# Homework Help: Finding the pressure of a gas in three identical balloons

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1. Oct 16, 2018

### Rain10399

1. The problem statement, all variables and given/known data
An adiabatic isolated system is formed of three identical balloons (of unknown volume). The balloons are joined by tubes of negligible volume. Each tube has a faucet/tap that is initially closed. The balloons have different quantities of the same ideal gas.
After opening only faucet R1, the system including balloons B and A reaches a state of equilibrium that has pressure p1 and temperature T1.
After opening only faucet R2, the system including balloons A and C reaches a state of equilibrium that has pressure p2 and temperature T2.
After opening only faucet R3, the system including balloons C and B reaches a state of equilibrium that has pressure p3 and temperature T3.
Find the initial pressure of the gas in each balloon (pA, pB and pC).

The drawing:

Also, the book that I'm using has the following results at the end:
pA=p2+p3-p1
pB=p3+p1-p2
pC=p1+p2-p3

2. Relevant equations
PV=nRT

3. The attempt at a solution
I tried to apply PV=nRT, but nothing that I did so far worked. I will edit the post if I have any ideas.

2. Oct 17, 2018

### ehild

What does adiabatic mean for the energy of the system? What is the formula for internal energy of an ideal gas?

3. Oct 17, 2018

### Staff: Mentor

After the valve between A and B is opened and the two chambers are allowed to equilibrate, how does the internal energy of the combined system compare with that of the two chambers initially?

4. Oct 17, 2018

### Rain10399

It means that the energy is transferred to the surroundings only as work, and the system doesn't transfer Q with its surroundings.
U = nCvT and U = -W (because Q=0).

U1=(nA+nB)CvT= nACvT + nBCvT= UA + UB

5. Oct 17, 2018

### Staff: Mentor

OK. Let V be the volume of the three chambers, let $n_A$ be the number of moles in chamber A, and let $T_A$ and $P_A$ be the initial temperature and pressure in chamber A respectively. Same for chambers B and C.

From the ideal gas law, in terms of $T_A$, $P_A$, and V what is $n_A$. Same for chambers B and C.

For the case where the valve between chambers A and B is opened, what is the first law energy balance on chambers A and B between the initial and final states in terms of $T_A$, $P_A$,$T_B$, $P_B$, V, and $T_1$?

What is the final pressure $p_1$ in terms of $T_A$, $P_A$,$T_B$, $P_B$, V, and $T_1$?

6. Oct 17, 2018

### ehild

The change of internal energy is equal to the negative of the work done by the system. But is there any work done on the surrounding if only a valve is opened between two balloons?

The initial temperatures in the balloons are not necessarily identical, neither are they the same as the final temperature in the joined balloons.

7. Oct 17, 2018

### Rain10399

nA=PAV/RTA
nB=PBV/RTB
nC=PCV/RTC

first law energy balance: U1=-W1
But I wasn't sure what formula to use for W here, so I searched on the internet for one.
nACvT1 + nBCvT1=-K[(2V)1-γ-(V)1-γ]/1-γ
I will assume that the balloons are filled with helium, which is a monatomic gas, so: (γ = 1.66)
nA1,5RT1 + nB1,5RT1=-K[(2V)-0,66-V-0,66]/-0,66

1,5PAVRT1/RTA + 1.5PBVRT1/RTB = -K[0,623V-0,66 -V-0,66]/-0,66
1,5PAVT1/TA+ 1.5PBVT1/TB=-K(-0.377V-0,66)/-0,66
1,5PAVT1/TA+ 1.5PBVT1/TB=0,571KV-0,66
2.626PAVT1/TA+ 2.626PBVT1/TB=-P1(2V)1.66V-0,66
2.626PAVT1/TA+ 2.626PBVT1/TB=-3.16P1V1.66V-0,66
0.831PAVT1/TAV+0.831PBVT1/TBV=-P1
P1=-0.831PAT1/TA-0.831PBT1/TB

I'm not sure if it's correct what I did.

Yes, I think so, because the walls of the system are not rigid.

Ohhh, yes, sorry! I got a bit carried away when I was solving the equation and for some reason I forgot about T1/TA/TB.
U1=(nA+nB)CvT1= nACvT1 + nBCvT1

Last edited: Oct 17, 2018
8. Oct 17, 2018

### ehild

Maybe balloon was not the right expression. They are containers or vessels with rigid walls. Is anything said about pistons ? And the pressure of the surrounding is not given, The adiabatic work formula refers to equilibrium processes. Opening a valve between two containers with different pressures is not that. So you can assume that the internal energy of A and B after having opened the valve and reaching equilibrium is the sum of the initial internal energies of A and B.

Write up the internal energy of the system consisting of vessels A and B before and after joining them.

9. Oct 18, 2018

### jbriggs444

Thank goodness. The pressure volume relationship in a balloon is not simple.

10. Oct 18, 2018

### Staff: Mentor

Correct so far.
This part was not correct, because the walls of the chambers are supposed to taken to be rigid, so no work is done on the surroundings.

That's more like it, but it is not quite correct yet. The equation should read:
$$(n_A+n_B)C_vT_1=n_AC_vT_A+n_BC_vT_B$$
What do you get if you substitute your equations for $n_A$ and $n_B$ into this equation?

11. Oct 18, 2018

### Rain10399

Ohhh. In the book it said balloons so I thought they actually were balloons. Thank you!
So because there's no heat exchanged and no work done the internal energy remains the same? I think I understand this part now:D
Then U1=UA+UB and (nA+nB)CvT1=nACvTA+nBCvTB, like Chestermiller said

Thank you! I tried again

$$(n_A+n_B)C_vT_1=n_AC_vT_A+n_BC_vT_B$$
$$(n_A+n_B)2P_1V/R(n_A + n_B)=n_AT_A+n_BT_B$$
$$2P_1V/R=P_AVT_A/RT_A + P_BVT_B/RT_B$$
$$2P_1V/R=(P_A + P_B)V/R$$
$$2P_1=P_A + P_B$$

Last edited: Oct 18, 2018
12. Oct 18, 2018

### Staff: Mentor

Great. This should be enough to enable you to complete the solution.

Last edited: Oct 18, 2018
13. Oct 18, 2018

### Rain10399

Yes, now I managed to solve the problem! Thank you so much to everyone who responded ^^. I love this forum.