Finding the pressure of a gas in three identical balloons

[Rain10399]

Homework Statement

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

Homework Equations

PV=nRT
PV^γ = constant (adiabatic process)

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.

 

[ehild]

Homework Statement

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:
View attachment 232287

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

Homework Equations

PV=nRT
PV^γ = constant (adiabatic process)

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.

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

 

[Chestermiller]

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?

 

[Rain10399]

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

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).

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?

U₁=(n_A+n_B)CvT= n_ACvT + n_BCvT= U_A + U_B

 

[Chestermiller]

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 $T_1$?

 

[ehild]

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).

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?

U₁=(n_A+n_B)CvT= n_ACvT + n_BCvT= U_A + U_B

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

 

[Rain10399]

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 $T_1$?

n_A=P_AV/RT_A
n_B=P_BV/RT_B
n_C=P_CV/RT_C

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.
n_ACvT₁ + n_BCvT₁=-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 + n_B1,5RT₁=-K[(2V)^-0,66-V^-0,66]/-0,66

1,5P_AVRT₁/RT_A + 1.5P_BVRT₁/RT_B = -K[0,623V^-0,66 -V^-0,66]/-0,66
1,5P_AVT₁/T_A+ 1.5P_BVT₁/T_B=-K(-0.377V^-0,66)/-0,66
1,5P_AVT₁/T_A+ 1.5P_BVT₁/T_B=0,571KV^-0,66
2.626P_AVT₁/T_A+ 2.626P_BVT₁/T_B=-P₁(2V)^1.66V^-0,66
2.626P_AVT₁/T_A+ 2.626P_BVT₁/T_B=-3.16P₁V^1.66V^-0,66
0.831P_AVT₁/T_AV+0.831P_BVT₁/T_BV=-P₁
P₁=-0.831P_AT₁/T_A-0.831P_BT₁/T_BI'm not sure if it's correct what I did.

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?

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

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

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=(n_A+n_B)CvT₁= nACvT₁ + nBCvT₁

 

[ehild]

n_A=P_AV/RT_A
n_B=P_BV/RT_B
n_C=P_CV/RT_C

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.

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

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.

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=(n_A+n_B)CvT₁= nACvT₁ + nBCvT₁

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

 

[jbriggs444]

Maybe balloon was not the right expression. They are containers or vessels with rigid walls.

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

 

[Chestermiller]

n_A=P_AV/RT_A
n_B=P_BV/RT_B
n_C=P_CV/RT_C

Correct so far.

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.
n_ACvT₁ + n_BCvT₁=-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 + n_B1,5RT₁=-K[(2V)^-0,66-V^-0,66]/-0,66

1,5P_AVRT₁/RT_A + 1.5P_BVRT₁/RT_B = -K[0,623V^-0,66 -V^-0,66]/-0,66
1,5P_AVT₁/T_A+ 1.5P_BVT₁/T_B=-K(-0.377V^-0,66)/-0,66
1,5P_AVT₁/T_A+ 1.5P_BVT₁/T_B=0,571KV^-0,66
2.626P_AVT₁/T_A+ 2.626P_BVT₁/T_B=-P₁(2V)^1.66V^-0,66
2.626P_AVT₁/T_A+ 2.626P_BVT₁/T_B=-3.16P₁V^1.66V^-0,66
0.831P_AVT₁/T_AV+0.831P_BVT₁/T_BV=-P₁
P₁=-0.831P_AT₁/T_A-0.831P_BT₁/T_BI'm not sure if it's correct what I did.

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.

(n_A+n_B)CvT₁= nACvT₁ + nBCvT₁

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?

 

[Rain10399]

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.

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=U_A+U_B and (n_A+n_B)CvT₁=n_ACvT_A+n_BCvT_B, like Chestermiller said

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?

Thank you! I tried again

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

 

[Chestermiller]

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=U_A+U_B and (n_A+n_B)CvT₁=n_ACvT_A+n_BCvT_B, like Chestermiller said
Thank you! I tried again

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

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

 

[Rain10399]

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