Identical ideal gas in two different volume containers

[keltik]

Homework Statement

Container A in Fig. 19-22 holds an ideal gas at a pressure of 5*10^5 [Pa] and a temperature of 300 [K].

It is connected by a thin tube (and a closed valve) to container B, with four times the volume of A.

Container B holds the same ideal gas at a pressure of 1*10^5 [Pa] and a temperature of 400 [K]

The valve is opened to allow the pressures to equalize, but the temperature of each container is maintained.

What is then the pressure?

(I hope you can make up the problem-figure by yourself, if not i can post a picture.)

Homework Equations

only the "ideal gas law":

P*V=n*R*T

The Attempt at a Solution

When the state is "valve closed" then the following values hold:
Information about container A:
Volume: V_{A} = \frac{V_{B}}{4} [m^3]
Temperature: 300 [K]
Pressure: 5*10^5 [Pa]

Information about container B:
Volume: V_{B} = 4*V_{A}
Temperature: 400 [K]
Pressure: 1*10^5 [Pa]

Ok, setting up the "ideal gas law" for container A gives equation:
P_{A}*V_{A}=n_{A}*R*T_{A}

Plugging in Temperature, Pressure and the gas constant R, yields:
5*10^5*V_{A}=n_{A}*8.3*300

Solving that for Moles in container A, yields:
n_{A}=200.8*V_{A}


Doing exactly the same for container B, yields:
n_{B}=30.1*V_{B}

in that equation exchanging V_{B} with 4*V_{A} gives:
n_{B}=30.1*4*V_{A}, which is in turn:
n_{B}=120.4*V_{B}

A sanity check on these Mole-numbers reveals that these values could be correct, since there is a higher pressure in container A with the lower Volume than there is pressure in container B. So these numbers make sense. But i still don't have any values for it, so i decided to look at the "valve opened and equalized"-state:


Equalized means that container A gives container B a certain amount of his Molecules, since it has more pressure (although the temperature is higher in container B, it cannot outweigh the overpressure of container A)

This happens isothermically, which means T_{A} = 300 [K] and T_{B} = 400 [K] remain as they are.

My forecast is that in container A the pressure is going to be less and in container B the pressure is going to be higher than in their respective "valve is closed"-state, but i can't give you numbers.

Also total volume is now:
V_{C} = V_{A} + V_{B}

And the total number of molecules is:
n_{C} = n_{A} + n_{B}

But plugging those values in does not cancel something out (or i was too stupid for it?), thus yielding no concrete values.

Can someone give me one or two hints?

Thanks for your time. (I know i could look into the solutions manual, but that would be too easy...)

 

[mfb]

I would still keep both containers as separate parts in the equation, as their temperature is different.
Assuming n*V_B moles flow from A to B, can you calculate the pressure in A and B?

 

[Chestermiller]

What you are missing is that in the final state, the pressures in the two chambers will be the same.

 

[keltik]

What you are missing is that in the final state, the pressures in the two chambers will be the same.

That is correct, i didnt write that down, but i am aware of that.

 

[keltik]

Thank you guys i looked into the solutions manual for hints.

The "trick" basically is:
- i know that the total molecules will not change (no leakage) after equalizing, so i can write:

n_{total, closed} = n_{total, equalized}

which can be written as:

n_{A,closed} + n_{B,closed} = n_{A,equalized} + n_{B,qualized}

the rest is "use the ideal gas law, plug everything in and insert it here, cancel out", out comes the "equalized pressure".

 

[Ahamed]

Hi, Can you tell me which textbook does this question comes from?

 

[mfb]

Welcome to physicsforums Ahamed.
This thread is 1.5 years old and keltik didn't make posts since then. I'll close the thread, for the unlikely case that he will ever see that he can send you a private message.

 

[berkeman]

Hi, Can you tell me which textbook does this question comes from?

Hi Ahamed,

Google "Container A in Fig. 19-22 holds" (in quotes), and you will get lots of hits to the problem... :-)