Does the pressure increase or decrease?


by zorro
Tags: decrease, increase, pressure
zorro
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#1
Oct17-10, 03:00 PM
P: 1,395
1. The problem statement, all variables and given/known data
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As the time passes the pressure of the gas in the cylinder-

a)Increases
b)Decreases
c)remains constant
d)increases or decreases depending on the nature of the gas

3. The attempt at a solution

The cylinder is metallic, so it is not an isolated system i.e. temperature remains constant.
As the volume is decreased, pressure must increase in accordance with Boyle's Law.
Unfortunately, the correct answer is b.
I have no clue
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fss
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#2
Oct17-10, 03:04 PM
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Quote Quote by Abdul Quadeer View Post
...it is not an isolated system i.e. temperature remains constant.
You might want to re-think this statement.
zorro
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#3
Oct17-10, 03:38 PM
P: 1,395
But temperature is lost to the surroundings due to conduction and convection processes.

ehild
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#4
Oct17-10, 03:52 PM
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Does the pressure increase or decrease?


"The piston is suddenly moved in to compress the gas"
What type of process is it?

ehild
zorro
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#5
Oct17-10, 04:01 PM
P: 1,395
Well I guess it will be an adiabatic process in that case and work will be done at the expense of its own internal energy and temperature rises. How do we relate pressure to it?
ehild
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#6
Oct17-10, 04:06 PM
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When the piston is suddenly pushed in, work is done on the gas, but it had not enough time to reach equilibrium with its surroundings. In equilibrium it will be at the same temperature as the surroundings, which is lower than the temperature just after the compression.

ehild
zorro
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#7
Oct17-10, 04:21 PM
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So what will happen to the pressure?
ehild
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#8
Oct17-10, 04:34 PM
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Quote Quote by Abdul Quadeer View Post
So what will happen to the pressure?
The volume is constant, amount of gas is the same, the temperature decreases, what happens to the pressure?

ehild
zorro
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#9
Oct19-10, 04:25 PM
P: 1,395
What I conclude from your explanation is that whenever we do work on a gas to compress it, its temperature increases. So in this case the temperature is increased and as the time passes, the temperature decreases in order to reach equilibrium with the surroundings. Since the volume is constant, pressure must decrease.

Is it right?

P.S.- Is there any equation to show that temperature increases on doing work on the gas? Moreover, we don't know whether the pressure increases/decreases before we fix the piston in that position after compressing.

Thanks for your anticipation.
ehild
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#10
Oct20-10, 01:26 AM
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What is thought by the undergraduate curses of Thermodynamics is really Thermostatics. It discusses the equilibrium states. The state variables are defined for equilibrium. Only those processes can be treated in the frames of Thermostatics which are so slow that practically go through equilibrium states. These processes are reversible. Other processes are treated in the frames of Non-equilibrium Thermodynamics or Irreversible Thermodynamics.

In this problem, the first stage of this process is very fast. The gas molecules are pushed by the piston and gain some extra momentum in the direction of the piston, so the distribution of velocity is not random any more unlike it is in equilibrium. If the gas is not in equilibrium with itself you can not define pressure and temperature or internal energy. After the piston stopped, the gas gets into equilibrium very fast and the First Law applies between the initial and final states: The change of the internal energy is equal to the added heat plus the work done on the gas by the external forces. No heat is added as the walls prevent fast heat transfer. The work done is not integral (-PdV) but the integral of the external force between the initial and final positions of the piston. This work is positive, so the gas gained energy. The amount of the gas stayed the same, and if it is ideal, U=Cv n T. If U increases, so does the temperature. Knowing the initial and final volumes and the work done by the external force, you can find the final temperature and pressure for the first stage of the process. The next stage is slow: heat transfer through the walls. The gas reaches equilibrium with its environment at the end, its temperature becomes equal to the external temperature. As PV/T is the same for the initial and the final states, and the final temperature is lower than the initial one, so is the pressure.

If the piston moves relatively slowly so the mass elements of the gas are in local equilibrium, for special problems you might determine the local volume, pressure and temperature during the compression, knowing how the piston moves and the heat transfer properties of the gas and the walls and you might determine the heat transferred. But this is the topics of Irreversible Thermodynamics.

ehild
zorro
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#11
Oct20-10, 02:44 AM
P: 1,395
Thanks for you detailed explanation!


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