Work done in adiabatic compression

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  • #1
so_gr_lo
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Homework Statement:
A piston in a cylinder has total heat capacity C, it is thermally isolated from the environment but in contact with nm moles of monatomic gas

T and V are the temperature and volume. The gas is compressed in a reversible way. The cylinder and gas are in thermal equilibrium.

In terms of Cv, C, R, T, V, dT and dV. What is the work done on the gas? You may not need all terms
Relevant Equations:
dW = -pdV

Cp = Cv + R
The equation I know for adiabatic work is W = P1V1((V1/V2)ϒ-1 - 1))/ϒ-1, but this involves ϒ, but I can use ϒ = Cp/Cv = Cv+R/Cv = 1 + Cv/R, does this seem correct? But I still have a P1
 
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  • #2
haruspex
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Homework Statement:: A piston in a cylinder has total heat capacity C, it is thermally isolated from the environment but in contact with nm moles of monatomic gas

T and V are the temperature and volume. The gas is compressed in a reversible way. The cylinder and gas are in thermal equilibrium.

In terms of Cv, C, R, T, V, dT and dV. What is the work done on the gas? You may not need all terms

The expression I know for isothermal compression is the one in relevant equations but I’m not sure how to eliminate nm, which I’m assuming is required by the question, without ending up with a term for pressure.
Relevant Equations:: W = nmRT * ln(V)

nm is number of moles

I know pV = nmRT but I’d be left with a P if I rearrange
I see no attempt to use dV or dT, yet those are the variables that tell you what was done.
 
  • #3
jbriggs444
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The title says "isothermal compression". However, this appears to be much more in the nature of an adiabatic compression to me.
 
  • #4
so_gr_lo
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Yeah I think it is adiabatic, I’ve changed it now
 
  • #5
kuruman
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I believe that here you must assume a starting temperature ##T_0## and volume ##V_0## for the initial state of the gas. If not, what symbols are you going to use on the right hand of the equation ##W= \dots~## for the work done on the gas as it goes from its initial to its final state?
 
  • #6
Delta2
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Actually there is a kind of funny and fishy way to answer this question. Assuming that the gas is ideal (not explicitly stated) so that it obeys ##PV=nRT##, we can write $$W=\int p dV=\int \frac {nRT}{V} dV$$ and leave it that way.

The "funny part "is that we do not perform the integral calculation (we just use the integral symbol but that is a mathematical operation, like +,-,.,/ are) and the fishy part is that of course in an adiabatic process T,V are not independent.
 
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  • #7
kuruman
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Actually there is a kind of funny and fishy way to answer this question. Assuming that the gas is ideal (not explicitly stated) and that it obeys ##PV=nRT##, we can write $$W=\int p dV=\int nRT dV$$ and leave it that way.

The "funny part "is that we do not perform the integral calculation (we just use the integral symbol but that is a mathematical operation, like +,-,.,/ are) and the fishy part is that of course in an adiabatic process T,V are not independent.
But the work done on the gas depends on the path and the path specified by the question is adiabatic. The expression you propose could be any path and needs to be modified specifically for the adiabatic path. After that is done, then I suppose one can leave it as an indefinite integral like you propose or integrate and throw in an integration constant ##C## or integrate over definite limits which is what I proposed.
 
  • #8
Delta2
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But the work done on the gas depends on the path and the path specified by the question is adiabatic. The expression you propose could be any path and needs to be modified specifically for the adiabatic path. After that is done, then I suppose one can leave it as an indefinite integral like you propose or integrate and throw in an integration constant ##C## or integrate over definite limits which is what I proposed.
Yes what you saying is part of what I call "the fishy thing". The path of that integral is the curve ##T(V)## for an adiabatic process. But the problem statement asks for the formula in terms of T and dV and that's what I did, I just leave ##T=T(V)## to float in the air, without fully defining it, not even mentioning that T and V are dependent.

The problem statement helps too, it just says ""T and V are the temperature and volume" it doesn't say "T and V are the temperature and volume of the initial or final state"...
 
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  • #9
TSny
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I'm also struggling with the wording of the question. The list of symbols for expressing the answer does not include the number of moles of gas. Is this just an oversight?

The list does include dV and dT. Does this mean they want an expression for the work done on the gas for just an infinitesimal step of the compression? But that would be too easy.

Anyway, the gas itself is not being compressed adiabatically since the gas will exchange heat with the piston.
 
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  • #10
haruspex
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Yeah I think it is adiabatic, I’ve changed it now
Not so fast. The question statement you posted specifies neither isothermal nor adiabatic; it merely says "reversible". Further, it notes that the gas and piston are at all times in thermal equilibrium, implying a gradual process, and that is essentially what you need for reversibility.
So I would try not to assume either isothermal or adiabatic.
 
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  • #11
haruspex
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The list of symbols for expressing the answer does not include the number of moles of gas. Is this just an oversight?
I think it must be an oversight since a variable name for the number of moles was given.
 
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  • #12
Delta2
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@TSny, @haruspex is there a way to determine T(V) or P(V), for this problem? You say its neither isothermal or adiabatic...
 
  • #13
haruspex
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@TSny, @haruspex is there a way to determine T(V) or P(V), for this problem? You say its neither isothermal or adiabatic...
There’s probably a better way, but it can be approached by considering two small steps. First, an infinitesimal adiabatic compression, then a flow of heat from the gas into the piston to restore thermal equilibrium.
 
  • #14
Delta2
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There’s probably a better way, but it can be approached by considering two small steps. First, an infinitesimal adiabatic compression, then a flow of heat from the gas into the piston to restore thermal equilibrium.
I don't think there is anything in the statement of the problem that forces this assumption. What if the process is isothermal btw, why it can't absorb heat from the piston and be compressed?
 
  • #15
haruspex
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I don't think there is anything in the statement of the problem that forces this assumption.
What assumption?
What if the process is isothermal
It cannot be isothermal. The system as a whole is thermally isolated, and compression occurs.
why it can't absorb heat from the piston and be compressed?
Do you mean the gas to become compressed as a result of heat flowing from the piston? At all times, piston and gas are at almost the same temperature.
 
  • #16
Delta2
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What assumption?
The assumption that this process consists of an adiabatic compression + a heat flow process.
It cannot be isothermal. The system as a whole is thermally isolated, and compression occurs.
But the gas exchanges heat with the piston, isn't this what you said before?
Do you mean the gas to become compressed as a result of heat flowing from the piston? At all times, piston and gas are at almost the same temperature.
If they are at the same temperature at all times, then how can they exchange heat?
 
  • #17
TSny
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I believe that the wording implies that the system consisting of the piston and gas is thermally isolated from the surroundings. But the surroundings can do work on the system by moving the piston. For the system as a whole, the process is reversible and adiabatic.

The OP gave a formula for the work done on ##n## moles of an ideal gas if the gas undergoes a reversible, adiabatic compression: $$W = \frac{P_1V_1}{\gamma -1} \left[\left(\frac{V_1}{V_2}\right)^{\gamma - 1} -1\right]= \frac{nRT_1}{\gamma -1}\left[\left(\frac{V_1}{V_2}\right)^{\gamma - 1} -1\right]$$
For this problem, the gas exchanges heat with the piston as the gas is compressed. Although the compression of the gas itself is no longer adiabatic, I believe you can derive a similar-looking expression for the work done on the gas. The result differs from the equation above in that ##\gamma## is replaced by an expression that includes the total heat capacity, ##C##, of the piston.
 
  • #18
Delta2
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Excuse me but if I get it right, you saying that the gas and the piston exchange heat, however the temperature of the gas and piston are equal at all times. I sense a mini paradox here but ok.
 
  • #19
TSny
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Excuse me but if I get it right, you saying that the gas and the piston exchange heat, however the temperature of the gas and piston are equal at all times. I sense a mini paradox here but ok.
Whenever heat is reversibly exchanged between two objects, the objects will be at essentially the same temperature. As the piston is pushed in, the piston and gas increase their temperatures together so that they are always at essentially the same temperature.
 
  • #20
Delta2
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But we forbid the isothermal process as a case because we say they can't exchange heat because they are always at the same temperature? For the same reason isothermal reversible processes could never happen.
 
  • #21
TSny
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But we forbid the isothermal process as a case because we say they can't exchange heat because they are always at the same temperature? For the same reason isothermal reversible processes could never happen.
The process is not isothermal. The temperature of the piston and the gas change during the process.
 
  • #22
Delta2
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The process is not isothermal. The temperature of the piston and the gas change during the process.
How do you know it is not isothermal. Fact 1) You accept that gas and piston exchange heat. Why can't this happen in an isothermal way?
 
  • #23
TSny
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When I say that the process is not isothermal, I mean that the temperatures of both the piston and the gas increase. But at any given point during the process, the temperature of the piston equals the temperature of the gas.
 
  • #24
haruspex
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How do you know it is not isothermal. Fact 1) You accept that gas and piston exchange heat. Why can't this happen in an isothermal way?
Isothermal does not mean merely that the components are always at the same temperature as each other; it means that the components remain at the same temperature over time.
 
  • #25
Delta2
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Guys, you both accept that the gas and the piston exchange heat. What forbids this heat flow to be done in an isothermal way?
 
  • #26
haruspex
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Guys, you both accept that the gas and the piston exchange heat. What forbids this heat flow to be done in an isothermal way?
The piston is incompressible and presumably contains no other reservoir of internal energy. If it exchanges heat then it is changing temperature, so the process is not isothermal.
 
  • #27
Delta2
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The piston is incompressible and presumably contains no other reservoir of internal energy. If it exchanges heat then it is changing temperature, so the process is not isothermal.
Ok, I find this explanation satisfactory. I guess its one of those silent assumptions we have quite often in physics problems.
 
  • #28
Delta2
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What about the other assumption of adiabatic process+heat flow process, can any process be broken down to these two steps (except isothermal e hehe).
 
  • #29
TSny
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What about the other assumption of adiabatic process+heat flow process, can any process be broken down to these two steps (except isothermal e hehe).
I don't understand the question. But, I need to get some shut-eye. It's well past midnight here.

So, I'll leave it for now. Enjoy.
 
  • #30
Delta2
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I don't understand the question. But, I need to get some shut-eye. It's well past midnight here.

So, I'll leave it for now. Enjoy.
And I was ready to ask you if that's how you viewed the process when writing post #17
 
  • #31
haruspex
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I don't understand the question. But, I need to get some shut-eye. It's well past midnight here.

So, I'll leave it for now. Enjoy.
@Delta2 is referring to the approach I proposed in post #13.
I don't know about thermodynamic processes in general, but it certainly seems reasonable here since it mirrors reality. If we move the piston in small steps, pausing after each, that should approximate a slow steady movement.
 
  • #32
Delta2
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@Delta2 is referring to the approach I proposed in post #13.
I don't know about thermodynamic processes in general, but it certainly seems reasonable here since it mirrors reality. If we move the piston in small steps, pausing after each, that should approximate a slow steady movement.
OK I understand that's what we should do if we want to find an explicit formula for the work, however what about my earlier post where I give the work as integral of T,V and dV, this satisfies all the requirements of the statement of the problem but I agree that it is funny and fishy.
 
  • #33
Delta2
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Oh unless of course the gas isn't ideal so we can't use ##PV=nRT##. But I think the problem can't be solved with your way too if we don't assume an ideal gas...
 
  • #34
haruspex
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OK I understand that's what we should do if we want to find an explicit formula for the work, however what about my earlier post where I give the work as integral of T,V and dV, this satisfies all the requirements of the statement of the problem but I agree that it is funny and fishy.
It is unclear whether an integral involving an undefined function would constitute an acceptable answer. If we can find a more definitive expression then I would think not.
 
  • #35
Delta2
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It is unclear whether an integral involving an undefined function would constitute an acceptable answer. If we can find a more definitive expression then I would think not.
yes well, though I don't like this very much, I think you might be right.
 

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