Temperature and volume physics problem

[M11]

A fixed amount of gas molecules are trapped in a box.
when the volume of the container is incresed slowely, the temperature decreases. However, when the volume is increased quickly the temperature don't change or decreases slightly.
Explain why the temperature decreased more when the volume of the box was increased slowely.

 

[Mapes]

Hi M11, welcome to PF. Please show what you've tried so far.

 

[M11]

I think when the volume is increased fast, the molecules does not have enough time to exchange heat with the surrounding environment but I am not sure.

 

[Mapes]

I agree that it's a question of energy exchange over time, but I don't think it's necessarily energy transfer via heat. The question doesn't mention the temperature (or any properties) of the surrounding environment, but it does state that the walls of the box are moving. What's another possibility for energy exchange?

 

[M11]

the question is based on this simulation : http://phet.colorado.edu/simulations/sims.php?sim=Gas_Properties

I don't really know, In one case T decreases and on the other it doesn't although Vo and Vf are the same for both cases.

 

[M11]

the question is based on this simulation : http://phet.colorado.edu/simulations/sims.php?sim=Gas_Properties

I don't really know, In one case T decreases and on the other it doesn't although Vo and Vf are the same for both cases.

 

[Mapes]

Looks like the box walls are insulated (adiabatic), so there's not going to be any heat exchange. How else can energy be exchanged?

 

[M11]

maybe pressure

 

[Count Iblis]

M11, let's think of an exteme example in which you know what happens. Suppose you have a gas in a container at extremely high pressure, say 1000 bar. Then we make hole in the container. The gas steams out at high velocity, right?

All that energy could be used in a useful way, if you let the gas steam out slowly. But then the kinetic energy of the gas would not increase. The gas would occupy a larger volume as it performs work.

The performed work is energy lost to the gas and the gas cools as a result.

If you let the gas stream out violently into a larger container, then all of the energy that could have been used to perform work goes into the kinetic energy of the gas, so it stays in the system. No energy is thus extracted from the gas and the gas therefore does not cool (in case of an ideal gas, otherwise the gas can cool or heat up slightly).

 

[M11]

thx

but, it is still not clear

 

[M11]

which agent will reduce the temperature in the first case, and why it is not present in the second case ??

can anyone explain more please, I am running out of time.

 

[M11]

anyone ?

 

[Count Iblis]

which agent will reduce the temperature in the first case, and why it is not present in the second case ??

First case:

Due to the pressure, the gas exerts a force on the walls of the box. If the volume is increased slowly then the gas performs work of an amount equal to the pressure times the volume increase. This is energy lost to the gas. You can convince yourself in different intuitive ways that this is really the case. E.g. you can consider the gas pressure being used to lift a weight. The gas can push a piston which pushes a weight upward. The increase in potential energy of the weight is then recisely the pressure times the volume increase of the gas.

Then, to the gas it doesn't matter what happens to the performed work (it doesn't know if weight is lifted using it or that it is dissipated outside the box).

Now, suppose that the gas pushes a piston which moves frictionlessly in the horizontal direction. Then you would expect that the piston will accelerate the piston. So, the gas performs work which goes into the kinetic energy of the piston.

If we instead move the piston very fast to a new position but we don't fix it to the new position, then we create a vacuum between the gas and the piston. The gas then expands very fast to fill the vacuum. It then bumps into the piston, accelerating it. In this case, you still have a similar outcome as in the first case.

But now suppose that we first move the piston very fast to the new position and then fix it there. Then, when the gas arrives at the new position of the piston, it bumpes into it but it cannot transfer any energy to it as it is fixed. So, the kinetic energy of the gas stays in the gas where it will be dissipated. The gas doesn't perform any work.

Because no energy is lost, the temperature stays (approximately) the same.

 

[M11]

thx, it is clear now

:)