopic balls - if I insert a (thin) partition halfway in between the balls while displacing as few as possible, does any little bit of the area of the box's surface suddenly have much more or much less crowding next to it than before? The same thing happens here.


As a complimentary answer to Aaron Stevens answer; when we talk about thermodynamics and its macroscopic properties we consider them in the Thermodynamical limit. In other words the system will be in the state that has the highest probability of happening. For a system with a huge amount of particles this state will have a far higher probability than the others.

This means that global quantities such as pressure can be used for the system as a whole as well as for "non-microscopic" subdivisions of the system. Here "non-microscopic" means that the subdivision still contains enough particles to fulfill the statistical limit.

In your question you are trying to relate an extensive quantity (the number of particles in the system) to pressure which is an intensive quantity. You can try this instead:

  1. In the thermodynamic limit we expect the density of the gas (particle density) to be uniform across the whole system. If we divide the system up into a sum of boxes each box should still have the same gas density in it.

    1. Pressure has two components: The force of each particle and the amount of collisions per unit area. The amount of collisions depends on the particle density and not solely on the number of particles in the box. Otherwise you would expect the same number of collisions from a (10,000-particle, 1 $m^3$) box as a (10,000-particle, 10 $m^3$) box.

    2. Particle density is an intensive quantity in contrast to "Number of particles" which is extensive. If you divide the system into many parts you expect the particle density to remain the same in each new box.

    3. As with the other answers; this is readily seen e.g. in the $PV = nRT$ law. Rewrite it as $P = \left(\frac{n}{V}\right)RT$. The expression in the paranthesis is readily seen as the particle density. It carries the "amount of collisions" part of pressure. The $RT$ part carries the expected energy, or the force, of each collision.

As in Aaron's answer. You are not only halving the number of particles $n$; you are also halving the volume $V$, and those two together ends up making the particle density $\left(\frac{n}{V}\right)$ stay the same. Using the particle density as a lens to see the problem through I hope it becomes clear to you.

It ended up being quite a length answer. Hopefully it is not too overbearing. :)


Yet another way of looking at it: if the container is in equilibrium, then in aggregate the particles on one side the partition are the same as the ones on the other side. Every side a particle hits the partition on side A, that particle stays on side A, but would have moved to side B if it weren't for the partition. But there's another particle on side B that would have moved to side A. So those particles "cancel out" (again, in the aggregate). The "missing" particles that aren't are on side B because of the partitions are replaced by particles that stay on side B. If the container is equilibrium, then by definition, all the regions of the container contain essentially the same particles, so it doesn't matter whether it exchanges its particles with the neighboring regions (which is what happens without a partition), or it keeps its own particles (which is what happens with a partition). Putting in a partition doesn't affect the macrostate (other than the partition itself).


It may be interesting to give an answer based on statistical mechanics. Consider the situation where the system is a cylinder of length $L$ with a piston of surface $S$. I'm sure you have seen pictures describing this. In this case the pressure on the piston is computed via

$$ P = \frac{1}{S} \left \langle \frac{\partial H}{\partial x} \right \rangle = \frac{1}{S} \frac{\partial}{\partial x} \left \langle H \right \rangle $$

where $H$ is the system Hamiltonian, $x$ the coordinate of the piston, and the brackets describe the statistical average. $\left \langle H \right \rangle$ is the system's energy.

Now imagine cutting the cylinder in two. We can achieve this via the scaling transformation that sends $x\mapsto \alpha x$ (with $\alpha =1/2$, but we'll keep it more general), such that the length of the cylinder becomes $L\mapsto \alpha L$. Under this transformation the volume is sent to $V\mapsto \alpha V$ and the energy to

$$ \left \langle H \right \rangle \mapsto \alpha \left \langle H \right \rangle \ \ \ \ \ (1) $$

(modulo surface terms) since the energy is extensive. The surface of the piston is clearly constant $S\mapsto S$, while $\partial / \partial x \mapsto (1/\alpha) \partial / \partial x$. All in all we have shrunk the volume by a factor $\alpha$ but the pressure

$$P\mapsto P$$

stays invariant, in other words is intensive.

Remark 1 This proof is valid also for interacting systems, and not only for the ideal gas in which the interactions are discarded.

Remark 2 The presence of the piston is obviously not necessary. It is there only to allow visualize things or to measure the force. Moreover the proof can be clearly adapted to other geometries. Then one realizes that

$$ S dx = d V $$

represents the change in the volume. The formula for the pressure then becomes

$$ P = \frac{\partial E}{\partial V} $$

familiar from thermodynamics ($E = \langle H\rangle$). From this latter expression it's even more obvious that the pressure is intensive (sending $V \mapsto \alpha V$ one has $E \mapsto \alpha E$).


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