# Ideal gases!

1. Jul 15, 2009

1. The problem statement, all variables and given/known data
A and B are two identical containers connected by a tap initially closed. A contains an ideal gas at a pressure P1 and temperature T1. B contains the same gas at a pressure P2 and a temperature T2. The tap is the opened. If the temperatures of the containers A and B remain constant at T1 and T2 respectively, the final pressure of the gas mixture will be:

A.(P1+P2)/2
B.(P1+P2)
C.(P1T1+P2T2)/(T1+T2)
D.(P1T2+P2T1)/(T1+T2)
E.(P1/T1 + P2/T2)(T1+T2)

2. Relevant equations
PV=nRT

3. The attempt at a solution
P1 X 1 = n1 RT1 P2 X 1=n2 RT2

:( thats all i can do

Last edited: Jul 16, 2009
2. Jul 16, 2009

### queenofbabes

From what I see in your attempt, are you assuming volume of B is twice that of A? The question says two identical containers.

The key is that the final pressure is the same in both A and B. Use PV = nRT for both containers, before and after opening the tap.

3. Jul 16, 2009

sorry that was my typing error. Ive tried but because the answer does not involve volume or moles i was clueless and did not know how to come to an answer :(

4. Jul 16, 2009

### queenofbabes

Show us what you've tried, then we can tell you what went wrong!

5. Jul 16, 2009

i figured the partial pressure inside of each container before the tap is open is P1=P2T1/T2 and P2=P1T2/T1 respectively, but i am really lost for wat the final pressure will turn out to be as there are many variables such as change of volume and the pressures of the gases interacting with each other

6. Jul 16, 2009

### queenofbabes

You're not supposed to assume both containers had the same amt of gas to begin with. The volume of each container doesn't change. The total amount of gas doesn't change. So there aren't that many unknown variables at all! As I said, use PV=nRT for both containers. Lemme do it here:

Let f denote final.

$$P_1 V = n_1 RT_1$$

$$P_2 V = n_2 RT_2$$

$$P_f V = n_{1f} RT_{1} = n_{2f} RT_{2}$$

This should get you started =)

Last edited: Jul 16, 2009
7. Jul 16, 2009

thnx but im a little confused here. Shudnt PfV=n1fRT1 + n2fRT2 ? i dont see how they can be equal. And how abouts wud i be able to get rid of the n, V and the R because the answer in the multichoice only has P and T? sorry im only studying AS physics now so i may be making many wrong assumptions plz correct me if im wrong.

8. Jul 16, 2009

this is what i have attempted to work out so far:

P1V=n1RT1 so n1=P1V/RT1
P2V=n2RT2 so n2=P2V/RT2

when tap s is opened total number of moles= n1+n2 =nf (n moles final)

since P=nRT/V,

Pf= ((P1V/RT1+P2V/RT2)RT)/V
Pf= (P1 + P2)/2

But the answers say (P1T2+P2T1)/(T1+T2) is the correct answer so im assuming i went wrong somewhere?

9. Jul 16, 2009

### queenofbabes

The reason why you cannot use a single equation for the final state is because both containers are held at different temperatures. Well, if the final temperature for both containers is the same you could say Pf = nfRTf / 2V --This equation treats the whole system (A+B) as a single container of volume 2V. Thus in your case, Pf= ((P1V/RT1+P2V/RT2)RT)/V is therefore incorrect. You have to consider each container in the final state separately.

10. Jul 16, 2009

hmm i see. I am really lost for ideas for what to do next. Somehow i also have to eliminate n and R to get the answer. I've spent a long time on this question can u tell me what to do?

11. Jul 16, 2009

### Q_Goest

Hi galdius. Welcome to the board. If we draw a control volume around the two identical containers, then in addition to the ideal gas law, we can also apply the first law of thermodynamics. If there is no exchange of heat or work with the surroundings, then what does that say about the total change in internal energy? Note that one container may pick up some mass and the other container may loose a bit, but the entire contents must come to equilibrium (constant pressure and temperature throughout).

Think of it this way, you have a container with a partition that has hot gas on one side and cold gas on the other. You remove the partition and they mix. The hot gas gets colder and the cold gas gets hotter.

To calculate changes in internal energy for an ideal gas, do you know what equation to use? Hint: Cv not Cp

12. Jul 16, 2009

hi Goest, sorry i dont know, im still in high school 6th form :( i understand that the 2 temperatures will mix but the question states that the temps are kept constant.

13. Jul 16, 2009

### queenofbabes

Yes, do exactly what you said: you have to eliminate n. This is more of an exercise in algebra than physics, really...

Remember that $$n_1 + n_2 = n_{1f} + n_{2f}$$
You can take V, R = 1 without any loss of generality since they appear in the same way in each of the stated equations in post#6 (if you're not convinced, repeat the whole process keeping them in place, you'll see that they cancel out)

So, we sub $$n_1 = \frac{P_1}{T_1} , n_2 = \frac{P_2}{T_2}$$
and $$n_{2f} = n_{1f} \frac{T_1}{T_2}$$ into the equation above.

I'll leave you to finish up the final bits of algebra =)

14. Jul 16, 2009

### Q_Goest

I'm sorry, I missed that part. My bad. I see queenofbabes came back and gave you a helpful responce, so I'll leave it at that.

15. Jul 16, 2009

i got: P1/T1 +P2/T2= n1f + n1fT1/T2
simplifies to 2n1f=P1T1/T1T2+P2T1/T2
n1f=(P1T1/T1T2+P2T1/T2)/2

sub into PfV=n1fRT1 assuming V, R =1
Pf=n1fT1
Pf=P1T1/2T2 + P2T1^2/2T2

sorry im horrible at this, where did i go wrong?

16. Jul 16, 2009

### queenofbabes

Your second line is wrong...check it again.

17. Jul 17, 2009

YES I THINKS I GOTS IT!

P1/T1 + P2/T2=n1f + n1fT1/T2

simplifies to n1f= ((P1T2/T1)+P2)/T1+T2

since Pf=n1fT1

Pf= ((P1T2/T1)+P2)/T1+T2 X T1
= (P1T2+P2T1)/(T1+T2)
= correct

Thank you very much queenofbabes for your help. Just wondering though, can u only assume R and V as 1? would this still work if u assumed it was 2? and how abouts do u come to think of the relevant equations needed to solve this question?.
Thanks again for your time and effort :)

18. Jul 17, 2009

### queenofbabes

If you did all your calculations with R and V as they were, you will find that they will all cancel out in the end. Thus it doesn't matter what values they were: they could be anything, but of course assuming R, V = 1 simplifies your working.

And really, if you examine the starting steps, there's nothing more to it than applying PV = nRT to each container, before and after opening the tap, then solving simultaneous equations.