# Thermal Physics Question

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Thread starter 8 months ago
#1
Can someone help me with the following question, thanks:

X and Y are two gas bottles that are connected by a tube that has negligible volume
compared with the volume of each bottle.
Initially the valve W is closed.
X has a volume 2V and contains hydrogen at a pressure of p.
Y has a volume V and contains hydrogen at a pressure of 2p.
X and Y are both initially at the same temperature.
W is now opened. Assuming that there is no change in temperature, what is the new gas
pressure?
0
8 months ago
#2
The final pressure exerted by the gasses, is it 4p/3?

This is a really weak topic of mine, I think I'm wrong ngl. But I still wanna improve so I'm gonna post.

I first found n in each container and then added them together to get the total n from the 2 containers; so I know the total amount number of moles of gasses present. At the end since the volume of the tube connecting the containers is negligible compared to the containers, this means that when the gasses are at equilibrium again they will be occupying a space of 3V. I added the number of moles from each container to get the total number of moles present, which was 4pV/RT(2). Then p(3V)=nRT(1), where this n is the total number of moles in both containers at the the new equilibrium. I subbed in my value of the total number of moles and rearrange to get my new pressure. The p in (1) is different to the p in (2).

This is probably wrong oh well.
1
8 months ago
#3
(Original post by Despaxir)
The final pressure exerted by the gasses, is it 4p/3?

This is a really weak topic of mine, I think I'm wrong ngl. But I still wanna improve so I'm gonna post.

I first found n in each container and then added them together to get the total n from the 2 containers; so I know the total amount number of moles of gasses present. At the end since the volume of the tube connecting the containers is negligible compared to the containers, this means that when the gasses are at equilibrium again they will be occupying a space of 3V. I added the number of moles from each container to get the total number of moles present, which was 4pV/RT(2). Then p(3V)=nRT(1), where this n is the total number of moles in both containers at the the new equilibrium. I subbed in my value of the total number of moles and rearrange to get my new pressure. The p in (1) is different to the p in (2).

This is probably wrong oh well.
The answer looks correct to me.
0
Thread starter 7 months ago
#4
Sorry for the late reply, but thank you for your help, I finally got the answer in the end.
(Original post by Despaxir)
The final pressure exerted by the gasses, is it 4p/3?

This is a really weak topic of mine, I think I'm wrong ngl. But I still wanna improve so I'm gonna post.

I first found n in each container and then added them together to get the total n from the 2 containers; so I know the total amount number of moles of gasses present. At the end since the volume of the tube connecting the containers is negligible compared to the containers, this means that when the gasses are at equilibrium again they will be occupying a space of 3V. I added the number of moles from each container to get the total number of moles present, which was 4pV/RT(2). Then p(3V)=nRT(1), where this n is the total number of moles in both containers at the the new equilibrium. I subbed in my value of the total number of moles and rearrange to get my new pressure. The p in (1) is different to the p in (2).

This is probably wrong oh well.
0
Thread starter 7 months ago
#5
(Original post by Stonebridge)
The answer looks correct to me.
Thank you for confirming the answer.
1
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