# Why does temperature rise with pressure?

## Main Question or Discussion Point

I am not a physics person. I am a lawyer. So bear with me:

I do not understand why the temperature of a gas or a liquid increases with pressure. I understand the Ideal Gas Law of PV=nRT, and I understand it mathematically, but I do not understand conceptually what is happening at the molecular level.

If I have a liter of oxygen in a sealed container that cannot expand, then I have a set number of molecules with a specific, set amount of average kinetic energy, temperature.

If I change the volume from a liter to a quarter liter, I understand that I have increased the frequency of collisions and therefore the pressure. But if the collisons with the walls of the container and with each other are perfectly elastic, doesn't the average kinetic energy (temperature) remain the same so that the increase in pressure does not increase the temperature?

If there is to be an increase in temperature, then there must be energy added to the average kinetic energy, and that must come either from an external source or potential energy, I believe.

If I am correct, then how does a diesel engine increase the temperature of the air by compression so that it and the fuel spontaneously combust under increased pressure? Is it the work of compression that imparts the additional kinetic energy to the gas molecules, or is my understanding set forth aboe wrong?

This is driving me nuts trying to figure it out.

Thanks so much for your help.

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HallsofIvy
Homework Helper
Yes, the work done to compress the gas is the energy that increases the temperature.

Gonzolo
Someone could come along with a better answer, but consider that when there are more collisions (due to higher pressure), the molecules change direction more often, so they move more. In my mind, that is what temperature is.

The heat that is emitted can be said to come from the energy used to compress the gas.

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if the amount of collisions increases with pressure increase, this means that the amount of EFFECTIVE collisions also increase, and also the rate at which the molecules move is also increased. This is the cause of temperature, an increase in speed of vibration. This is why temperature increases.

Chi Meson
Homework Helper
Temperature is proportional to the average kinetic energy of the molecules of the gas. THe previous posts do not explain where this additional kinetic energy comes from (well Hallsofivy does, but if your not familiar with the concept of Work, it might not be meaningful).

Think of a piston with a gas inside that cannot escape. As a force is exerted on the shaft of the piston, the piston head in turn applies a force on each of the gas molecules as they collide with the piston head.

Just as a ping pong ball will bounce away from a paddle with greater speed after it is struck, the gas molecules will rebound from the piston with greater speed if the piston is being compressed. THis force that is exerted on the molecules through a (rather tiny) distance is the quantity known as work.

If you were to forceably expand a gas by pulling the piston out, the rebound speed of the molecule will be less than the incoming speed. This is kind of like a bunt in baseball, where the bat absorbs some of the kinetic energy of the ball.

DoobleD
russ_watters
Mentor
But if the collisons with the walls of the container and with each other are perfectly elastic, doesn't the average kinetic energy (temperature) remain the same so that the increase in pressure does not increase the temperature?
If the walls of the container are moving inward (which they must do to increase the pressure), they add energy to any particle they hit just like a baseball bat adds energy to the baseball it hits.

DoobleD
Thanks for all of the replies and so quick.

I understand then that the extra kinetic energy and therefore the increase in temperature are a function of the work of compression (the piston (or walls) moving and imparting energy to the molecules) and not the function of pressure per se.

Am I correct then in understanding that pressure is certainly affected by temperature but temperature is not affected by pressure (absent some external force)?

LURCH
Temperature is effected not only by pressure, but more directly by compression. You can think of it this way; in the liter of gas from your original post, there is a certain amount of energy (heat). That energy is spread out over the entire volume of the container pretty much evenly. The amount of energy found within a given sample of that gas (e.g., 1 cm3) is what determines the temperature.

Compress the container to 1/4 its original size, and you have not removed any energy from the gas. Indeed, as has already been stated elsewhere, you have actually added some energy through the work done in compressing the gas. However, even without this additional energy, the temperature would increase. This is because the same amount of energy is now contained in a smaller space, so the amount of heat found in that same 1cm3 is now much greater. So, just as the density of the gas increases because there are now more gas molecules per cm3, the temperature also increases because there is more heat energy per cm3. Temperature can be thought of as the density of energy.

Thanks, Lurch.

I had thought that temperature was the average amount of kinetic energy of the molecules irrespective of pressure or volume.

It seems to be that if the statement "This is because the same amount of energy is now contained in a smaller space, so the amount of heat found in that same 1cm3 is now much greater" is correct, the Ideal Gas Law of pV=nRT would be knocked out of kilter, and Charles Law (volume increases proportionally with temperative) would not work.

Am I wrong? Is temperature the average kinetic energy of the molecules per unit of volume or pressure?

Thanks for the debate; I love it.

russ_watters
Mentor
tomtraxler said:
Am I correct then in understanding that pressure is certainly affected by temperature but temperature is not affected by pressure (absent some external force)?
You cannot change pressure without applying an external force just as you cannot change temperature without applying external heat.

cepheid
Staff Emeritus
Gold Member
LURCH said:
Temperature is effected not only by pressure, but more directly by compression. You can think of it this way; in the liter of gas from your original post, there is a certain amount of energy (heat). That energy is spread out over the entire volume of the container pretty much evenly. The amount of energy found within a given sample of that gas (e.g., 1 cm3) is what determines the temperature.

Compress the container to 1/4 its original size, and you have not removed any energy from the gas. Indeed, as has already been stated elsewhere, you have actually added some energy through the work done in compressing the gas. However, even without this additional energy, the temperature would increase. This is because the same amount of energy is now contained in a smaller space, so the amount of heat found in that same 1cm3 is now much greater. So, just as the density of the gas increases because there are now more gas molecules per cm3, the temperature also increases because there is more heat energy per cm3. Temperature can be thought of as the density of energy.
Umm...is this correct? I'm not trying to be confrontational, but you seem to be saying that if one could somehow magically decrease the volume of the container without compressing the gas thereby doing work on it , then the temperature would still increase, even though the overall thermal energy (not heat) contained in the gas has not changed. That seems wacky. As far as I know, temperature is proportional to the average kinetic energy of the particles in the substance, regardless of the volume they occupy. BTW tomtraxler... it's cool to see someone with such an inquisitive mind, and it must serve you very well. The lawyer's approach to physical inquiry is useful in the sense that you take a 'law' or definition in physics and ask someone..."if I am to take this prinicple at face value, then please explain to me how 'such and such' phenomenon could be possible?". However, I would question how much room there is for debate in basic thermodynamic principles. The definition of temperature is certainly not open to debate. As far as our scientific models describing nature go, all that is required to invalidate a physical 'law' such as the ideal gas law is for it to be violated even once in a real experiment. By the way, as I'm sure you know an ideal gas is an idealisation (of course). If you're interested (perhaps you've come across this already), there is another equation called the van der Waals equation which more accurately describes the behaviour of real gases since it takes into account that the particles in the gas have non-zero volume (they themselves take up space) and that there are forces between the particles. It's just the ideal gas law with a few correction factors introduced and is not difficult to understand at all.

krab
tomtraxler said:
Thanks, Lurch.

I had thought that temperature was the average amount of kinetic energy of the molecules irrespective of pressure or volume.
You are correct.

tomtraxler said:
It seems to be that if the statement "This is because the same amount of energy is now contained in a smaller space, so the amount of heat found in that same 1cm3 is now much greater" is correct, the Ideal Gas Law of pV=nRT would be knocked out of kilter, and Charles Law (volume increases proportionally with temperative) would not work.
Again, you are correct.

Conversely, if you allow a gas to suddenly expand, it will cool, and in the process do work on the mechanism allowing it to expand. This is how mechanical energy is extracted from a gas in for example the power stroke in the internal combustion engine. Microscopically, what happens is that the molecules striking the piston are slowed as the piston is receding.

Microscopically the force on a container wall with some (ideal) gas in it is proportional to the average velocity (determines the rate of collisions) and momentum (determines the 'impact' of the collisions) of the molecules constituting the gas.

As you might know, pressure is a force per area. Kinetic energy is half the product of momentum and velocity. So you can easily see that pressure is proportional to the average kinetic energy and thus the temperature...

LURCH
cepheid said:
Umm...is this correct? I'm not trying to be confrontational, but you seem to be saying that if one could somehow magically decrease the volume of the container without compressing the gas thereby doing work on it , then the temperature would still increase, even though the overall thermal energy (not heat) contained in the gas has not changed. That seems wacky.
Yes, I'm afraid this is exactly what I am saying. Somebody tell me if I'm way off hear, but if you decrease the amount of area in which a set amount of energy is contained, without adding any new energy, but also without removing any, then the same amount of thermal energy is concentrated into a smaller space. The average amount of thermal energy at any point within that smaller space must therefore be higher. (Theoreticaly, of course, since in practice this would be impossible to do.)

As an illustration, take the effect of a magnifying glass. It can take the sunlight from a certain area of space, and concentrate it into a smaller area. The glass does not add any energy to the sunlight, in fact it takes some energy away by reflecting it off the surface of the glass, or deflecting it (through internal imperfections) to some location outside the intended target area, or by heating up the glass itself. Nevertheless, most of the energy that was going to reach the Earth's surface in an area of several inches is now concentrated into an area of a fraction of an inch. The temperature in that smaller area is much higher than the temperature of the larger area the sunlight was reaching. Same amount of energy x smaller space = higher temperature.

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LURCH
I'm back with a question to clarify my own understanding. If we apply pressure to a fluid that does not compress easily, like water, what happens? We have expended energy in the attempt to compress the fluid, but its volume has not decreased any significant amount. Does the temperature of the water go up? I think it does not, but is that because the energy expended does not count as "work done"?

Just a "controll groupe" kind of question, we can better understand what happens when pressure is applied to compress a fluid if we can say what happens when pressure is applied without compressing a fluid, yes?

russ_watters
Mentor
LURCH said:
Yes, I'm afraid this is exactly what I am saying. Somebody tell me if I'm way off hear, but if you decrease the amount of area in which a set amount of energy is contained, without adding any new energy, but also without removing any, then the same amount of thermal energy is concentrated into a smaller space. The average amount of thermal energy at any point within that smaller space must therefore be higher. (Theoreticaly, of course, since in practice this would be impossible to do.)
That's all true, Lurch, but that's not the definition of temperature. If it were, it'd be awfully hard to get water at the same temperature as steam: water's density is many orders of magnitude higher than steam's.

Temperature is average particle kinetic energy: particle density is irrelevant.

In the sunlight analogy, you are taking a constant amount of energy and applying it to a smaller number of particles - not just a smaller area.

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I think, it is more simple.
Tomtraxler, you are absolutely correct in your reasoning.
You entangle in "But if the collisons with the walls of the container and with each other are perfectly elastic, doesn't the average kinetic energy (temperature) remain the same so that the increase in pressure does not increase the temperature?" Forget.
Ask yourself "What is the temperature?" It's measure of molecular velocity (frequency of collisions). So you say "I understand that I have increased the frequency of collisions and therefore the pressure." It's key. Increase the pressure =>increase the frequency of collisions=> increase the temperature

krab
da_willem said:
Microscopically ...
Goed zo. That's the coolest explanation I've seen.

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krab
LURCH said:
I'm back with a question to clarify my own understanding. If we apply pressure to a fluid that does not compress easily, like water, what happens? We have expended energy in the attempt to compress the fluid, but its volume has not decreased any significant amount. Does the temperature of the water go up? I think it does not, but is that because the energy expended does not count as "work done"?
If you apply pressure, and it does not compress, then no work is done, no energy expended. Work is force time distance, or pressure times change in volume.

krab
LURCH said:
Yes, I'm afraid this is exactly what I am saying. Somebody tell me if I'm way off hear, but if you decrease the amount of area in which a set amount of energy is contained, without adding any new energy, but also without removing any, then the same amount of thermal energy is concentrated into a smaller space. The average amount of thermal energy at any point within that smaller space must therefore be higher.
OK. You're way off here. Temperature is not a measure of energy density. Temperature is a measure of average energy per particle (actually, per degree of freedom, but let's not get too technical in this very basic discussion).

Okay, I'm a bit confused here. Perhaps a detail was left out of the scenario. If a cylinder of gas is compressed by an external force, say to one forth it's original volume, the pressure will increase. If the pressure increases to four times it's original value, wouldn't the temperature remain the same? PV=nRT so in this case, increased pressure does not mean increased temperature. But when the cyliner is rapidly compressed (say to one forth what it was before) and that added energy makes the gas hotter, the pressure has to go up by more than a factor of four.

The cylinder can be compressed slowly so that the temperature remains constant (infinitely slowly if no heat is exchanged with the surrounding environment) as pressure and volume change.

Maybe this isn't the right way to say it, but isn't PV=nRT an equation of state and not fully able to describe a dynamic situation?

krab
Bob3141592 said:
Okay, I'm a bit confused here. Perhaps a detail was left out of the scenario. If a cylinder of gas is compressed by an external force, say to one forth it's original volume, the pressure will increase. If the pressure increases to four times it's original value, wouldn't the temperature remain the same? PV=nRT so in this case, increased pressure does not mean increased temperature. But when the cyliner is rapidly compressed (say to one forth what it was before) and that added energy makes the gas hotter, the pressure has to go up by more than a factor of four.

The cylinder can be compressed slowly so that the temperature remains constant (infinitely slowly if no heat is exchanged with the surrounding environment) as pressure and volume change.

Maybe this isn't the right way to say it, but isn't PV=nRT an equation of state and not fully able to describe a dynamic situation?
You are correct. You need more than just the equation of state to tell you what happens during compression. Implicit in the above discussion was that the compression occurred fast compared with the heat transfer. This is usually called adiabatic. The equation that governs this is

$$PV^\gamma=\mbox{ constant}$$

where $\gamma$ is a constant related to the number of degrees of freedom; it is 7/5 for air. Combine this with PV=nRT, (i.e. substitute nRT/V for P) and you find how much the temperature increases when volume is decreased.

LURCH
So 100 joules of energy contained in 4m3 of a gas will have the same temperature as 100 joules of energy contained in 1m3 of that same gas?

So, what is actually going on in Bob's scenario? As I see it, the volume decreases, the pressure increases, work was done in compressing it - so the average KE of the particles must be higher - therefore higher T.

I'm not doubting you - but I just don't quite get this bit. Maybe I'm missing an assumption here? Is the system exchanging energy with the surroundings as the gas is compressed? If that were happening at a rate which compensates for the work done on the gas, I think I get it.

If not I'm completely perplexed....

russ_watters
Mentor
LURCH said:
So 100 joules of energy contained in 4m3 of a gas will have the same temperature as 100 joules of energy contained in 1m3 of that same gas?
If you mean strictly kinetic energy, then yes.