Effects of decreasing volume and temperature for ideal gases

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  • #1
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How does decreasing the volume increase the temperature of the gases?

I was doing an experiment today and when i decreased the volume of the gas from 65ml at 1atm to 20ml the temperature detected an increase of 0.5°C. However, in Boyle's Law temperature is a constant.

So would this mean that I would have to combine Boyle's Law with the pressure law or to use the ideal gas law (PV=nRT) to get my actual pressure reading? But concept wise, I have a question about this occurrence.

When i decrease the volume, the only thing that changes is that the frequency of collision and the total surface area of the container. So assuming that all collision are elastic, when the gas molecules collide with the walls of the container won't they have the same velocity as before they collide. And since temperature is the average kinetic energy of the gas molecules, won't the temperature of the gas remain constant?

Lastly, so in tutorial questions where they are testing on Boyle's Law but don't state that temperature is constant, should we assume that the temperature remains the same? And only if they give us the final temperature, then we should use the ideal gas law to get the actual pressure?

Thanks for the help :smile:
 

Answers and Replies

  • #2
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Decreasing the volume of a gas is a compression- think of a piston. You have to do work on the gas to compress it, and the energy you put in goes towards increasing the rms kinetic energy of the gas. This energy is transferred via the collisions between the molecules and the inwardly-moving boundary.

Boyle's law is a special case of the ideal gas law when nRT is constant. If you detected a change in temperature then Boyle's law is inapplicable, and P1V1 = P2V2 will not match your experimental results. Instead, wait until the temperature of the compressed gas drops to its original value (room temp?) and then measure the pressure. This is the pressure that will satisfy P1V1 = P2V2.
 
  • #3
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Decreasing the volume of a gas is a compression- think of a piston. You have to do work on the gas to compress it, and the energy you put in goes towards increasing the rms kinetic energy of the gas. This energy is transferred via the collisions between the molecules and the inwardly-moving boundary.

Boyle's law is a special case of the ideal gas law when nRT is constant. If you detected a change in temperature then Boyle's law is inapplicable, and P1V1 = P2V2 will not match your experimental results. Instead, wait until the temperature of the compressed gas drops to its original value (room temp?) and then measure the pressure. This is the pressure that will satisfy P1V1 = P2V2.

Oh so the temperature increases because of the added energy to the system. But when the piston moves down then the energy starts getting transfered to the gas molecules right? So during that time would thr collisions be elastic?

And also, actually if we waited for the temperature to drop, how would the heat be given off? Because now that we've stopped decreasing the volume, the temperature would stop increasing. So by the elastic collision assumption of the gas how would the energy be given off? Because if the collisions are elastic then the velocities of the molecules should remains the same.

Similarly, when applying heat to the gas the energy would get to the gas molecules by conduction. So that would mean that after the collision the speed of the molecules increase. So again won't this mean that the collision isn't perfectly elastic?

Thanks for the help :)
 
  • #4
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Anyone has an ideas for this?
 
  • #5
jbriggs444
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Similarly, when applying heat to the gas the energy would get to the gas molecules by conduction. So that would mean that after the collision the speed of the molecules increase. So again won't this mean that the collision isn't perfectly elastic?

If you adopt a microscopic view, the collisions are elastic. But if the average kinetic energy of a gas molecule is greater than the average kinetic energy of a molecule in the wall then, on average, an elastic collision will take energy away from the gas molecule and deposit it in the wall molecule. [Note that this is a somewhat over-simplified view].

From a macroscope point of view and only looking at the behavior of the gas, the average effect is as if the collisions with the wall were not perfectly elastic -- at least until the temperatures equalize.
 
  • #6
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If you adopt a microscopic view, the collisions are elastic. But if the average kinetic energy of a gas molecule is greater than the average kinetic energy of a molecule in the wall then, on average, an elastic collision will take energy away from the gas molecule and deposit it in the wall molecule. [Note that this is a somewhat over-simplified view].

From a macroscope point of view and only looking at the behavior of the gas, the average effect is as if the collisions with the wall were not perfectly elastic -- at least until the temperatures equalize.

Hi so if we were to consider the gas as an ideal gas then how would moving piston's collision with the gas molecules conserve the momentum and kinetic energy of them? Because I'm thinking that the kinetic energy of the gas increases so how can the KE and momentum remain the same?

Thanks for the help :)
 
  • #7
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Anyone able to explain this? Thanks
 
  • #8
nasu
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Hi so if we were to consider the gas as an ideal gas then how would moving piston's collision with the gas molecules conserve the momentum and kinetic energy of them? Because I'm thinking that the kinetic energy of the gas increases so how can the KE and momentum remain the same?

Thanks for the help :)
Conservation of KE does not mean that each object involved in collision maintains its original KE.
You may have transfer of KE even during an elastic collision. Think about the extreme case of a billiard ball hitting another ball at rest. The first ball stops and the second one starts moving.
The first ball lost all its KE during an elastic collision.
If the second ball is not at rest, there will be just a partial transfer of KE.

Same for the collisions with the moving piston: the conservation laws applies to the system molecule-piston and not just for one component (gas molecule).
 
  • #9
morrobay
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Another way to look at this : What happens with temperature during increase in volume and
expansion of the gas ? The gas absorbs heat from the surroundings,to overcome attractions between gas atoms or molecules, and the temperature decreases.
So what has to happen when the gas is compressed in terms of conservation of energy.
 
  • #10
Andrew Mason
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Hi so if we were to consider the gas as an ideal gas then how would moving piston's collision with the gas molecules conserve the momentum and kinetic energy of them?
When an ideal gas is compressed, a force must be applied through a distance, so the force does work on the gas (ie. FΔs = (F/A)AΔs = PΔV where P can be just a tad more than the pressure of the gas in a slow compression). What does this work do? It does work on the gas molecules by increasing the kinetic energy of the molecules, so the temperature goes up (assuming heat flow out of the gas is prevented).

Because I'm thinking that the kinetic energy of the gas increases so how can the KE and momentum remain the same?

An elastic collision simply means that the total energy of the objects involved in the collision does not decrease. It does not mean that the kinetic energy one of the objects does not change. A baseball hitting a bat may be modeled as an almost elastic collision. The moving bat causes the ball to bounce off the bat with greater kinetic energy than does a stationary bat. That is the difference between a home-run and a bunt.

In the case of gas molecules, the force is greater when the wall is moving inward than when it is stationary. The force is less when the wall is moving outward than when it is stationary. So, in the case of an inwardly moving wall, the collision of a molecule with the wall causes the molecule to recoil with greater energy than its incident energy because the wall is doing work on the molecule while it is in contact with it.

AM
 

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