# Finding the pressure of a gas in three identical balloons

• Rain10399
In summary, the problem involves an adiabatic isolated system consisting of three identical balloons (of unknown volume) connected by tubes with faucets/taps that can be opened or closed. The initial state of the system includes different quantities of the same ideal gas in each balloon, and after opening only certain faucets, the system reaches equilibrium states with different pressures and temperatures. The task is to find the initial pressure of the gas in each balloon, represented by pA, pB, and pC. The solution involves using the ideal gas law and the first law of thermodynamics to determine the values of the initial pressures in terms of the initial temperatures, pressures, and volume, as well as the final state temperatures reached after opening the different
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
PVγ = 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.

#### Attachments

• k0GRKIm.png
5.1 KB · Views: 723
Rain10399 said:

## 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
PVγ = 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?

Rain10399
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
ehild said:
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).

Chestermiller said:
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=(nA+nB)CvT= nACvT + nBCvT= UA + UB

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##?

Rain10399
Rain10399 said:
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?

Rain10399 said:
U1=(nA+nB)CvT= nACvT + nBCvT= UA + UB
The initial temperatures in the balloons are not necessarily identical, neither are they the same as the final temperature in the joined balloons.

Rain10399
Chestermiller said:
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##?
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/TBI'm not sure if it's correct what I did.

ehild said:
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.

ehild said:
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=(nA+nB)CvT1= nACvT1 + nBCvT1

Last edited:
Rain10399 said:
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.

Rain10399 said:
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.

Rain10399 said:
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
Write up the internal energy of the system consisting of vessels A and B before and after joining them.

Rain10399
ehild said:
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.

Rain10399 said:
nA=PAV/RTA
nB=PBV/RTB
nC=PCV/RTC
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.
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/TBI'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.

(nA+nB)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
ehild said:
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=UA+UB and (nA+nB)CvT1=nACvTA+nBCvTB, like Chestermiller said

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

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

Last edited:
Rain10399 said:
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

$$(n_A+n_B)C_vT_1=n_AC_vT_A+n_BC_vT_B$$
$$(n_A+n_B)2P_1V/R(n_A + n_B)=n_AT_A+n_BT_B$$
$$2P_1V/R=P_AVT_A/RT_A + P_BVT_B/RT_B$$
$$2P_1V/R=(P_A + P_B)V/R$$
$$2P_1=P_A + P_B$$
Great. This should be enough to enable you to complete the solution.

Last edited:
Rain10399
Yes, now I managed to solve the problem! Thank you so much to everyone who responded ^^. I love this forum.

berkeman

## 1. How do you calculate the pressure of a gas in three identical balloons?

The pressure of a gas in three identical balloons can be calculated by using the ideal gas law, which states that pressure is equal to the product of the number of moles of gas, the universal gas constant, and the temperature divided by the volume of the balloons.

## 2. What is the ideal gas law and how does it relate to finding the pressure of a gas in three identical balloons?

The ideal gas law is a mathematical equation that describes the relationship between pressure, volume, temperature, and the amount of gas present. It is commonly used to calculate the pressure of a gas in a closed system, such as three identical balloons, by using the values of the other variables.

## 3. Is the pressure of a gas in three identical balloons affected by the size of the balloons?

Yes, the pressure of a gas in three identical balloons is affected by the size of the balloons. According to Boyle's Law, the pressure of a gas is inversely proportional to its volume, meaning that as the volume of the balloons increases, the pressure of the gas inside will decrease.

## 4. How does temperature affect the pressure of a gas in three identical balloons?

According to Charles's Law, the pressure of a gas is directly proportional to its temperature, meaning that as the temperature of the gas inside the balloons increases, the pressure will also increase. This is because as the gas particles gain more kinetic energy, they will collide with the walls of the balloons more frequently, resulting in a higher pressure.

## 5. Can the pressure of a gas in three identical balloons be changed without changing the temperature or the number of moles of gas present?

Yes, the pressure of a gas in three identical balloons can be changed by changing the volume of the balloons. As mentioned before, according to Boyle's Law, pressure and volume are inversely proportional, so by increasing or decreasing the volume of the balloons, the pressure of the gas inside will also change.

• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
904
• Introductory Physics Homework Help
Replies
1
Views
627
• Introductory Physics Homework Help
Replies
3
Views
319
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
4K