Thermo - PV work and mechanical work

[LoopInt]

Hello,

I`m a bit confused about PV work. Here is the situation I am facing.

δw = P dV and dw = F dx

δw is an inexact differential, because it's path dependent. I read some articles that equal δw with dw.

saying that

P dV = F dx

For what path is that True? In other words, When δw=dw?

Thanks in advance!

 

[Chestermiller]

Let's rewrite the equation as follows:
$$PdV=Fdx=\frac{F}{A}(Adx)$$
If A is the area of the piston, then Adx = dV

So
$$PdV=Fdx=\frac{F}{A}(Adx)=\frac{F}{A}dV$$
Therefore, $$F = PA$$
For a more detailed discussion of how to properly calculate the work at the interface between the system and surroundings, see my Blog at https://www.physicsforums.com/member.php?u=345636

Chet

 

[DrDu]

δw = P dV and dw = F dx
!

Instead of dw=F dx you should rather write δw=F dx. Work alone is in general not a total differential.

 

[LoopInt]

Thanks for the replies.
Chestermiller, I did these steps but I couldn`t figure out why Pdv=Fdx if they are path dependent.
And yes DrDu you are right. dw is only the case when the force is conservative, right?

dw=PdV is right when the process is adiabatic (the force will be conservative now)
so, if the process is adiabatic I can say dw=dw, and therefore PdV=Fdx
Am I right?

 

[Chestermiller]

Thanks for the replies.
Chestermiller, I did these steps but I couldn`t figure out why Pdv=Fdx if they are path dependent.

They're both different ways of writing exactly the same thing. Unless you are dealing with a conservative force field, F is path dependent.

Did you read my blog?

Chet

 

[LoopInt]

Yes, I did. The confusion I'm making is the following:

I'm considering work the area under the p-v diagram curve. If I have a piston and pressurize it, the force will be F=P.A.
Now for the work point of view. w=\integral P.dv which may vary depending on the path. Therefore consider 2 different paths that done w1 and w2.
w1 = F1.dx
w2 = F2.dx
if w1 > w2 then F1 >F2. But for a given pressure on the piston, the force is F regardless of the path.

More precisely, I'm trying to figure out the force that a membrane exerts. For that I am using F=Pdv/dx (dv/dx is the rate of which the volume changes when I pull it dx). But I am afraid I am calculating the wrong F, because it may depend on which process I do that.

For a given pressure, the membrane should exert the same force regardless of how it reached that state (pressure and volume).

 

[Chestermiller]

Yes, I did. The confusion I'm making is the following:

I'm considering work the area under the p-v diagram curve. If I have a piston and pressurize it, the force will be F=P.A.
Now for the work point of view. w=\integral P.dv which may vary depending on the path. Therefore consider 2 different paths that done w1 and w2.
w1 = F1.dx
w2 = F2.dx
if w1 > w2 then F1 >F2. But for a given pressure on the piston, the force is F regardless of the path.

More precisely, I'm trying to figure out the force that a membrane exerts. For that I am using F=Pdv/dx (dv/dx is the rate of which the volume changes when I pull it dx). But I am afraid I am calculating the wrong F, because it may depend on which process I do that.

For a given pressure, the membrane should exert the same force regardless of how it reached that state (pressure and volume).

As I said in my blog, for an irreversible path between the initial and final equilibrium states, the work is not the area under the equilibrium curve on the p-v diagram. The pressure within the system for an irreversible path is typically not even uniform. So what pressure do you use? The correct pressure to use is the pressure at the interface with the surroundings (which differs from the equilibrium pressure at the volume v). Only for a reversible path is the work equal to the area under the equilibrium curve on the p-v diagram. This is because, for a reversible path, the pressure within the system is uniform, and equal to the pressure at the interface with the surroundings.

To help solidify your understanding, why don't you dream up a specific problem that we can solve together.

Chet

 

[LoopInt]

Sure, that's very nice of you.

Let's consider a balloon that is fixed in one end at the ceiling and the other end has a load F=mg. The balloon has volume V1 and initial pressure P1. The mass is at height 0 (reference point). We then blow air into the balloon and the pressure inside it increases. The mas then raises a distance x. What is that distance? What about if the pressure is kept constant and take the load off instead?

I draw something to make it clear.

 

[Chestermiller]

Sure, that's very nice of you.

Let's consider a balloon that is fixed at the ceiling and the other end has a load F=mg. The balloon has volume V1 and initial pressure P1. The mass is at height 0 (reference point). We then blow air into the balloon and the pressure inside it increases. The mas then raises a distance z. What is that distance? What about if the pressure is kept constant and take the load off instead?

I'm requesting that you think of a simpler problem than this one (one that is also capable of illustrating your issue). If we're dealing with a balloon, then we have to consider the stress/deformation behavior of the balloon material (which affects the force, pressure, and movement of the mass). We would also have to consider the non-linear kinematics of the balloon deformation, which is going to involve large changes in balloon curvature and large inhomogeneous deformations of the balloon membrane. I would like to avoid this complexity.

Might I suggest some sort of simple problem involving a gas, piston, and cylinder, with possibly surroundings pressure or force external to the piston.

Chet

 

[LoopInt]

Sure, sorry!

Then imagine a piston at the floor pointing up (vertical position). It has a mass at the top. Pressure inside is P1 and volume V1. We pressurize it and the mass goes up by x. The new pressure is now P2 and V2. The piston is a cylinder. What is the distance x? And what happens if we take the load out keeping the pressure P1?

 

[Chestermiller]

Sure, sorry!

Then imagine a piston at the floor pointing up (vertical position). It has a mass at the top. Pressure inside is P1 and volume V1. We pressurize it and the mass goes up by x.

How exactly do we pressurize it? Increase the temperature? Add more gas to the cylinder?
What is the condition on the other side of the piston, air at atmospheric pressure?

And what happens if we take the load out keeping the pressure P1?

How exactly do we take the load off while maintaining the pressure at P1, add a force in place of the weight?

Chet

 

[LoopInt]

How exactly do we pressurize it? Increase the temperature? Add more gas to the cylinder?
What is the condition on the other side of the piston, air at atmospheric pressure?

We put more gas in it. Outside it is atmospheric pressure. Temperature is kept constant.

How exactly do we take the load off while maintaining the pressure at P1, add a force in place of the weight?

While we take the load off we put gas in, so that pressure stays constant.

 

[Chestermiller]

We put more gas in it. Outside it is atmospheric pressure. Temperature is kept constant.

You are not going to be able to increase the pressure to P2 this way. The weight of the mass is mg, and the outside pressure is 1 atm., so, at final equilibrium, the pressure will be P2=P1 = mg/A + 1 atm and the weight will rise by ΔV/A.

While we take the load off we put gas in, so that pressure stays constant.

Same answer for this case.

 

[LoopInt]

You are right.

Then initially its P1. Then we increase the mass until there is P2 in the cylinder. What's the distance x it dropped?

I found that

x= (L(A(P1-P0)-(m+M)g))/(P0A+(m+M)g)

where L is the distance between the bottom and the top of the piston, m is the mass at the beginning, M is the adicional mass.

The way I did it was:

- The sum of forces are 0, because it is in equilibrium;
- As the temperature is constant, P1V1=P2V2
- Isolate P2 for the above equation
- Consider V1=LA and V2=(L+x) I should had put (L-x) because P2 is higher, but that's not so relevant because I can choose x to increase downwards.
- Substitute V1 and V2 in the P2 equation
- Sum of forces when the mass is increased = 0
- Substitute P2 in the sum of forces equation
- Isolate x

If that is correct, consider two separate cases. We know that when we have P1 and V1 and press the piston by x, the force that the piston exerts is (m+M)g. If we press very fast and stop when P2 is reached, the force we apply to keep the system in P2 (equilibrium) is (m+M)g? In other words, I can do whatever I want to the system in any way, and at the state P2, V2 and the load will be the same always?

 

[Chestermiller]

You are right.

Then initially its P1. Then we increase the mass until there is P2 in the cylinder. What's the distance x it dropped?

I think you mean "decrease the mass", correct?

I found that

x= (L(A(P1-P0)-(m+M)g))/(P0A+(m+M)g)

where L is the distance between the bottom and the top of the piston, m is the mass at the beginning, M is the adicional mass.

The way I did it was:

- The sum of forces are 0, because it is in equilibrium;
- As the temperature is constant, P1V1=P2V2
- Isolate P2 for the above equation
- Consider V1=LA and V2=(L+x) I should had put (L-x) because P2 is higher, but that's not so relevant because I can choose x to increase downwards.
- Substitute V1 and V2 in the P2 equation
- Sum of forces when the mass is increased = 0
- Substitute P2 in the sum of forces equation
- Isolate x

This analysis was not done correctly. Even if the temperature is constant, you decreased the number of moles. According to the ideal gas law, P1(V2-V1)=ΔnRT. So, here again the pressure doesn't change, but the volume does. For this case:
x = (V1-V2)/A=ΔnRT/(P1A)=ΔnRT/(mg+P0A)

Chet

 

[Chestermiller]

OK. In the previous post, I misinterpreted what you were saying. I thought by mass, you were talking about the mass of gas in the cylinder. I will re-evaluate what you did, and get back with you.

Chet

 

[Chestermiller]

You are right.

Then initially its P1. Then we increase the mass until there is P2 in the cylinder. What's the distance x it dropped?

I found that

x= (L(A(P1-P0)-(m+M)g))/(P0A+(m+M)g)

where L is the distance between the bottom and the top of the piston, m is the mass at the beginning, M is the adicional mass.

The way I did it was:

- The sum of forces are 0, because it is in equilibrium;
- As the temperature is constant, P1V1=P2V2
- Isolate P2 for the above equation
- Consider V1=LA and V2=(L+x) I should had put (L-x) because P2 is higher, but that's not so relevant because I can choose x to increase downwards.
- Substitute V1 and V2 in the P2 equation
- Sum of forces when the mass is increased = 0
- Substitute P2 in the sum of forces equation
- Isolate x

Aside from the sign, I get the same answer. I've re-expressed it as:
$$x=\frac{MgL}{((M+m)g+P_0A)}$$

If that is correct, consider two separate cases. We know that when we have P1 and V1 and press the piston by x, the force that the piston exerts is (m+M)g. If we press very fast and stop when P2 is reached, the force we apply to keep the system in P2 (equilibrium) is (m+M)g? In other words, I can do whatever I want to the system in any way, and at the state P2, V2 and the load will be the same always?

Yes, as long as you do what you need to to make sure that the final temperature is the same.

 

[LoopInt]

Ok, I got it! Thanks!
Now, concerning that case of the ballon. The rubber would make it more complex, but is it the same case? I can do whatever I want, and still have same results since P, V and T are same?

 

[Chestermiller]

Ok, I got it! Thanks!
Now, concerning that case of the ballon. The rubber would make it more complex, but is it the same case? I can do whatever I want, and still have same results since P, V and T are same?

Sure, as long as the balloon material is perfectly elastic.
Now I have a problem for you:

Suppose you attach a handle to the top of the mass M so you can lower the mass by hand. If you lower it gradually, how much work does piston do on the gas (assuming that the temperature is held constant, and you lower it slowly enough for the process to be reversible)? How much work does the piston do on your hand as you lower the weight? How much work is done on the piston in this process?

Chet

 

[LoopInt]

The work the piston does on the gas is

W = A(nRT Ln(L/(L-d))-P0d)

For lowering the weight I do work of Mgd. Work done on the piston... It is all the time in equilibrium, so There is no net force. Work should be 0 on the piston.

I do work of -Mgd, gravity does Mgd. The mass does A(nRT Ln(L/(L-d))-P0d) to the piston, P0 does P0Ad to the piston and gas does -AnRT Ln(L/(L-d)) to the piston.

 

[Chestermiller]

The work the piston does on the gas is

W = A(nRT Ln(L/(L-d))-P0d)

For lowering the weight I do work of Mgd. Work done on the piston... It is all the time in equilibrium, so There is no net force. Work should be 0 on the piston.

I do work of -Mgd, gravity does Mgd. The mass does A(nRT Ln(L/(L-d))-P0d) to the piston, P0 does P0Ad to the piston and gas does -AnRT Ln(L/(L-d)) to the piston.

This is a pretty good start. You recognized that this is a reversible process, so the work on the gas is the area under the p-v curve. I got a slightly different answer:
$$W=nRT\ln{\left(\frac{L}{L-x}\right)}= nRT\ln{\left(\frac{V_1}{V_2}\right)}=nRT\ln{\left(\frac{P_2}{P_1}\right)}$$
Now let's continue addressing the rest of my questions. We're going to start out by doing a force balance on the combination of piston m and mass M. Since the system is always just slightly removed from equilibrium, we won't have a mass times acceleration term. Here is a list of the forces acting on the combination at some point during the lowering of the mass M:

F = upward force exerted by your hand on the top of M
P₀A = downward force exerted by the atmospheric pressure on the top of M
PA = upward force exerted by the gas on the bottom of m
w = (M+m)g = downward force of gravity on m+M

Please write out this force balance. Also write out what this force balance reduces to time t = 0, when the mass M just begins to get lowered, m and M are just starting to kiss, and the volume of gas has not yet changed.

Please bear with me. I know that doing a force balance seems a little hinkey, but it will help us determine the amount of work being done.

Chet

 

[LoopInt]

I didn't do that with only 1 FBD, but with 2 separate cases instead.
But following your thinking,
Doing the force balance on the piston gives me:

-Fatm+Fhand-Fmass+Fgas=0

Initially, my hand is lifting the whole mass so Fhand=mg, and therefore Fgas=Fatm.
Now, as I lower the weight

Fgas=Fatm+Fmass-Fhand

Where Fhand goes from mg to 0.
The work the piston does on the gas is the same as the work the gas does on the piston but with opposite signs, right?
So I can calculate the work of this expression
\integral F(x) dx = \integral (Fatm+Fmass-Fhand) dx
Now I need to express how the force varies with x
I will use PV=nRT to do that, because this equal the gas force.
P=nRT/V = nRT/(LA)
the force, then, is:
F=PA = nRT/L
But as I lower the weight by x, the height is L-x so,
F=nRT/(L-x)
Carrying the integration from x varying from 0 to d (maximum distance)
W = nRT Ln(L/(L-d))

 

[Chestermiller]

I didn't do that with only 1 FBD, but with 2 separate cases instead.
But following your thinking,
Doing the force balance on the piston gives me:

-Fatm+Fhand-Fmass+Fgas=0

Initially, my hand is lifting the whole mass so Fhand=mg, and therefore Fgas=Fatm.
Now, as I lower the weight

Fgas=Fatm+Fmass-Fhand

Where Fhand goes from mg to 0.
The work the piston does on the gas is the same as the work the gas does on the piston but with opposite signs, right?
So I can calculate the work of this expression
\integral F(x) dx = \integral (Fatm+Fmass-Fhand) dx
Now I need to express how the force varies with x
I will use PV=nRT to do that, because this equal the gas force.
P=nRT/V = nRT/(LA)
the force, then, is:
F=PA = nRT/L
But as I lower the weight by x, the height is L-x so,
F=nRT/(L-x)
Carrying the integration from x varying from 0 to d (maximum distance)
W = nRT Ln(L/(L-d))

Your force balance(s) do not look correct. Let C be the downward contact force exerted by the mass M on the piston. Then the force balance on the piston of mass m is:
$$PA-mg-C=0$$
And the force balance on the mass M is:
$$C-Mg+F-P_0A=0$$
At time zero, C=P₀A and P=P₁, so

$P_1A-P_0A=mg$ at time t = 0 and
$F=Mg$ at time t = 0
If we add the force balance on m to the force balance on M, we get the overall force balance on the "stack:"
$$PA-mg-Mg+F-P_0A=0$$
or
$$F=(m+M)g-PA+P_0A$$
Now, since downward movement is taken as positive, the differential work that the hand does when the stack moves down dx is -Fdx. Therefore, let's multiply the previous equation by -dx:
$$dW_F=-Fdx=-(m+M)gdx-PdV+P_0dV$$
where dV=-Adx
Now, the next step is to integrate this equation from x = 0 to x = x (the final x) to get W_F the work that the hand does on the stack, and -W_F the work that the stack does on the hand.

After we finish this problem, we can do the irreversible problem of letting the mass M spontaneously fall from its initial position just kissing the top of the piston until it reaches its final equilibrium position. The results will not be the same. Now, won't that be fun?

Chet

 

[LoopInt]

What I did was do the overall force balance directly. It equal to your force balance, but I considered the piston to be massless, wrongly.
The work done by the piston on the gas should be nRT Ln(L/(L-d)), right?
What you calculate is the work your hand is doing, isn't it?

 

[Chestermiller]

What I did was do the overall force balance directly. It equal to your force balance, but I considered the piston to be massless, wrongly.

Oh. OK. You had mentioned something about 2 force balances, so I got confused. So I think we now have the same force balance, correct?

The work done by the piston on the gas should be nRT Ln(L/(L-d)), right?

Correct.

What you calculate is the work your hand is doing, isn't it?

As I said, "W_F the work that your hand does on the stack, and -W_F the work that the stack does on the hand." So the answer to this question is Yes.

I think you've gotten what I wanted you to get out of this problem. I leave it up to you to complete the other integrations on the right hand side of the equation. But, just as a matter of summerizing, one good (practically fool proof) way of figuring out the amounts of work done (so that you get all the signs correct) is to start out with the force balance and then integrate the equation over the displacement.

If you are interested in continuing, we can do the irreversible problem in which, rather than lowering the weight gradually by hand, we just release the mass M and let it drop. We would be focusing on determining the work done by the gas on the piston, which is different for this case from what we got when we lowered the mass gradually. Any interest in continuing? If not I understand.

Chet

 

[LoopInt]

Yea, I sure want to continue! I need to learn this to model my system. Before we continue, may I ask some questions.
I am doing an experiment of my system. If I do that very slowly, so it's in quasi-equilibrium, and the system is in contact with the ambient, the system temperature may raise but as I wait long enough, the ambient equalizes the temperature of the system, so Tamb=Tsys. Therefore it's an isothermal process, right?
Else, if I do that very fast, there will be no time to the system exchanges heat, and so Tamb may not be Tsys, then I consider the process to be adiabatic, right?
There is no way a process can be isothermal AND adiabatic at same time, is there?
So If I am modeling my system to match experimental static data, I should use isothermal and If I test the system dynamically, I should use adiabatic. Is that correct?

EDIT: I need to correct one thing. I didn't mention I test my system adding gas to it. So it can't be adiabatic.

We can sure continue now with this! And by the way, Thanks A LOT!

 

[Chestermiller]

Yea, I sure want to continue! I need to learn this to model my system. Before we continue, may I ask some questions.
I am doing an experiment of my system. If I do that very slowly, so it's in quasi-equilibrium, and the system is in contact with the ambient, the system temperature may raise but as I wait long enough, the ambient equalizes the temperature of the system, so Tamb=Tsys. Therefore it's an isothermal process, right?

Yes. If you do it slowly enough, and your system is not insulated, the system temperature will always be close to the ambient temperature.

Else, if I do that very fast, there will be no time to the system exchanges heat, and so Tamb may not be Tsys, then I consider the process to be adiabatic, right?

Yes. In the limit of very rapid deformation.

There is no way a process can be isothermal AND adiabatic at same time, is there?

In certain cases it's possible if there is significant viscous heating in the system, particularly in the case of a flow system. For example, if an ideal gas is forced through an adiabatic valve, its temperature won't change. The expansion cooling will be canceled by the viscous heating.

So If I am modeling my system to match experimental static data, I should use isothermal and If I test the system dynamically, I should use adiabatic. Is that correct?

Both are ideal approximations to the real system behavior.

EDIT: I need to correct one thing. I didn't mention I test my system adding gas to it. So it can't be adiabatic.

Yours is a so called open system. Look up how to apply the first law to open systems.

We can sure continue now with this! And by the way, Thanks A LOT!

OK. Let's continue. In the case we are going to be looking at next, the force F exerted by our hand is going to be zero, and the mass M is going to be dropping spontaneously, along with the piston. So there is going to be a mass times acceleration term in the force balance.

In addition, since the deformation is going to be irreversible, the gas pressure within the cylinder is going to be non-uniform. So we cannot apply the ideal gas law to the gas within the cylinder. Also, since the gas pressure is varying with position, rather than calling P the pressure throughout the gas within the cylinder, we are now going to call P the pressure of the gas on the bottom surface of the piston (this is the interface of the gas with its surroundings). By Newton's third law, this pressure will also be the force per unit area exerted by the piston on the gas. This interface pressure is what we will use to calculate the work done by the piston on the gas. Initially, P = P₁, and finally, P=P₂ (since in the initial and final equilibrium states, the pressure of the gas is uniform).

So. Please take the force balance that we derived for the quasi-static case, and modify it appropriately to obtain the force balance for this case.

Chet

 

[LoopInt]

The force balance is

-P0A+PA-(M+m)g=m1a

Initially

-P0A+P1A-(M+m)g=0

then

-P0A+P12A-(M+m)g=m1a

and finally

-P0A+P2A-(M+m)g=0

How can we proceed?

 

[Chestermiller]

The force balance is

-P0A+PA-(M+m)g=m1a

What's m1? Also, I'd like "a" to be measured downward so that a = d²x/dt²

Initially

-P0A+P1A-(M+m)g=0

This equation is not correct, because there is instantaneous downward acceleration as soon as the mass M is released. However, initially, -P0A+P1A-mg=0

and finally

-P0A+P2A-(M+m)g=0

How can we proceed?

So what I get for the force balance is:
$$(m+M)\frac{d^2x}{dt^2}=(m+M)g+P_0A-PA$$
In this "dropping mass" problem, I'd like to confine attention to the case where P₀=0 so that there is vacuum within the room surrounding the cylinder. This gives:
$$(m+M)\frac{d^2x}{dt^2}=(m+M)g-PA$$
For the initial state at t = 0, P=P₁=mg/A
and, for the final state, at t →∞, P=P₂=(m+M)g/A
So, P₂/P₁=(m+M)/m.
Now, what we are going to try to do is to use the force balance equation to determine the work that the gas does on the combination of masses, or the work that the masses do on the gas, from time t = 0 until the system re-equilibrates at the new equilibrium state. As in the previous problem, this is going to involve an integration. To get the ball rolling, I would like you to multiply both sides of the equation by dx/dt, and then to recognize that the resulting left side of the equation is an exact differential. (I hope this makes sense). If you recognize this exact differential, I would like you to rewrite that term in terms of the exact differential. If not, I'll fill in this step.

Chet

 

[LoopInt]

The m1 was the mass I didn't know what it was. Now that you wrote, it makes sense its m+M (total mass involved).

This equation is not correct, because there is instantaneous downward acceleration as soon as the mass M is released. However, initially, -P0A+P1A-mg=0

Yes, I forget that in that initial case M was not involved.

Since (m+M) d^2 x/dt^2 = (m+M)g−PA we need to multiply that by dx to get the work

\delta W = (m+M)g dx−PAdx

but we need to integrate over time so dividing by dt and multiplying by dt (chain rule)

\delta W = \int ((m+M)g dx/dt−PAdx/dt) dt

This is strange... I don't know how to proceed the integration

However, since dE=\delta W-\delta Q
for the case of the mass dropping fast, the system don't have time to exchange heat with the surroundings, so \delta Q = 0

dE=\delta W implies that dE=dW, thus an exact differential. Is that correct?

 

[Chestermiller]

Since (m+M) d^2 x/dt^2 = (m+M)g−PA we need to multiply that by dx to get the work

\delta W = (m+M)g dx−PAdx

but we need to integrate over time so dividing by dt and multiplying by dt (chain rule)

\delta W = \int ((m+M)g dx/dt−PAdx/dt) dt

This is strange... I don't know how to proceed the integration

However, since dE=\delta W-\delta Q
for the case of the mass dropping fast, the system don't have time to exchange heat with the surroundings, so \delta Q = 0

dE=\delta W implies that dE=dW, thus an exact differential. Is that correct?

Good try, but no. What you are calling ΔW is really the change in kinetic energy of the masses. We want to determine the work done by the gas on the bottom face of the piston. This is the work W done by the "system" on the "surroundings" (since we regard the gas as the system, and everything else as the surroundings). That work W is given by:
$$W=\int{PdV}=-\int{PAdx}=-\int_{t=0}^{t=∞}{PA\left(\frac{dx}{dt}\right)dt}$$
We can find this work by first multiplying the force balance by dx/dt to obtain:
$$(m+M)\frac{d^2x}{dt^2}\left(\frac{dx}{dt}\right)=(m+M)g\left(\frac{dx}{dt}\right)-PA\left(\frac{dx}{dt}\right)=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)$$
The left hand side of this equation is an exact differential, so we can rewrite the previous equation as:

$$\frac{d}{dt}\left(\frac{(m+M)}{2}\left(\frac{dx}{dt}\right)^2\right)=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)$$
Rearranging this equation gives:
$$P\left(\frac{dV}{dt}\right)=\frac{d}{dt}\left(\frac{(m+M)}{2}v^2\right)+\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)$$
where v is the velocity of the combined masses dx/dt. The left hand side is rate at which the gas is doing work on the two masses, the first term on the right hand side is the rate of change of kinetic energy of the two masses, and the second term on the right hand side is the rate of change of potential energy of the two masses. If we integrate this equation from time t = 0 to arbitrary time t, we obtain:
$$\int_{V_1}^{V(t)}{PdV}=\left(\frac{(m+M)}{2}v^2\right)+\frac{(m+M)g}{A}(V(t)-V_1)$$
where we have made use of the initial conditions that v = 0 and V = V₁ at t = 0.

Let's discuss what's happening here. After we release the masses, they will speed up, reach a maximum downward velocity and then slow down until they reach a maximum downward displacement where they come to a stop. But when they get to this bottom location, there is now a net upward force on them, so they start moving back upward again. They speed up, reach a maximum upward velocity, and then slow down until they reach a maximum upward displacement. But, when they get to the upper location, there is now a net downward force on them, so they start moving down again. The net result of all this is that the masses will oscillate up and down, very much as if they were in simple harmonic motion.

However, even with the piston being frictionless, the amplitude of the oscillation will decrease over time as a result of viscous dissipation within the gas. The details of how this happens is unimportant for our purposes, except that, as time progresses, the velocity of the masses will progressively decrease, and at very long times, it will become equal to zero. Therefore, at final steady state of the system, we will have:
$$W=\int_{V_1}^{V_2}{PdV}=\frac{(m+M)g}{A}(V_2-V_1)$$
Amazingly, the work from the initial to the final equilibrium state reduces to this very simple equation.

I'm going to stop here an let you digest what I've presented, as well as let you ask any question you might have. When you're ready, we can continue. We will be looking at both adiabatic and isothermal compression resulting from releasing the mass M.

 

[LoopInt]

I'm sorry, I didn't follow how you got from (m+M)d²xdt²(dx/dt) to d/dt((m+M)2(dx/dt)^2).

Also, you said this will be in harmonic motion, so isn't it "strange" to make the conditions for the final equilibrium stage without specifying how the energy is being lost? How do you see that this makes sense?

 

[Chestermiller]

I'm sorry, I didn't follow how you got from (m+M)d²xdt²(dx/dt) to d/dt((m+M)2(dx/dt)^2).

I started out by saying:

"We can find this work by first multiplying the force balance by dx/dt to obtain:

$(m+M)\left(\frac{dx}{dt}\right)\frac{d^2x}{dt^2}=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)$"

If we make the substitutions v = dx/dt, and dv/dt = d²x/dt²l on the left hand side, we get
$$(m+M)\left(\frac{dx}{dt}\right)\frac{d^2x}{dt^2}=(m+M)\frac{vdv}{dt}=(m+M)\frac{d(\frac{v^2}{2})}{dt}$$

Also, you said this will be in harmonic motion, so isn't it "strange" to make the conditions for the final equilibrium stage without specifying how the energy is being lost? How do you see that this makes sense?

This isn't exactly what I meant. I meant that the motion would be similar to simple harmonic motion, but damped. There is no energy being lost. Part of the mechanical energy (the kinetic energy) is being converted into internal energy and heat. Even with the piston being frictionless, you wouldn't expect the masses to continue bobbing up and down forever, would you? Neither would I. The gas in the cylinder has a physical property called viscosity. Have you heard of this property before? Even though the viscosity of the gas is very low, it is not zero. The viscosity acts to convert mechanical energy to heat when the gas deforms. So, not only does the gas act "elastically" like a spring, it also partially acts like a damper. This is analogous to a car shock absorber where you have a spring and a damper in series. When you try to oscillate the front end of the car up and down, the deforming viscous oil in the damper causes the amplitude of the oscillation to decay with time. The same kind of thing happens in our system until all the kinetic energy is converted. It is possible to formulate and solve the gas dynamics problem describing all the details of what is happening within the cylinder as a function of both spatial position and time, but this is not necessary for our purposes, because we are mainly interested in the initial and final equilibrium states of the system. So all we really need to know for our purposes is that the kinetic energy gets damped out over time (but this energy does not just vanish). I hope this makes better sense.

Chet

 

[LoopInt]

Ok, let me see if I get it. We are not specifying what is this damping factor and how it interacts with the system in detail, but we know that this exists and eventually all the kinetic energy will be converted to internal energy and heat, and so v final will be 0, which is all that matters to calculate the work.

If that's correct, let's move on!

 

[Chestermiller]

Ok, let me see if I get it. We are not specifying what is this damping factor and how it interacts with the system in detail, but we know that this exists and eventually all the kinetic energy will be converted to internal energy and heat, and so v final will be 0, which is all that matters to calculate the work.

If that's correct, let's move on!

Yes. That's right on.

So, the work equation we've derived so far makes no assumptions about whether the compression is isothermal or adiabatic (or anything else for that matter). Which case would you prefer to consider first? Isothermal or adiabatic?

Chet