Find the equilibrium position of this piston

[utkarshakash]

Homework Statement

A portion of helium gas in a vertical cylindrical container is in thermodynamic equilibrium with the surroundings. The gas is confined by a movable heavy piston. The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished. After that, the container is insulated and then the piston is released. After the piston comes to rest, what is the new equilibrium position of the piston?

The Attempt at a Solution

Let the piston initially be situated at a distance l from bottom of cylinder and let P1 be the pressure of the gas inside. Let M be the mass of piston. Doing force balance on piston gives

Mg+P₀A = P₁A

where A =area of cross-section and P₀ = atmospheric pressure

When the piston is taken to height H let pressure inside be P₂.

$\dfrac{P_1 l}{T_0} = \dfrac{P_2 (l+H)}{T_2}$

After cylinder is insulated and piston is released, the process will be adiabatic
Let final height of piston from bottom of cylinder be l'.

$P_2 (l+H)^{\gamma} = P_1 l'^{\gamma}$

But how do I eliminate T2 and l?

 

[sankalpmittal]

Homework Statement

A portion of helium gas in a vertical cylindrical container is in thermodynamic equilibrium with the surroundings. The gas is confined by a movable heavy piston. The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished. After that, the container is insulated and then the piston is released. After the piston comes to rest, what is the new equilibrium position of the piston?

The Attempt at a Solution

Let the piston initially be situated at a distance l from bottom of cylinder and let P1 be the pressure of the gas inside. Let M be the mass of piston. Doing force balance on piston gives

Mg+P₀A = P₁A

where A =area of cross-section and P₀ = atmospheric pressure

When the piston is taken to height H let pressure inside be P₂.

$\dfrac{P_1 l}{T_0} = \dfrac{P_2 (l+H)}{T_2}$

After cylinder is insulated and piston is released, the process will be adiabatic
Let final height of piston from bottom of cylinder be l'.

$P_2 (l+H)^{\gamma} = P_1 l'^{\gamma}$

But how do I eliminate T2 and l?

Errm, i am myself a student, but...

Why do you take temperature into account. Isothermal process means ΔT=0, but since they first stated that process is infinitesimally slow, it is an approximate isothermal process. So you apply combination of Boyle's law and Net force equilibrium.

In second set up you use adiabatic process formula.

 

[Chestermiller]

Sankalpmittal is right about the first part of the problem. The required relationship should be
$$P_1l=P_2(l+H)$$

The second part of the problem is more complicated. Let P(t) be the pressure exerted by the gas on the bottom face of the piston at time t during the deformation and let m be the mass of the piston. Apply Newton's 2nd law to determine the equation of motion of the piston during this deformation (in terms of P(t), m, P₀, g, A, and x (the distance that the piston moves downward from position l + H. What do you get? We are next going to use this equation to determine the cumulative amount of work done by the piston on the gas when the system reaches final equilibrium.

Chet

 

[Yellowflash]

I am new here, so I do not know how things work, so forgive my stupidity.

I am getting 0.6H

 

[utkarshakash]

I am new here, so I do not know how things work, so forgive my stupidity.

I am getting 0.6H

It's the correct answer.

 

[utkarshakash]

Sankalpmittal is right about the first part of the problem. The required relationship should be
$$P_1l=P_2(l+H)$$

The second part of the problem is more complicated. Let P(t) be the pressure exerted by the gas on the bottom face of the piston at time t during the deformation and let m be the mass of the piston. Apply Newton's 2nd law to determine the equation of motion of the piston during this deformation (in terms of P(t), m, P₀, g, A, and x (the distance that the piston moves downward from position l + H. What do you get? We are next going to use this equation to determine the cumulative amount of work done by the piston on the gas when the system reaches final equilibrium.

Chet

Can't we directly use the adiabatic formula?

 

[Chestermiller]

Can't we directly use the adiabatic formula?

No. This is an irreversible compression. The "adiabatic formula" applies only to reversible expansions and compressions.

Chet

 

[utkarshakash]

No. This is an irreversible compression. The "adiabatic formula" applies only to reversible expansions and compressions.

Chet

Thanks for clarifying this. OK, here's what I'm getting

$mg+(P_0-P(t))A = mv \frac{dv}{dx}$

 

[Chestermiller]

Thanks for clarifying this. OK, here's what I'm getting

$mg+(P_0-P(t))A = mv \frac{dv}{dx}$

Nice. Now, if we rearrange this equation, we get:

$P(t)A=mg+P_0A - mv \frac{dv}{dx}$

If we integrate this equation with respect to x, we get:
$$-W(t)=\int{PAdx}=(mg + P_0A)x(t) + \frac{mv^2(t)}{2}$$
where W(t) is the cumulative amount of work that the gas does on the piston between time t = 0 and time t=t, x(t) is the downward displacement of the piston at time t, and v(t) is the downward velocity of the piston at time t.

After a very long period of time, the piston will stop oscillating and will come to rest. This will tell us the cumulative amount of work that has been done by the gas on the piston at final steady state (infinite time).
$$W(∞)=-(mg + P_0A)x(∞)$$
The reason that the piston comes to rest (i.e., loses all its kinetic energy) is that viscous dissipation has been occurring within the cylinder.

Since this compression takes place adiabatically, Q = 0. Therefore,
$$ΔU = MC_v(T(∞)-T_0)=(mg + P_0A)x(∞)$$
where M is the number of moles of gas in the cylinder. From the ideal gas law, in terms of P₁, T₀, l, and A, what is M?
Combine this with the previous equation and with the equation for P₁ in terms of mg, A, and P₀ to solve for T as a function of T₀ and x(∞)/l.

 

[Chestermiller]

I am new here, so I do not know how things work, so forgive my stupidity.

I am getting 0.6H

Hi Yellowflash. Welcome to Physics Forums!

The way things work here is that, rather than giving the answer (or every detail of how to solve the problem) we try to help the original poster (OP) work through the problem on his own, by keeping him pointed in the right direction. (I personally have not worked through the problem to completion, but, my gut feeling is that yours is the correct answer: x = H/γ)

Chet

 

[utkarshakash]

Nice. Now, if we rearrange this equation, we get:

$P(t)A=mg+P_0A - mv \frac{dv}{dx}$

If we integrate this equation with respect to x, we get:
$$-W(t)=\int{PAdx}=(mg + P_0A)x(t) + \frac{mv^2(t)}{2}$$
where W(t) is the cumulative amount of work that the gas does on the piston between time t = 0 and time t=t, x(t) is the downward displacement of the piston at time t, and v(t) is the downward velocity of the piston at time t.

After a very long period of time, the piston will stop oscillating and will come to rest. This will tell us the cumulative amount of work that has been done by the gas on the piston at final steady state (infinite time).
$$W(∞)=-(mg + P_0A)x(∞)$$
The reason that the piston comes to rest (i.e., loses all its kinetic energy) is that viscous dissipation has been occurring within the cylinder.

Since this compression takes place adiabatically, Q = 0. Therefore,
$$ΔU = MC_v(T(∞)-T_0)=(mg + P_0A)x(∞)$$
where M is the number of moles of gas in the cylinder. From the ideal gas law, in terms of P₁, T₀, l, and A, what is M?
Combine this with the previous equation and with the equation for P₁ in terms of mg, A, and P₀ to solve for T as a function of T₀ and x(∞)/l.

$T_{\infty} = T_0 \left\{ (\gamma -1) \lambda +1 \right\}$
where λ=x(∞)/l

If this is correct, what should be my next step?

 

[Chestermiller]

$T_{\infty} = T_0 \left\{ (\gamma -1) \lambda +1 \right\}$
where λ=x(∞)/l

If this is correct, what should be my next step?

Nice work.

The hard part is over. You know the final temperature in terms of x, you can express the final volume in terms of A, l, H, and x, and, you know that the final pressure has to be the same as the original pressure P₁ (since the same equilibrium force balance applies again). You also know that the number of moles haven't changed. This is enough information to calculate x from the ideal gas law.

Chet

 

[Saitama]

No. This is an irreversible compression. The "adiabatic formula" applies only to reversible expansions and compressions.

Chet

Hello Chet!

I am at a loss in concluding the above from the problem statement. Can you please explain how it is irreversible? Thanks!

 

[haruspex]

Hello Chet!

I am at a loss in concluding the above from the problem statement. Can you please explain how it is irreversible? Thanks!

It's because heat was allowed to flow in after the piston moved up. That increases entropy.

 

[utkarshakash]

Nice work.

The hard part is over. You know the final temperature in terms of x, you can express the final volume in terms of A, l, H, and x, and, you know that the final pressure has to be the same as the original pressure P₁ (since the same equilibrium force balance applies again). You also know that the number of moles haven't changed. This is enough information to calculate x from the ideal gas law.

Chet

Thanks a lot! I get x=H/γ ≈ 0.6H which is correct.

 

[Chestermiller]

Hello Chet!

I am at a loss in concluding the above from the problem statement. Can you please explain how it is irreversible? Thanks!

Because the problem statement says "and then the (heavy) piston is released." Before the piston was released, you were supporting it (and it was in force equilibrium). Once you let go, it is no longer in force equilibrium, and it rapidly drops. This results in an irreversible adiabatic compression of the gas. In order for the compression to be reversible, you would have to lower the piston gradually (i.e., quasistatically).

Chet

 

[utkarshakash]

It's because heat was allowed to flow in after the piston moved up. That increases entropy.

@Chestermiller: What do you have to say about this?

 

[Chestermiller]

@Chestermiller: What do you have to say about this?

I would say that the process of raising the piston was done close to reversibly, and the process of dropping the piston took place irreversibly. Of course, it doesn't matter how raising the piston took place, because you are told the initial and final equilibrium conditions of the system for this case. However, to determine the final equilibrium conditions for dropping the piston, you need to take into account that the piston was released spontaneously, and the compression took place irreversibly.

Chet

 

[haruspex]

I would say that the process of raising the piston was done close to reversibly, and the process of dropping the piston took place irreversibly. Of course, it doesn't matter how raising the piston took place, because you are told the initial and final equilibrium conditions of the system for this case. However, to determine the final equilibrium conditions for dropping the piston, you need to take into account that the piston was released spontaneously, and the compression took place irreversibly.

Chet

Sorry, Chet, but I believe you have it backwards. Adiabatic processes, i.e. where heat does not flow down a temperature gradient, are reversible. The entropy does not change (much). The slow rise of the piston and the subsequent pause allowing temperatures to equalise are the irreversible steps. Entropy has increased and cannot be made to go back down again.

 

[Chestermiller]

Sorry, Chet, but I believe you have it backwards. Adiabatic processes, i.e. where heat does not flow down a temperature gradient, are reversible. The entropy does not change (much). The slow rise of the piston and the subsequent pause allowing temperatures to equalise are the irreversible steps. Entropy has increased and cannot be made to go back down again.

Hi haruspex. I guess we have a disagreement between experts here. My understanding is that adiabatic processes are reversible only if they are carried out reversibly. Spontaneously allowing the piston to drop in the compression step is not a reversible process.

The wording for the expansion step (raising the piston) is not too clear. Whether it was carried out reversibly or irreversibly is not an issue, since, in either case the entropy of the system increased between the initial and final states. However, just because the entropy of the system increased does not mean that the process was irreversible; the entropy of a system increasing or decreasing is not a criterion for reversibility or irreversibility of a process. Irrespective of whether raising the piston was reversible or irreversible, the change in entropy of the gas was MR ln[(H+l)/l], where M is the number of moles of gas. The process was reversible if the sum of the entropy changes for the system plus surroundings was zero.

The problem statement is ambiguous regarding the expansion step: "The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished." Raising the piston slowly has a suggestion that the step was close to reversible, but saying that it was kept in the elevated position long enough for the thermodynamic equilibirum to be reestablished suggests a certain degree of irreversibility. That's why, in my earlier post, I said "the process of raising the piston was done close to reversibly.

In any event, this has nothing to do with the compression step which is clearly irreversible. The final answer to the problem is consistent with this.

Chet

 

[Yellowflash]

Hi Yellowflash. Welcome to Physics Forums!

The way things work here is that, rather than giving the answer (or every detail of how to solve the problem) we try to help the original poster (OP) work through the problem on his own, by keeping him pointed in the right direction. (I personally have not worked through the problem to completion, but, my gut feeling is that yours is the correct answer: x = H/γ)

Chet

Thanks, sorry for not following the rules.

 

[Chestermiller]

Thanks, sorry for not following the rules.

No problem. We were all newbies once.

Chet

 

[haruspex]

My understanding is that adiabatic processes are reversible only if they are carried out reversibly.
Spontaneously allowing the piston to drop in the compression step is not a reversible process.

Mine is that if it's adiabatic and isentropic then it's reversible: http://en.wikipedia.org/wiki/Reversible_adiabatic_process.
I don't think it is altered by the fact that the piston is merely allowed to drop. What does change things is this line in the OP which I overlooked:

After the piston comes to rest

If the system were completely insulated and frictionless, this would not happen. The piston would oscillate forever, reaching its original height each cycle. The fact that it comes to rest implies there are losses, so entropy does increase, and as you say, this phase is not reversible either.

just because the entropy of the system increased does not mean that the process was irreversible; the entropy of a system increasing or decreasing is not a criterion for reversibility or irreversibility of a process.

The same Wikipedia link disagrees with you on that. If system entropy increases, how can the process be reversed?

The problem statement is ambiguous regarding the expansion step: "The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished." Raising the piston slowly has a suggestion that the step was close to reversible, but saying that it was kept in the elevated position long enough for the thermodynamic equilibirum to be reestablished suggests a certain degree of irreversibility. That's why, in my earlier post, I said "the process of raising the piston was done close to reversibly.
Chet

You're right that raising the piston is not necessarily irreversible. If it is done so slowly that it's effectively isothermal then there's negligible increase in system entropy. But it's the phrase

then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished.

that makes it irreversible. It implies a temperature difference arose.

 

[Chestermiller]

Hi haruspex. I feel like we are reaching consensus.

Mine is that if it's adiabatic and isentropic then it's reversible: http://en.wikipedia.org/wiki/Reversible_adiabatic_process.

Yes. I agree, but all I was saying was that not all adiabatic processes are isentropic and reversible.

I don't think it is altered by the fact that the piston is merely allowed to drop.

I don't think I implied that the expansion phase was affected by the compression phase. The two process steps are completely separate.

What does change things is this line in the OP which I overlooked:

If the system were completely insulated and frictionless, this would not happen. The piston would oscillate forever, reaching its original height each cycle. The fact that it comes to rest implies there are losses, so entropy does increase, and as you say, this phase is not reversible either.

Yes.

The same Wikipedia link disagrees with you on that. If system entropy increases, how can the process be reversed?

With all due respect to Wikipedia, if I expand a gas isothermally and reversibly (adding whatever heat necessary to hold the temperature constant), then its entropy increases. I can then reverse the process by compressing it isothermally and reversibly (removing whatever heat necessary to hold the temperature constant), and return it to its original state, with its entropy decreasing by the same amount that it increased during the expansion. The net effect of the forward and reverse paths is no change in entropy.

You're right that raising the piston is not necessarily irreversible. If it is done so slowly that it's effectively isothermal then there's negligible increase in system entropy. But it's the phrase
that makes it irreversible. It implies a temperature difference arose.

Yes. As I said, it's probably close to reversible (but, again, in my judgement the wording is very imprecise and ambiguous).

Chet

 

[haruspex]

Hi haruspex. I feel like we are reaching consensus.

Yes indeed.

I don't think I implied that the expansion phase was affected by the compression phase. The two process steps are completely separate.

No, I was referring to this observation:

Spontaneously allowing the piston to drop in the compression step is not a reversible process.

The irreversibility has nothing to do with the piston's dropping under its own weight. What makes it irreversible is that there are losses of some sort - whether heat flow or friction - that lead it to stop oscillating eventually.

if I expand a gas isothermally and reversibly (adding whatever heat necessary to hold the temperature constant), then its entropy increases. I can then reverse the process by compressing it isothermally and reversibly (removing whatever heat necessary to hold the temperature constant), and return it to its original state, with its entropy decreasing by the same amount that it increased during the expansion.

True, but I said system entropy. In what you describe above, entropy is merely moving between parts of the system, not increasing overall. If the expansion phase is isothermal, no system entropy increase. But it appears it was not quite isothermal, so not quite reversible. I think we agree on that, and no conflict with Wikipedia.

 

[Chestermiller]

No, I was referring to this observation:

The irreversibility has nothing to do with the piston's dropping under its own weight. What makes it irreversible is that there are losses of some sort - whether heat flow or friction - that lead it to stop oscillating eventually.

What I was referring to was that the non-quasistatic nature of the rapid deformation prevents the process from being reversible by creating pressure and temperature non-uniformities within the gas during the deformation. These non-uniformities die out as a result of viscous dissipation and heat conduction as the final equilibrium state is approached.

True, but I said system entropy. In what you describe above, entropy is merely moving between parts of the system, not increasing overall. If the expansion phase is isothermal, no system entropy increase. .

The system I am referring to is the gas alone. So entropy is not merely moving between parts of the system. If I compress the gas isothermally and reversibly by doing quasistatic work on it and exchanging heat with the surroundings to hold its temperature constant (using only differential temperature driving forces), its entropy increases. But, once completed, this process can be reversed such that both the system and the surroundings are restored to their original states without any traces of the changes that once occurred. That's what a reversible process is.

Chet

 

[Chetty]

Is it possible that this topic can still be discussed further?

 

[Chestermiller]

Is it possible that this topic can still be discussed further?

No problem. What are your thoughts?

 

[Chetty]

The problem as stated was an equalized gas cylinder with a piston had the piston lifted up and held at a distance h and allowed to again equalize by allowing heat conduction through the cylinder to bring the gas temperature back to ambient. At this point, the cylinder was insulated and the the piston was allowed to drop. In the workup, if I remember correctly, the work done was (mg+Po*A)*x and this was set equal to the internal energy temperature rise. But in addition to the heat generated by the oscillations, there is a change in the value of the pdV integral which it seems should be accounted for and not all of the work done attributed to just a change in the internal energy. Looking at the starting condition before the piston was raised and the final equalized condition, the pressures remain the same because they support the same piston and external pressure but the volumes are different. Did I miss something?

 

[Chestermiller]

The problem as stated was an equalized gas cylinder with a piston had the piston lifted up and held at a distance h and allowed to again equalize by allowing heat conduction through the cylinder to bring the gas temperature back to ambient. At this point, the cylinder was insulated and the the piston was allowed to drop. In the workup, if I remember correctly, the work done was (mg+Po*A)*x and this was set equal to the internal energy temperature rise. But in addition to the heat generated by the oscillations, there is a change in the value of the pdV integral which it seems should be accounted for and not all of the work done attributed to just a change in the internal energy. Looking at the starting condition before the piston was raised and the final equalized condition, the pressures remain the same because they support the same piston and external pressure but the volumes are different. Did I miss something?

Before the piston is raised, the force balance on the piston is $$P_1A=mg+P_{atm}A$$where m is the mass of the piston, A is the area of the piston, $P_1$ is the gas pressure in the initial state, and $P_{atm}$ is the atmospheric pressure. After the piston is raised, and the system re-equilibrates, by the ideal gas law, the gas pressure in the intermediate state is $P_2$, given by: $$P_2=\frac{l}{(l+H)}P_1=\frac{l}{(l+H)}\left(P_{atm}+\frac{mg}{A}\right)$$

After the piston is released (i.e., during the irreversible adiabatic compression of the gas by the piston), the force balance on the piston is given by $$PA-mg-P_{atm}A=m\frac{dv}{dt}$$where P(t) is the pressure of the gas against the piston at time t during the compression and v is the (upward) velocity of the piston.

OK so far?

 

[Chetty]

Looks good to me.

 

[Chestermiller]

Looks good to me.

I corrected a couple of errors in the force balance on the piston. Did you notice?

If I multiply the left side of the force balance by dx/dt and the right side of the force balance by its equivalent, v, I obtain:$$PA\frac{dx}{dt}-mg\frac{dx}{dt}-P_{atm}A\frac{dx}{dt}=mv\frac{dv}{dt}$$

Are you OK with this now?

 

[Chetty]

Looks good. My concern is when you get to the point of writing :
delta U= MCv(Tf-T0)=(mg+P0 A) xf

This seems to say that the work done only impacts the temperature increase due to the oscillation damping and no account for the gas compression that also takes place from when the piston is held at l+h and then released. Should there be a term for the adiabatic compression as well as the excitation for the damping?

 

[Chestermiller]

Looks good. My concern is when you get to the point of writing :
delta U= MCv(Tf-T0)=(mg+P0 A) xf

This seems to say that the work done only impacts the temperature increase due to the oscillation damping and no account for the gas compression that also takes place from when the piston is held at l+h and then released. Should there be a term for the adiabatic compression as well as the excitation for the damping?

We'll get to that soon.

So, if I write the previous equation a little differently, I get:

$$PA\frac{dx}{dt}=(mg+P_{atm}A)\frac{dx}{dt}+mv\frac{dv}{dt}$$
If I integrate this with respect to time, I get:$$\int_0^t{PA\frac{dx}{dt}dt}=(mg+P_{atm}A)[x(t)-(l+H)]+m\frac{v^2(t)}{2}$$or equivalently
$$W_g(t)=(mg+P_{atm}A)[x(t)-(l+H)]+KE_P(t)$$where $W_g(t)=\int_0^t{PA\frac{dx}{dt}dt}$ is the work done by the gas on its surroundings (the piston) up to time t, and $KE_P(t)=m\frac{v^2(t)}{2}$ is the kinetic energy of the piston at time t.

OK so far?

 

[Chetty]

The sign of Patm must agree with the sign of mg. I see you have taken x as measured from the bottom of the cylinder making the terms mg + Patm A always negative which makes sense for work done by the gas but am wrestling a bit with the positive piston kinetic energy but find no other issue with the equation. Interesting approach.