Final pressure when two hydrogen bottles equalize

[zanyzoya]

Homework Statement

X and Y are two gas bottles that are connected by a tube that has negligible volume compared with the volume of each bottle. There is a valve in the tube that is initially closed. X has a volume of 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 initially at the same temperature. The valve between the two bottles is now opened, Assuming that there is no change in temperature, what is the new gas pressure?

Homework Equations

pV=nRT

The Attempt at a Solution

rearranging R = pV/nT so
(PxVx)/(Tx nx) = (PyVy)/(Ty ny)
T's cancel so
(PxVx)/nx = (pY Vy)/ny
Vx = 2Vy and Px = 1/2 Py so...
1/2Py x 2Vy / nx = Py Vy / ny
PyVy/nx = PyVy/ny
hence nx = ny

therefore PxVx/Tx = PyVy/Ty
temperatures cancel so
PxVx = PyVy
Vx = 2Vy and Px = 1/2 Py so...
1/2 Pyx 2Vy = PyVy
PyVy = Py Vy ?!?

 

[BvU]

You want to use the gas law, apparently before and after opening. How does that show in your working ?

 

[Chestermiller]

If T is the temperature, how many moles are present in each container to begin with (in terms of p, V, and T)?

 

[zanyzoya]

If T is the temperature, how many moles are present in each container to begin with (in terms of p, V, and T)?

2PV/RT moles in each container. Therefore a total number of 4PV/RT moles in the system.
Hence After opening
3P2= 4P
I.e. P2= 4/3 P
Am I right?

 

[Chestermiller]

2PV/RT moles in each container. Therefore a total number of 4PV/RT moles in the system.
Hence After opening
3P2= 4P
I.e. P2= 4/3 P
Am I right?

You know you are.

 

[zanyzoya]

Thanks for the help

 

[Falcon99]

Just got a question about this question . I'm using Boyle's law for this but am getting the new pressure as 2p/3 ?

Both bottles are at the same temperature and so will both have same constants of 2Vp.
So once the valve is opened total volume is now 3V however there is no change in temperature so boyle's constant will be the same . Equating this with P being new pressure : 3VP=2Vp ... P=2p/3 ? Is my theory wrong here?

 

[Chestermiller]

Just got a question about this question . I'm using Boyle's law for this but am getting the new pressure as 2p/3 ?

Both bottles are at the same temperature and so will both have same constants of 2Vp.
So once the valve is opened total volume is now 3V however there is no change in temperature so boyle's constant will be the same . Equating this with P being new pressure : 3VP=2Vp ... P=2p/3 ? Is my theory wrong here?

This is not correct. The OP analyzed the problem correctly in post #4. What made you think you could use Boyle's law on this?

 

[Falcon99]

This is not correct. The OP analyzed the problem correctly in post #4. What made you think you could use Boyle's law on this?

because bottle X:p,2V . Y:2p, V . Both x and y at same temperature , so constant for both is 2Vp . When valve opened new volume is 3V but the constant stays the same because temperature stays the same no ?

 

[mjc123]

The "constant" is proportional to the number of moles (as in PV = nRT). 1 mole of gas will not have the same value of PV as 10 moles.

 

[Chestermiller]

because bottle X:p,2V . Y:2p, V . Both x and y at same temperature , so constant for both is 2Vp . When valve opened new volume is 3V but the constant stays the same because temperature stays the same no ?

Initial number of moles = $\frac{p(2V)}{RT}+\frac{(2p)V}{RT}=\frac{4pV}{RT}$

Final number of moles = $\frac{P(3V)}{RT}$

Initial number of moles = Final number of moles