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Will temperature of gas increase or decrease, from the formula PV=nRT?

by lord1412
Tags: decrease, formula, increase, pvnrt, temperature
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lord1412
#1
Jan14-14, 11:33 PM
P: 2
Dear all,

I would like to ask:

Given a control volume (V1), at an initial temperature (T1), the pressure (P1) in the volume (V1) is decreased by sucking out air. The suction hole is then closed to achieve a state of P2 * V2 = n2 * R * T2 .

What would happen to the temperature in the control volume? (ie would T2 > T1 or T2 < T1)

From my understanding:
"
By sucking out air it means reducing the number of moles (n) of air from the control volume.

P1 * V1 = n1 * R * T1
P2 * V2 = n2 * R * T2

so given that R is constant,

T2 = ( P2 / P1 ) * ( n1 / n2 ) * T1

so if ( P2 / P1 ) * ( n1 / n2 ) < 1, then T2 < T1, and otherwise is true.
"

Is my understanding correct? and if yes, how do i determine whether ( P2 / P1 ) * ( n1 / n2 ) > 1 or ( P2 / P1 ) * ( n1 / n2 ) < 1 ?

Thanks and best regards,
Andrew.
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Andrew Mason
#2
Jan14-14, 11:55 PM
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Quote Quote by lord1412 View Post
Dear all,

I would like to ask:

Given a control volume (V1), at an initial temperature (T1), the pressure (P1) in the volume (V1) is decreased by sucking out air. The suction hole is then closed to achieve a state of P2 * V2 = n2 * R * T2 .

What would happen to the temperature in the control volume? (ie would T2 > T1 or T2 < T1)

From my understanding:
"
By sucking out air it means reducing the number of moles (n) of air from the control volume.

P1 * V1 = n1 * R * T1
P2 * V2 = n2 * R * T2

so given that R is constant,

T2 = ( P2 / P1 ) * ( n1 / n2 ) * T1

so if ( P2 / P1 ) * ( n1 / n2 ) < 1, then T2 < T1, and otherwise is true.
"

Is my understanding correct? and if yes, how do i determine whether ( P2 / P1 ) * ( n1 / n2 ) > 1 or ( P2 / P1 ) * ( n1 / n2 ) < 1 ?

Thanks and best regards,
Andrew.
This is a first law problem. Since there is no heat flow involved here, if work is done on the gas that remains, what happens to the internal energy (hence temperature) of the gas? (hint: apply the first law where Q = 0). What happens if work is done BY the gas that remains?

So the questions are: Does the process of removing air molecules result in the surroundings doing work on the air inside the container? / Does the process result in the air that remains inside the container doing work on the surroundings?

AM
indianya
#3
Jan15-14, 05:02 PM
P: 3
Yes T2 < T1.

From my experience, if the valve on top of a can of compressed air is pressed to release the air, after a while can of air feels cold.

Zentrails
#4
Jan15-14, 05:58 PM
P: 98
Will temperature of gas increase or decrease, from the formula PV=nRT?

Quote Quote by Andrew Mason View Post
This is a first law problem. Since there is no heat flow involved here, if work is done on the gas that remains, what happens to the internal energy (hence temperature) of the gas? (hint: apply the first law where Q = 0). What happens if work is done BY the gas that remains?

So the questions are: Does the process of removing air molecules result in the surroundings doing work on the air inside the container? / Does the process result in the air that remains inside the container doing work on the surroundings?

AM
I would also ask what the OP considers to be "air" and how it deviates from an ideal gas.
Andrew Mason
#5
Jan15-14, 10:26 PM
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Quote Quote by indianya View Post
Yes T2 < T1.

From my experience, if the valve on top of a can of compressed air is pressed to release the air, after a while can of air feels cold.
Is that a case of the air being "sucked out"? Does that make any difference?

The question is not clear how this sucking occurs. If the air was removed by placing a completely empty chamber (vacuum) at an opening, would the air exiting the vessel do any work? What effect would that have on temperature?

AM
lord1412
#6
Jan16-14, 06:09 AM
P: 2
Thank you all for replying.

I think i will consider Andrew Mason's advice of looking for the change in internal energy.

The process of removing air is by simple vacuuming and then isolating the control space.

Zentrails, i know air is not ideal. But how else should i calculate for gases such as air?
dauto
#7
Jan16-14, 12:46 PM
Thanks
P: 1,948
The answer may depend on the details of the pump operation. for most situations the temperature will drop. But it is also possible for it to rise. Read up on Joule–Thomson effect for an explanation.
Andrew Mason
#8
Jan16-14, 02:49 PM
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Quote Quote by dauto View Post
The answer may depend on the details of the pump operation. for most situations the temperature will drop. But it is also possible for it to rise. Read up on Joule–Thomson effect for an explanation.
We are dealing with air. Dry air is close to an ideal gas and has almost no Joule Thomson effect. If there was a significant Joule-Thomson effect, I don't think it would result in heating. Rather the opposite would occur.

The only way temperature could rise is if work is done on the air in the process of removing it. That would require compression. I don't see any compression here. The question is really whether the air cools or whether it stays at the same temperature, and that will depend on whether the air in the container does any work in the process.

AM
russ_watters
#9
Jan16-14, 05:09 PM
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I don't see how there can be anything but a drop in temperature: this looks like a basic PV/T=k situation. The air in the container does work in pushing some air out.
Andrew Mason
#10
Jan16-14, 06:15 PM
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Quote Quote by russ_watters View Post
I don't see how there can be anything but a drop in temperature: this looks like a basic PV/T=k situation. The air in the container does work in pushing some air out.
Suppose the air is "sucked out" by sealing a large empty chamber (vacuum) around an outlet valve to the container and then opening the valve and leaving it open. That would be a free expansion. When everything returns to equilibrium there would be no work done on the surroundings so there would be no change in internal energy and no change in temperature (if the air is assumed to be an ideal gas). The wording of the question allows for such a scenario.

AM
russ_watters
#11
Jan16-14, 08:48 PM
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P: 22,219
But the air inside the chamber isn't undergoing free expansion, is it? Only the air that is released from the container. The air inside has to be constrained by something in order to release only the right amount of air: a valve. The air released is accelerated by the air that stays in the container, giving it kinetic energy and doing work. And the valve dissipates additional energy. The air inside the container is never truly unconstrained.

[edit]
Erp, maybe not: I was under the impression that air that stays in a container can't undergo free expansion, but it looks like it can (first diagram): http://en.wikipedia.org/wiki/Free_expansion

I'm not sure why the air slamming against the wall of the chamber doesn't cause a loss of energy, but in any case, the diagram points to a way the OP's question could be satisfied: Move the piston faster than the expansion (faster than the speed of sound) into a new position that provides the desired pressure. Then insert a new partition in place of the old one. Not certain if that satisfies the OP's constraints or not.
Andrew Mason
#12
Jan16-14, 11:26 PM
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Quote Quote by russ_watters View Post
But the air inside the chamber isn't undergoing free expansion, is it? Only the air that is released from the container. The air inside has to be constrained by something in order to release only the right amount of air: a valve. The air released is accelerated by the air that stays in the container, giving it kinetic energy and doing work. And the valve dissipates additional energy. The air inside the container is never truly unconstrained.

[edit]
Erp, maybe not: I was under the impression that air that stays in a container can't undergo free expansion, but it looks like it can (first diagram): http://en.wikipedia.org/wiki/Free_expansion

I'm not sure why the air slamming against the wall of the chamber doesn't cause a loss of energy, but in any case, the diagram points to a way the OP's question could be satisfied: Move the piston faster than the expansion (faster than the speed of sound) into a new position that provides the desired pressure. Then insert a new partition in place of the old one. Not certain if that satisfies the OP's constraints or not.
You are quite right that the air does work when it produces a directed air flow (through the valve) - but it does work on itself, not on the surroundings. When it settles down and reaches an equilibrium state all of that dynamic kinetic energy becomes random again and forms part of the internal energy of the gas, in a larger volume.

AM


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