Ideal gas law says that $pV = nRT$. So with compression, one way to add pressure which makes the volume smaller, can you actually increase the temperature?

It does not make sense to me that you can increase temperature, that is, add to the kinetic energy of the molecules, simply by increasing pressure, pressure meaning the force exerted on the gas from its surroundings. Can someone please explain this contradiction?

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    $\begingroup$ pressure does not mean the force exerted on the gas by the surroundings, it means the force exerted by the gas on the walls of the container divided by the area of the wall. $\endgroup$ – orthocresol Oct 17 '15 at 23:52
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    $\begingroup$ Ok fair enough feed back about wording $\endgroup$ – user22003 Oct 18 '15 at 5:37
  • $\begingroup$ Karl, are you saying that, in the adiabatic compression of an ideal gas in a closed system (e.g., an insulated cylinder with a piston), there is no change in the gas internal energy or temperature? $\endgroup$ – Chet Miller Oct 18 '15 at 16:58
  • $\begingroup$ Really. Then please comment on the following for the case of an adiabatic reversible volume change: $dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$. So, $d\ln T=-\frac{R}{C_v}d\ln V$. So, when the volume decreases, the temperature increases. $\endgroup$ – Chet Miller Oct 18 '15 at 18:12
  • $\begingroup$ I mixed up free expansion and reversible compression/expansion. $\endgroup$ – Karl Oct 19 '15 at 19:39

I think what you are asking is "why can the temperature of the gas increase when you compress it, even if the cylinder is adiabatic so that no heat can enter the gas?" When you move the piston to compress the gas, you are doing work on the gas at the interface with the piston. The piston is moving toward the gas, and the molecules of gas that collide with the piston leave with a greater average velocity than when they arrived. So their average kinetic energy is increasing. If expansion were occurring, such that the piston were moving away from the gas, the colliding molecules would leave with lower average kinetic energy.

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  • $\begingroup$ Who said anything about compression rate? In the end, it is only the total amount of work that determines the change in temperature, but that's just equal to the integral of the force per unit area at the piston face times the rate of change of volume (compression rate) dt. You know, for an adiabatic process, $\Delta U = -\int{P_{ext}dV}$, where, for an ideal gas U = U(T). $\endgroup$ – Chet Miller Oct 18 '15 at 16:53
  • $\begingroup$ I stand by what I said. $\endgroup$ – Chet Miller Oct 18 '15 at 18:13
  • $\begingroup$ Are you talking about an ideal gas or air? For air at least part of the reason is that the molecules attract each other and it is not ideal. Would an ideal gas increase in temperature? (Either way PV=nRT, does not say wither T changes.) A good question not answered here or anywhere else I could find. $\endgroup$ – Tuntable Jun 24 at 4:30
  • $\begingroup$ @Tuntable I am talking about both real gases and ideal gases. Of course, an ideal gas would also increase in temperature. PV-nRT is not the only characteristic of an ideal gas that matters. The first law of thermodynamics also comes into play here, and, for an ideal gas, the internal energy is a function of temperature. Did you read my comments to the OP's post? $\endgroup$ – Chet Miller Jun 24 at 11:01
  • $\begingroup$ It is not at all clear that an ideal gas would increase in temperature, at least not significantly. Sure, increasing the pressure increases enthalpy, but enthalpy is T+PV. If you are sure that it increase temperature of an ideal gas, then by how much? Do you have a formula or a reference? $\endgroup$ – Tuntable Jun 26 at 1:17

If you had a way to increase pressure with no volume change, then yes, temperature would increase by the ideal gas law. In reality, most compression take place by reducing volume or increasing N, so the temperature effect is hard to see directly because other things are changing too.

The pressure in PV=nRT is the force exerted by the gas on the walls of the container. As the temperature increases, the particles move faster, and therefore have greater speeds, so greater momentum and therefore greater force when they collide with the walls, so the pressure increases.

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  • $\begingroup$ I understand what youre saying and I agree. Yes the ideal gas law says it happens in theory but does it actually happen in reality without changing the volume or the number of atoms? $\endgroup$ – user22003 Oct 18 '15 at 5:39
  • $\begingroup$ How can you compress a gas without changing its volume? Compressing means decreasing its volume. $\endgroup$ – Chet Miller Oct 18 '15 at 13:02
  • $\begingroup$ I was commenting on the idea of changing pressure without volume, not compression. $\endgroup$ – user22003 Oct 18 '15 at 17:50
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    $\begingroup$ At constant volume, you need to add heat to raise the temperature so that the pressure can increase. The temperature rise is the cause and the pressure rise is the effect, rather than the other way around. $\endgroup$ – Chet Miller Oct 18 '15 at 19:14
  • $\begingroup$ Ok, yes thats the idea I was getting at in the previous comment. Thanks! $\endgroup$ – user22003 Oct 18 '15 at 22:35

We all know, Solids have defenite size and obviously have defenite volume. Liquid has defenite volume but no shape. Gases have neither shape nor volume. Gas will occupy the available volume of the container. Molecules make use of the available free space for their movement.

Thus in gases, you can externally change the degree of freedom of its molecules. When you increase the container volume, you are increasing the degree of freedom of the gas molecules. And conversely also it is true.

Coming to the question, when you decrease the degree of freedom of molecules (by decreasing the container volume), due to the limitation in their mobility, the excess residual energy has to be given out (All system trend to minimize its energy state). Naturally the gas becomes hot in a big to exchange the excess energy to the surrounding. (Most of the natural energy exchange is done by heat energy).

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