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Thermodynamics - Freely moving piston and rigid cylinder ideal gas problem

  • #1
I'm a third year chemical engineering major and this is my first real thermodynamics class, so I'm not entirely sure what I'm doing. Any help is greatly appreciated!

Homework Statement


A rigid horizontal cylinder contains a freely moving piston. Initially, it divides the cylinder into equal volumes, and each side of the piston contains 1 mole of an ideal gas at 5°C and 1 bar. An electrical resistance heater is installed on side A (left side) of the piston, and is energized to slowly heat the gas on side A to 170°C. If the tank and the piston are perfect insulators, calculate the heat added to the system by the resistance heater. The molar heat capacities of the gas are: Cv = (3/2) R and Cp = (5/2)R. (Hint: choose your system wisely.)


Homework Equations


[tex]PV=nRT[/tex] [tex]\Delta U = W + Q[/tex] [tex]\Delta U = nC_v\Delta T[/tex] [tex]P = P_i (v_i/v_f)^\lambda[/tex]


The Attempt at a Solution


First I started off by converting the known data into the proper units and calculated the initial volume of each side using the ideal gas law.
[tex]T_Ai = T_Bi = 278 K[/tex] [tex]P_Ai = P_Bi = 1*10^5 N/m^2[/tex] [tex]n_Ai = n_Bi = 1 mol[/tex] [tex]T_Af = 443K[/tex] [tex]V_Ai = V_Bi = 0.023 m^3[/tex]
Now onto choosing a system. I'm torn on whether to choose A or B. Side B is adiabatic so I could use this equation [tex]P = P_i (v_i/v_f)^\lambda[/tex], but I don't know have a way to find the final volume of B. Side A is not adiabatic so I'm left with [tex]\Delta U = W + Q[/tex] and [tex]\Delta U = nC_v\Delta T[/tex], but like side B I don't have a way to find the final volume to calculate the work being done.

A push in the right direction would be awesome. The assigned textbook for this class has proven to be very unhelpful.
 

Answers and Replies

  • #2
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The way I would go about analyzing would be applying the first law to both sides, essentially two systems which will be connected by a constraint. You do have a way to find the final volume of A which can be used to find the final volume of B.

EDIT: Sorry I looked a bit closer at your problem and can see I was wrong at first glance. One of the hitches of the problem I believe is that since the gas in the left chamber (A) is doing moving boundary work on the right side (B), that the left side process may be isobaric? It's been a short while since thermo for myself so I'd like to see someone confirm this for us, but I feel that is probably it. This would give you a way to solve for the final state of side A and therefore find the final state of side B by using the isentropic relation you've listed.
 
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  • #3
rude man
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This is an ambiguous question. It depends on what the "system" is. If the system is the gas only, there is one answer. If the system includes the piston (tank) and cylinder, the answer is quite different.
 
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  • #4
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Hmmmm, I'm not sure what you are getting at just yet but, heat is the form of energy transferred due to a temperature difference?
 
  • #5
Well, according to the first law of thermo, heat is everything that is not work, but how does that help?
 
  • #6
rude man
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Hmmmm, I'm not sure what you are getting at just yet but, heat is the form of energy transferred due to a temperature difference?
That is exactly right.

So - if the system is the gas, is there heat transferred to the gas?

But - if the system includes the tank & everything in it - and the tank is insulated thermally, remember - is there heat transferred to the system?
 
  • #7
rude man
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OK, I believe the convention is to call the gas the system.

Otherwise, since there is no heat transferred from the environment to the tank's inside, Q would equal zero.

So let's proceed accordingly: take the given hint & look at the gas on the left: what is the expression for work done on a gas in general? What is the state equation for an ideal gas of 1 mole? what is the relationship between p and V for an ideal gas under adiabatic process?

Also: for an ideal gas with change of temperature ΔT, what is the change in internal energy?
 
  • #8
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I think what you're getting at is that taking the gas as the system means that overall there is no change in volume indicating no work done on the gas. This means that the heat transferred to the system is simply equal to the change in internal energy?
 
  • #9
rude man
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I think what you're getting at is that taking the gas as the system means that overall there is no change in volume indicating no work done on the gas. This means that the heat transferred to the system is simply equal to the change in internal energy?
I hinted that you should concentrate on the gas on the left-hand side. That volume will certainly increase as the heater heats up the gas, don't you think?


You need to look at the state equation for an ideal gas, i.e. the relationship between p, V and T, remembering you have 1 mole of the gas on the left. Then the special relationship between p and V for an adiabatic process. Then the change in U for a given temperature change. Finally, let's not forget the 1st law. All the above assumes an ideal gas, which you have.

You need to invoke all four of those equations.
 
  • #10
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I see I misunderstood I thought you were saying to consider the gas in both chambers as the system but I see that won't work. For the left chamber I don't see how you can apply the adiabatic/isentropic relations, the gas clearly has heat being transferred to it by the electric resistance heater. If it was adiabatic the temperature change would have to occur from work being done on the left side which there is none.

Through first law analysis I was able to see that the heat transfer is from the sum of internal energy changes for both chambers since the work against the right chamber is equal to the work done by the left chamber. But the key missing here is solving for the final temperature of the right side. I know the final volume can be found because since the container is rigid the increase in volume on the left is the decrease in volume on the right but even with isentropic relations on the right ( the only side I can see them being applied) there's still a missing variable in terms of the second pressure. Let me go through it again but I still can't seem to get past that point. I'll post my work up in a bit.
 
  • #11
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Ok, so I'm getting stuck at the same spot as before.

First law for both sides:

[tex]Q_{iA}=W_{oA}+\Delta U_A[/tex] No work being done to A, no heat leaving A
[tex]W_{iB}=\Delta U_B[/tex] B is adiabatic and only has work being done to it

Since the work being done on B is being performed by A, therefore:

[tex]W_{oA}=W_{iB}[/tex]

And thus:

[tex]Q_{iA}=\Delta U_A+\Delta U_B[/tex]

Applying the internal energy change for an ideal gas and noting that the number of moles each side are equal, the gas on each side is the same, and the initial temperature on each side is equal:

[tex]Q_{iA}=nc_v(T_{2A}-2T_1+T_{2B})[/tex]

Here, we know all information except [itex]T_{2B}[/itex]. We do know since the container is rigid that the final volume of side B is given by:

[tex]V_{2B}=V_1-\Delta V=V_1-(V_{2A}-V1)=2V_1-V_{2A}[/tex]

Since side B is adiabatic, the isentropic relation applies:

[tex]\frac{T_{2B}}{T_1}=(\frac{V_1}{V_2})^{k-1}[/tex]

[tex]T_{2B}=(2-\frac{V_{2A}}{V_1})^{1-k}[/tex]

From here we can find the ratio [itex]\frac{V_{2A}}{V_1}=\frac{T_{2A}}{T_1}\frac{P_1}{P_{2A}}[/itex] by the ideal gas law on chamber A. However I am unsure how to progress past:

[tex]T_{2B}=(2-\frac{T_{2A}}{T_1}\frac{P_1}{P_{2A}})^{1-k}T_1[/tex]

Because I am unsure on how to find the final pressure for side A. Everything else is known in both the heat transfer relation and the final temperature in side B with the exception of the final pressure in side A. Where do we go from here?
 
  • #12
rude man
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I see I misunderstood I thought you were saying to consider the gas in both chambers as the system but I see that won't work. For the left chamber I don't see how you can apply the adiabatic/isentropic relations, the gas clearly has heat being transferred to it by the electric resistance heater. If it was adiabatic the temperature change would have to occur from work being done on the left side which there is none.
You're absolutely right, I failed to see that, and it shoots my analysis down. I have struggled with this for some time since & found no way out yet.

Through first law analysis I was able to see that the heat transfer is from the sum of internal energy changes for both chambers since the work against the right chamber is equal to the work done by the left chamber. But the key missing here is solving for the final temperature of the right side. I know the final volume can be found because since the container is rigid the increase in volume on the left is the decrease in volume on the right but even with isentropic relations on the right ( the only side I can see them being applied) there's still a missing variable in terms of the second pressure. Let me go through it again but I still can't seem to get past that point. I'll post my work up in a bit.
[/quote]
Realize that pressures are always equal in both compartments.

Right, pVγ only applies to the right-hand compartment. I of course tried to use that fact, plus VB = V0 - VA where V0 is the total (constant) volume and A and B refer to left & right compartments.

Looking forward to your post & maybe we can get to the bottom of this!
 
  • #13
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Hmmmm are the pressures always equal? I didn't think they would be...

Besides that point I just realized that since the side B is an isentropic process, it should also mean that it is a polytropic one where the exponent n=k. This might allow us to use a convenient evaluation for the boundary work on side B which would be equivalent to the internal energy change on B. I'm investigating this right now.
 
  • #14
rude man
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I seem to have lost my last post.

I was arguing the reason why p has to be the same on both sides: the heating has to be done slowly, to maintain quasi-equilibrium always, otherwise the state equation pV = RT does not apply some of the time, and any integral would be meaningless. So pretend you're slowly heating, and then at some point you stop heating. The piston would be stationary, meaning that p has to be the same on both sides, otherwise the piston would accelerate.

More tomorrow, I hope. Am looking forward to your further explorations.
 
  • #15
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You know, I've been thinking more and more about this problem and correct me if I am wrong but I believe it is an impossible system.....

The interaction here is between two separated system with the following conditions:

System A:

-Has heat transferred into the system
-Does work on System B by virtue of moving the piston
-Has no work done on it (nothing to motivate work against it from system B)
-Has no heat loss from the condition that the tank and cylinder are perfect insulators meaning that no heat is transferred to system B or the surroundings

System B
-Is adiabatic, no heat transferred in or out from the insulator conditions
-Does no work on System A
-Has work done on it by System A

What I'm thinking here is System A violates the second law of thermodynamics considering the insulator conditions. It is supplied heat and produces work on system B but since there is no expulsion of waste heat, there is a 100% conversion from heat to work which violates the second law. I'm not sure this is why this problem is so difficult but just an observation.
 
  • #16
rude man
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Don't think so. Some of the heat Q is used to increase internal energy of system A = UA & some is used to perform work on system B, increasing UB also.

Anyway, I contend to have solved the problem, so if you dare go where angels fear to tread (namely, my physics):

What is the work done? Hint: use the p-V relationship applying to side B.
 
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  • #17
rude man
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Addendum: the entropy of the 'universe' is increased due to the amount of heat transferred from the heater to the A system. I can pull this right out of a table in my textbook (Zemansky, Heat & Thermodynamics). It is ΔS = Cpln(T2/T1) & we already know T1 and T2. It applies to all irreversible processes where work is adiabatically dissipated into heat.

Of course, the problem asks for Q, not ∫dQ/T ....

I know it sounds a bit weird, but work as provided by the electricity powering the heater is 100% transferred into internal energy increase to systems A and B: ΔUA + ΔUB = W since Q = 0 for the tank. Think of that work as a hand-cranked generator generating the electricity to power the heater. Some of that work is transferred as work done on gas B but all of it goes into increasing the internal energies of both sides according to ΔU = Cv{ΔTA + ΔTB}.

Anyway, the math turns out to be a lot simpler than I first thought so go ahead why don't you ...
 
  • #18
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Hmmm, can you elaborate a bit on the work bit with side B? I did at one point attempt to use the moving boundary work formula (especially because as a poly tropic process, the formula is simple) but if I recall correctly information about the second state on side B is required which we know none of the specifics regarding. I'll have to do some checking when I get home but that route ended up a dead end for me the first time too.
 
  • #19
rude man
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Well, what is the general formula for work done in going from volume 1 to volume 2? And, can you exploit your known relationship between V and p in gas B?

PS you're right about not being given the final volume in gas B but I'll try to help you get that too.
 
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  • #20
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The entire state of B at the end of the process is what's really troubling. No matter how I exploit the isentropic gas relations, the work done by a polytropic process, or use of the ideal gas law, I've been unable to get anything worked out for the heat delivered by the wire that does not have an unknown variable in it, either from the unknown end state variables of side A or the entirety of state B. I'm beginning to think that there is a continuity condition missing or something about the system I'm not realizing...

Have you worked out the mathematics on it? I'd like to see what you have done. Also, has the OP had any luck?
 
  • #21
rude man
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Have you worked out the mathematics on it? I'd like to see what you have done. Also, has the OP had any luck?
Yes I have, cm, and I'm trying to guide you in the right direction, but the Forum doesn't allow just giving you the math. So take my hints one by one if you will, and we can do this!

Funny - by now I thought you were the OP. Well, makes no difference.

If you knew VB2, what would you do with it?
 
  • #22
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If I had [itex]V_{B2}[/itex] I could easily find the final volume at side A as well which in turn would allow a solution to the final pressure and therefor the final temperature for B.
 
  • #23
rude man
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If I had [itex]V_{B2}[/itex] I could easily find the final volume at side A as well which in turn would allow a solution to the final pressure and therefor the final temperature for B.
Sorry, I meant VA2, not VB2. Did you get my private note?

I did not consider TB2 in my approach. Not necessary.

Wonder what happened to our OP? :confused:
 
  • #24
uart
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Wonder what happened to our OP? :confused:
Yep it looks like the OP is long gone.

If you don't mind rudeman, could you check if my answer is correct. I'm very rusty on this stuff so I'd been watching this thread trying to learn something. :smile:

Anyway, I got an answer of [itex]W \simeq 2.51[/itex] kJ for this and I was just wondering if that agreed with what you had.
 
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  • #25
rude man
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The increase in U on the A side alone is greater than your answer: ΔUA = cvΔT = (5*8.317/2)*(175 - 5) = 3535J! Q = ΔU + W done on the B side, so your answer can't be right.

I am sending you another private note for some of my results. The OP may still be lurking in the shadows ... :smile:
 

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