If we have empty container, and open the valve, and let the air in, what will be the temperature of air inside

Question

The container is initially empty (vacuum).
The walls are rigid, they are perfect thermal insulators, and have zero specific heat. There is also a small valve.

When the valve was open, the air rushed inside the container with a lot of hissing. After some time the pressure inside the containter became equal to atmospheric pressure on the outside.

Temperature of air in the room is 300K.
What is temperature of air inside the container?

The answer is 420K.

In the end n moles of gas are in the container.
Originally this amount of gas occupied volume
PoVo = nRTo,
and had internal energy
Eo = nCvTo.

This amount of gas did not lose any heat, and did not do any work. Instead work equal to
Work = PoVo = nRTo
was done on this gas by ambient pressure 1atm while pushing the gas into the container.

In the end internal energy of the gas is
E = Eo + Work =
Eo + PoVo =
(Cv + R)/Cv Eo =
Cp/Cv Eo

Final temperature of the gas is
T = Cp/Cv To = 7/5 x 300K = 420K

Answer 1

The air would cool. To tell how much, you need a book with either (a) the Joule-Thompson expansion effect for air at 300K, or (b) a PV chart for the same. [Sorry, I don't have these resources available].

Because air is composed of mainly nitrogen and oxygen, both of which are below the inversion point at room temperature, they will cool when doing a throttled expansion. If you talked with a high voice because you live in an atmosphere of Hydrogen and Helium, your air would get warmer when it expanded.

Why some gases cool and some gases heat up seems irrational at first. Kind'a like being voted best answer 105% of the time. lol

Really good question. Next time I get a Helium balloon for my kids, I am going to feel if it is warmer than atmosphere. It should be.


>>>>>>>>>>>>>>>>>> Edit

I found the following formula:

t.initial - t.final = mu * (p.initial - p.final) (273/(273+t))^2
where mu = 2.5 deg/MPa

ref: http://books.google.com/books?id=LxhQPdMRfVIC&pg=PA12&lpg=PA12&dq=joule+thomson+expansion+air&source=web&ots=ViobFqZiHH&sig=tTbYHvDGEBnJTVSlns2GN6YPxjQ

Here, p.initial = 0.1 MPa
p.final will vary between 0 to 0.1MPa
fortunately the relationship between t and p is linear therefor the temperature drop will be the average of p.final =0 and p.final = .1 MPA

therefore,

dT = mu (p.initial -.5 *p.initial) * (273/300)^2
=2.5 * (.1 MPa - .05MPa) * .83
= .104 deg
therefore new T=299.896 K

* I had thought the temperature drop would have been more. But looking at the equations, the reason is the drop in pressure from atmosphere to vacuum (or half-vacuum average) is not that much. To refrigerate gas, they really compress the stuff. I also note that scuba tanks are compressed to 20 MPa, which means that air coming out of them would drop ~ 40 K in temperature.

Answer 2

This is an adiabatic expansion. It takes energy for a gas to expand into a vacuum, but no energy is allowed in from outside (insulating walls). Hence the air will cool.

The amount of energy taken for a gas to expand into a vacuum is just PV, where V is the volume of the space (this follows from energy = force x distance = pressure x area x diatance). The total mass of gas involved is rho.V (rho = density of air at atmospheric pressure), so with c as the specific heat you get

c.rho.V.dT = PV

Or

dT = P/(c.rho)

Calculatig the temperature chage is tricky because the specific heat, c, will not be the same throughout the process and cannot easily be derived because both P and V are changing. However, you can get an estimate of the temperature change by using cp for air. And note that the temperature drop does not depend on volume.

You will find the result is surprisingly large - indeed adiabatic expansion is how gases are liquified.

Answer 3

Initially, as the air goes through the valve, the pressure will increase, so it will get slightly hotter. But as it enters the container, it will expand, and then lose this extra heat. Once the pressure is the same as the outside, the temperature should also be the same (if we ignore such minor things as any light/heat source outside like a lamp or sunlight that may make the outside air sightly warmer than inside the (assumed) dark container).

Answer 4

I've never heard this question before. This is really interesting.

I know when you let air out of a pressurized tank into the atmosphere it cools, but I think that's because it does work on the atmosphere's gases as it expands. Letting gas out into a vacuum is different -- there's nothing to do work on, so it will maintain its temperature.

Letting air into an empty tank looks like a situation that gradually changes from one scenario to the other. The first gas molecules to enter the tank won't do any work, but the later ones will do work on the ones that are already there.

This has me wondering .... it's gonna keep me thinking for quite a while. Good question!

Answer 5

Once the pressure has equalized, the temperature inside the container should be the same as the temperature outside of the container.

Answer 6

It'll have The Room Temperature.

Answer 7

Around 299.999K, depending on the size of the room.