Changes of gas in a vessel with a tap

Jun 2019
10
0
"A vessel of volume 2.0*10^-2 m^3 contains an ideal gas at a pressure of 1.50*10^5 Pa. A tap attached to the vessel is opened and the gas expands adiabatically until its pressure equals the atmospheric pressure of 1.01*10^5 Pa. The tap is then closed. The gas is allowed to return to its original temperature. In the final equilibrium state, the pressure of the gas that remains in the vessel is 1.13*10^5 Pa.

What is the volume of the gas that remains in the vessel under the initial pressure of 1.50*10^5 Pa and temperature?"

I do not fully understand the question in the end. I also could not picture what the vessel look like...
Could someone please explain...?
 
Jun 2016
1,198
565
England
Don't concern yourself over-much about the vessel
(if you need an image, its probably just a gas canister as used for camping gas or similar)
the key point is the phrase "the gas expands adiabatically"
There are standard equations relating Pressure, Volume and (crucially for this question) Temperature for an adiabatically expanding gas.

Note also that after the tap is closed the pressure in the closed container goes up,
Since the volume is fixed (the valve has been closed) this must mean that the temperature has gone up.
Since it ends up at the same temperature as it started, the temperature must have gone down during the initial expansion.

Ah, I have just re-read the final part of the question again, it is a bit odd isn't it!
So we have to work out how much gas is left in the vessel, and what volume that equates to at 150 KPa.
 
Jun 2019
10
0
Don't concern yourself over-much about the vessel
(if you need an image, its probably just a gas canister as used for camping gas or similar)
the key point is the phrase "the gas expands adiabatically"
There are standard equations relating Pressure, Volume and (crucially for this question) Temperature for an adiabatically expanding gas.

Note also that after the tap is closed the pressure in the closed container goes up,
Since the volume is fixed (the valve has been closed) this must mean that the temperature has gone up.
Since it ends up at the same temperature as it started, the temperature must have gone down during the initial expansion.

Ah, I have just re-read the final part of the question again, it is a bit odd isn't it!
So we have to work out how much gas is left in the vessel, and what volume that equates to at 150 KPa.
I have a question..
How does the gas allowed to return to its original temperature?
And... Why does the pressure increase when the tap is closed?
 
Jun 2016
1,198
565
England
The original temperature is "room" temperature (or the temperature of the environment outside the gas canister)
As the gas is allowed to escape, the remaining gas within the canister will get colder (adiabatic expansion equations).
After the valve is closed the canister is allowed to warm back up to "room" temperature.
 
Jun 2019
10
0
The original temperature is "room" temperature (or the temperature of the environment outside the gas canister)
As the gas is allowed to escape, the remaining gas within the canister will get colder (adiabatic expansion equations).
After the valve is closed the canister is allowed to warm back up to "room" temperature.
So it's like allowing the temperature to return to "room" temperature by removing its insulation?
 
Jun 2016
1,198
565
England
What Insulation?
I think you are overthinking again.
These questions often bear only the slimmest relationship to the "real" world.
You sometimes seem to have to invent an imaginary "teachers" world,
where all influences deemed extraneous to the question can be assumed not to exist.

An alternative view in this case might be that
One can rationalise the assumptions implicit within the question
by having very different timescales in the separate parts of the question.

So the initial adiabatic expansion happens quickly enough that the thermal conductivity of the gas canister is negligible
(the amount of heat energy that is transmitted through the metal shell is tiny compared to the adiabatic temperature change).

Then the canister is left for a long time (perhaps the rest of the day) to very slowly get back to room temperature.
 
Jun 2019
10
0
Oh... Okay. I get it.
Thanks a bunch for replying.