Finding Thermal Equilibrium Temperature

[Bizkit]

Homework Statement

A 1.5 m³ tank containing air at 35°C and 750 kPa is connected by a valve to another tank containing 8.25 kg of air at 65°C and 265 kPa. Now the valve is opened and the entire system is allowed to reach thermal equilibrium with the surroundings, which is at 25°C. Determine the final equilibrium temperature.

Homework Equations

I know that for simple thermal equilibrium problems involving only two substances, the equation I would use is [mcΔT]₁ + [mcΔT]₂ = 0. I've read that when 3 substances are in thermal equilibrium with each other, I would simply need to expand on the previous equation and make it [mcΔT]₁ + [mcΔT]₂ + [mcΔT]₃ = 0, but since I'm not given the mass of the surroundings, and I don't know enough to figure that out, I can't use this equation. Because of this, I'm not sure what equation to use.

The Attempt at a Solution

Since I'm not sure what equation to use, I can't attempt to find the solution.


If someone could please help me, I would really appreciate it. Thanks.

 

[kuruman]

Homework Statement

A 1.5 m³ tank containing air at 35°C and 750 kPa is connected by a valve to another tank containing 8.25 kg of air at 65°C and 265 kPa. Now the valve is opened and the entire system is allowed to reach thermal equilibrium with the surroundings, which is at 25°C. Determine the final equilibrium temperature.

Homework Equations

I know that for simple thermal equilibrium problems involving only two substances, the equation I would use is [mcΔT]₁ + [mcΔT]₂ = 0. I've read that when 3 substances are in thermal equilibrium with each other, I would simply need to expand on the previous equation and make it [mcΔT]₁ + [mcΔT]₂ + [mcΔT]₃ = 0, but since I'm not given the mass of the surroundings, and I don't know enough to figure that out, I can't use this equation. Because of this, I'm not sure what equation to use.

The Attempt at a Solution

Since I'm not sure what equation to use, I can't attempt to find the solution.


If someone could please help me, I would really appreciate it. Thanks.

What is your understanding of thermal equilibrium? What must be true when two systems are said to be "in thermal equilibrium"?

 

[Bizkit]

I know that when two substances are said to be in thermal equilibrium, they must have the same temperature.

I also know that, by the Zeroth Law of Thermodynamics, if two substances are in thermal equilibrium with another substance, then those two substances are in thermal equilibrium with each other.

Due to the Zeroth Law of Thermodynamics, I was thinking that I could treat this problem as a simple, two substance thermal equilibrium problem between the two tanks, and kind of ignore the surroundings, but I'm not sure if that would work.

If it does work, I'd get this:

m_tank1 = [PV/(RT)]_tank1 = 265*1.5/(.287*(35+273)) = 397.5/(.287*308) = 397.5/88.4 = 4.5 kg

[mc_v(T₂-T₁)]_tank1 + [mc_v(T₂-T₁)]_tank2 = 0
4.5*.718(T₂-35) + 8.25*.718(T₂-65) = 0
3.231(T₂-35) = 5.9235(65-T₂)
3.231T₂ - 113.085 = 385.0275 - 5.9235T₂
9.1545T₂ = 498.1125
T₂ = 54.41°C

Is this right, or at least part right? If it isn't, then I'm not sure how to solve this problem.

 

[ideasrule]

Haven't checked your numbers, but your equations look right: heat energy is conserved (or at least, assumed to be conserved), so the total increase in heat is 0. You can also think of it as mcv(T2-T1)]tank1 = [mcv(T1-T2)]tank2: the heat gained by tank 1 is equal to that lost by tank 2.

Note that the Cv cancels out, so the answer is independent of the gas involved.

 

[Bizkit]

So, am I correct, then, in saying that I can ignore the surroundings due to the Zeroth Law of Thermodynamics? Do the surroundings make a difference in the calculation?

 

[ideasrule]

OH. Forget what I wrote; I didn't see the part about being in thermal equilibrium with the surroundings.

The gas isn't just in thermal equilibrium with each other; it's in thermal equilibrium with its surroundings. That means that its temperature must be...

 

[kuruman]

Homework Statement

A 1.5 m³ tank containing air at 35°C and 750 kPa is connected by a valve to another tank containing 8.25 kg of air at 65°C and 265 kPa. Now the valve is opened and the entire system is allowed to reach thermal equilibrium with the surroundings, which is at 25°C. Determine the final equilibrium temperature.

It seems to me that if the entire system reaches thermal equilibrium with the surroundings and the surroundings are at 25 ^oC, then the equilibrium temperature must be 25 ^oC. I know it is trivial, but I suspect the intended question is to determine the equilibrium pressure not temperature.

 

[Bizkit]

That is trivial. I was thinking that might be the answer, but it just seemed too easy. Anyway, as for what you said about equilibrium pressure, that's actually the next part of the question. I didn't post it because I wanted to take one step at a time. The question is:

If a 2 m³ tank was added to the system that had 2.5 kg of air at 15°C, determine the pressure of this tank to maintain an equilibrium pressure of 345 kPa.

I've never had to deal with equilibrium pressure, so I'm not sure how to solve the problem.

 

[kuruman]

Use the ideal gas law. You know the final volume and final temperature. All you need is the number of moles. How can you find these?

 

[Bizkit]

Here's what I have:

V_tank2 = [mRT/P]_tank2 = 8.25*.287*(65+273)/265 = 2.36775*338/265 = 800.3/265 = 3.02 m³

P_1-2 = (m_tank1+m_tank2)*R*T_1-2/(V_tank1+V_tank2) = (4.5+8.25)*.287*(25+273)/(1.5+3.02) = 12.75*.287*298/4.52 = 241.25 kPa

Is this right so far? If it is, I'm not sure what to do next. We never went over any problems on Equilibrium pressure in class, and the book doesn't have any problems on it, so I'm kind of in the dark here.

 

[kuruman]

Please don't get intimidated by the use of the term "equilibrium pressure". The problem involves understanding of the ideal gas law, pure and simple.

What form of the ideal gas law are you using? Specifically, what is constant R? Your calculations would be more transparent if you used n, the number of moles, to denote the amount of gas, not the mass of the gas. I suggest that you first calculate the number of moles in each tank before the valve is opened.

 

[Bizkit]

I'm not necessarily intimidated by it, I'm just frustrated that I don't know what to do because it seems as though it should be really easy to figure out.

It is the specific ideal gas law that I am using. It is in terms of mass, not moles, and R is the specific gas constant (which is different for each substance) rather than the universal gas constant. Both my teacher and the book use it more, so that's why I use it. It gets you the same results, but I'll do it your way this time.

I Just realized that I accidentally used the wrong pressure to calculate the mass in tank 1. The mass should actually be 12.73 kg, which is a lot different than the 4.5 kg that I had gotten before.

n_tank1 = [m/M]_tank1 = 12.73/.02897 = 439.42 moles

n_tank2 = [m/M]_tank2 = 8.25/.02897 = 284.78 moles

n_tank3 = [m/M]_tank3 = 2.5/.02897 = 86.3 moles

Ok, now what do I need to do?

 

[kuruman]

I was not sure if you were using the specific gas constant because the 4.5 kg (which should have been 12.73) confused me. OK, now we are on the same page. You know how many moles (or kilograms) of air you have, you know the volume this air occupies and you know its temperature. Can you find its pressure?

 

[Bizkit]

If you mean the pressure of the first two tanks combined, I can find that. I'll just use the same equation I used before, except I'll insert the correct mass for tank 1.

P_1-2 = (m_tank1+m_tank2)*R*T_1-2/(V_tank1+V_tank2) = (12.73+8.25)*.287*(25+273)/(1.5+3.02) = 20.98*.287*298/4.52 = 396.98 kPa

 

[kuruman]

I see now what your problem with equilibrium pressure is.

Think of it this way. If the pressure in the two tanks were not the same, then gas will flow from to tank with higher pressure to the tank with lower pressure until the pressure on each side becomes the same. When this happens, the gas will have no pressure difference to push it one way or the other so it will stay where it is. So "equilibrium pressure" means "the same pressure everywhere" much like "equilibrium temperature" means "the same temperature everywhere." Pressure difference pushes mass around and temperature difference pushes heat around.

I didn't put in the numbers, but your calculation looks correct. I trust you have the right R for air.

 

[Bizkit]

I hate to say this, but I'm still completely in the dark. I know that this problem is probably really easy, and the answer is staring me right in the face, but I just don't get. I must have drank some stupid juice or something. I'm getting really frustrated. If you could just tell me how to solve this problem so I can relax, I would really appreciate it. Either that, or I'll just have to say that I don't know how to solve this part of the problem.

 

[kuruman]

Like I said, your calculation looks correct and the number you got, 397 kPa is probably correct. I say probably because I did not put in the numbers myself to check it.

pV = nRT -> p = nRT/V. Plug in.

You did all that and there is nothing more to be said.