# Ideal gas expansion between two tanks (Thermodynamics)

1. Sep 6, 2014

### ThermoThing

1. The problem statement, all variables and given/known data
Two interconnected tanks are of equal volume. One is filled with methane at 500bar and 293K, the other is initially evacuated. A valve connecting the two tanks is opened only long enough to allow the pressures to equilibrate. If there is no heat transfer between the gas and the tanks, what are the temperature and pressure of the gas in each tank after the valve has been shut? Assume the methane is an ideal gas.

I'm going to call the tank initially filled with gas "Tank A" and make all associated variables have an "A" subscript. Tank B is the evacuated one.

$P_At1 = 500 bar$

$T_At1 = 293K$

2. Relevant equations

$P V = N R T$

$ΔU_A = ΔU_B$

$V_A = V_B = V$

$P_{At2} = P_{Bt2} = P_{t2}$

$P_{At1} = P_t1$

$N_1 ∫C_v ^* dT = N_2 ∫C_v ^* dT$

$ΔS_sys = C_p ^* \cdot ln(T_{t2} / T_{t1}) - R \cdot ln(P_{t2} / P_{t1})$

Unknowns:

$P_{At2}, P_{Bt2}, T_{At2}, T_{Bt2}$

3. The attempt at a solution

We've tried multiple different approaches. The fatal flaw in my last attempt is that I assumed that because the pressures were at equilibrium, then the temperatures would be, too. Apparently this system does not uniformly expand, though. In an almost identical problem (with different-sized tanks), the temperature of Tank A lowers while that of Tank B rises.

Here's what my last attempt was:

Control Volume: Volume of gas remaining in Tank A after the expansion

Material balance:

$ΔN_{sys} = 0$

Energy balance:

$ΔU_{sys} = -∫P dv$

Entropy balance (not sure if this is needed, because the problem similar to it is in a chapter before entropy is introduced):

$ΔS_{sys} = ΔS_{gen} = 0$

Ideal gas law simplifies to:

$\frac{T_{At2}}{T_{At1}} = \frac{2P_{t2}}{P_{t1}}$

I then used the entropy balance to say:

Substituting the ideal gas law simplification in for the temperature part:

$ΔS_{sys} = 0 = C_p ^* \cdot ln(\frac{2 P_{t2}}{P_{t1}}) = R \cdot ln(\frac{P_{t2}}{P_{t1}})$

I solved for the pressures and found that

$e^{\frac{C_p ^* \cdot ln(2)}{R - C_p ^*}} = \frac{P_{t2}}{P_{t1}}$

I got the value of pressure which I found to be 203 bar, and used that to find the value of temperature that I found to be 267 K. These are both incorrect. Following the same logic of the example problem, the pressure should be exactly halved, and the temperatures would be inequal. The example problem has no information other than the values of the solution, though, so I have absolutely no idea how they got to that point. The best lead I have is that maybe Tank A is a free expansion, but the gas flowing into Tank B has a velocity/pressure gradient, so it generates entropy. I have no idea other than that, though. Can anyone help? It is direly needed.

2. Sep 16, 2014

### Greg Bernhardt

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

3. Sep 16, 2014

### Staff: Mentor

I think I can help.

The pressures will not necessarily be halved.

Let V represent the volume of tank A, and let n represent the initial number of moles in tank A. Using the ideal gas law, how are V and n related?

Let nA= number of moles remaining in tank A in the final state of the system
Let nB=number of moles in tank B in the final state of the system
Let TA=temperature in tank A in final state of system
Let TB=temperature in tank B in the final state of the system
Let P = pressure in both tanks in final state of system

How are nA and nB related to n?

Let's focus on the gas remaining in tank A at the final state of the system. If the gas leaving tank A seeps through the valve very slowly, it is reasonable to assume that the gas that remains in A has experienced an adiabatic reversible expansion. For such an expansion the temperature TA can be related to to the final pressure P by what equation? (The equation contains the initial temperature and initial pressure also). So, if you know the final pressure, you know the final temperature in tank A. The main unknown we will be solving for will be the final pressure. In terms of the final pressure and the tank volume, we can now use the ideal gas law to express the final number of moles in tank A nA in terms of V, P, and the initial conditions. What do you get doing this?

Since you now know the final number of moles in tank A, you can determine the final number of moles in tank B (algebraically, in terms of P and V). What do you get?

If we take as our system the total gas in both tanks, how much work does this gas do on the rigid walls of the tanks? Since the tanks are adiabatic, how much heat enters the system through the walls of the tanks. What does this imply in terms of the change in internal energy for the gases contained in the two tanks? In terms of our algebraic parameters, what is the final change in internal energy of the gas in tank A?

Using nB, P, and V, use the ideal gas law to express TB in terms of nB, P, and V.

I want to stop here and see what you come up with.

Chet

Last edited: Sep 17, 2014
4. Sep 16, 2014

5. Oct 1, 2014

### Staff: Mentor

Is anybody out there still interested in how to solve this problem? It is a very interesting and instructive problem. The OP never responded to any of the posts. If there is interest in how to solve this problem, I will provide the solution. Otherwise, it can fade into oblivion.

Chet

6. Jan 19, 2016

### Cora

I know it's been a while, but I'm interested in continuing with this problem. I am not familiar with working in moles however, so switching the ideal gas law to PV = ZmRT? In addition, there is no shutting of the valve at the end of my problem. The two tanks reach a final equilibrium state with final properties T2 and P2 shared between them.

My approach to find the temperatures would be to use conservation of internal energy between initial and final states.

U1 = U2
U1 = UA1 + UB1
U1 = uA1*mA1 + uB1*mA1
where tank A is initially evacuated, so mA1 = 0, therefor
U1 = uB1*mB1

and then for U2

U2 = uA2*mA2 + uB2*mB2
assuming from conservation of mass: mB1 = mA2 + mB2
and that since PA2=PB2 and TA2 = TB2, mA2 = mB2 = mB1/2​
U2 = (mB1/2)*(uA2 + uB2)
assuming uA2 = uB2 = .5*u2
U2 = (mB1/2)*(u2)
U2 = .5*mB1*u2

now equating U1 and U2:

uB1*mB1 = .5*mB1*u2
uB1 = .5*U2 = uB2 = uA2

so since I have uB1, TB1 and have equated the final specific internal energies to uB1, I can find the final temperature and move on. Yes?

Last edited: Jan 19, 2016
7. Jan 19, 2016

### Staff: Mentor

The problem statement implies that the tanks and the gases within them are thermally isolated from one another during the process: "If there is no heat transfer between the gas and the tanks, what are the temperature and pressure of the gas in each tank after the valve has been shut?" Even though the pressures in the two tanks must be equal after the system equilibrates, the temperatures will not be. If they were (i.e., if the tanks could thermally equilibrate also), the final temperature in the two tanks would be equal to the initial temperature.

Chet

8. Jan 19, 2016

### Cora

Okay, so my original problem statement was: "Two well-‐‐insulated rigid tanks of equal volume, tank A and tank B, are connected via a valve. Tank A is initially empty. Tank B has 2 kg of Argon at 350 K and 5000 kPa. The valve is opened and the Argon fills both tanks. State 2 is the final equilibrium state. The temperature and pressure of the room in which the tanks sit are 300 K and 100 kPa, respectively. Perform a closed system analysis.
a. Determine the volume of tank B.
b. Determine the final temperature.
c. Determine the final pressure.
d. Determine the entropy produced in the process.
e. Determine the exergy (or availability) destroyed in the process.
i. Using anexergy balance.
ii. Using the entropy produced.f. Determine the total exergy in tank B initially.​

I am stuck on this problem parts b/c. I thought the b/c were similar enough to the OP's problem that I would continue here rather than creating a new thread, but maybe that is not the case? As I continued to work through the problem, it seemed like the initial and final temperatures may be the same, although that is not intuitive to me.

I guess I'll create a new thread for this.

9. Jan 19, 2016