Work in isothermal PV process. Which method is correct?

[barryj]

Homework Statement

Isothermal process![/B]
I have a cylinder with base area of 0.01m^2. I have a piston that is 0.1 meters from the bottom giving a volume of 0.001m^3. I have a weight of 1500 Newtons. This gives a pressure of 1.5E5 Pa. Now, I remove 500 Newtons and the piston moves upwards. Volume increases, pressure decreases. What is the work done by the gas on the weight as the piston moves upwards?

Homework Equations

PV = PV, work = Force X distance, Work = (PV)ln(V2/V1)

The Attempt at a Solution

See attached solution. Method 1 gives me 50 Joules of work, Method 2 gives 60 Joules of work. Where am I making a mistake? Both methods seem OK to me.

 

[Chestermiller]

Method 1 is correct, and method 2 is incorrect.

The work done by the system on the surroundings, whether the process is reversible or irreversible, is always given by dw = P_IntdV, where P_Int represents the force per unit area at the interface between the system and the surroundings. For the irreverisble process you described, if you do a force balance on the piston, you get $P_{Int}A-mg=m\frac{dv}{dt}$, where v is the velocity of the piston and A is the area of the piston. If you multiply this equation by v = dx/dt (where x is the displacement of the piston), and integrate with respect to time from t = 0 to t = t, you get:

$$W(t)=\int_0^t P_{Int}(t)A\frac{dx}{dt}dt = \frac{mg}{A}[V(t)-V(0)] + m\frac{v^2(t)}{2}$$

where W(t) is the cumulative work done by the gas on the weight up until time t, and V(t) is the volume of the gas during the irreversible expansion at time t. I hope you realize that, even if the piston is frictionless, the weight will oscillate up and down about its final equilibrium position, but eventually it will come to rest because of viscous damping by the gas. So, at very long times, the total amount of work done by the piston on the surroundings will be:
$$W(∞)= \frac{mg}{A}[V(∞)-V(0)] $$
But mg/A is the final pressure exerted by the piston on the gas after the system has re-equilibrated, so
$$W(∞)= P(∞)[V(∞)-V(0)] $$
This confirms your result from Method 1.

Method 2 is not correct because the gas force per unit area on the piston face during the reversible expansion will not be equal to the thermodynamic pressure determined by the ideal gas law. This is because, in a rapid irreversible expansion, viscous stresses will also be present in the gas, which will alter the force per unit area at the piston face. At a given gas volume, for an irreversible expansion, the force per unit area will be less than one would be calculate from the ideal gas law.

For more on this, please see my recent Physics Forums Insights article at https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/, particularly the section on the first law of thermodynamics. There are also two open threads that other members and I have been involved with to analyze problems like this one, and which may interest you:
https://www.physicsforums.com/threa...eversible-expansion-mean.812804/#post-5104268

and

https://www.physicsforums.com/threa...n-adiabatic-process-and-heat-capacity.796304/
See posts beginning with #15.

Chet

 

[barryj]

Thanks for the reply Chet. I will definitely go look at the links you show. Does it matter if the process is rapid expansion or slow expansion? In method 1, I assumed that you instantly released the piston and after the oscillations (if any) died out the net work would be the P(V2-V1). On the other hand, I was looking at something on Kahn academy website and the work was described as the integral from V1 to V2 of K/v dv where k was the initial and final pv (since the process is isothermal). Both methods seem logical to me so I thought I made some sort of silly error. I will keep studying. thanks

 

[barryj]

I am really confused now. I went to a website http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/isoth.html#c3
and found this applet. After entering my numbers I found that method 2 seems to be correct ?

 

[Chestermiller]

Thanks for the reply Chet. I will definitely go look at the links you show. Does it matter if the process is rapid expansion or slow expansion? In method 1, I assumed that you instantly released the piston and after the oscillations (if any) died out the net work would be the P(V2-V1). On the other hand, I was looking at something on Kahn academy website and the work was described as the integral from V1 to V2 of K/v dv where k was the initial and final pv (since the process is isothermal). Both methods seem logical to me so I thought I made some sort of silly error. I will keep studying. thanks

You can only use the second method if the expansion is reversible. You can't use the ideal gas law for the work unless the process is reversible. One of the characteristics of an irreversible expansion is that it takes place rapidly, and one of the characteristics of an reversible expansion is that it is carried out exceedingly slowly. The Khan academy is apparently doing the problem for a reversible expansion. My Insights article should answer all your questions. And the threads that I referenced show the practical application of all this.

Chet

 

[Chestermiller]

I am really confused now. I went to a website http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/isoth.html#c3
and found this applet. After entering my numbers I found that method 2 seems to be correct ?

As I said, for your irreversible expansion problem, method 2 is incorrect, and method 1 is correct. With all due respect to your applet, I stand by what I said. When you use an applet, you need to understand its inherent assumptions. Unfortunately, in this applet, they are not telling you that this only applies to a reversible expansion. With online applets, caveat emptor.

Chet

 

[barryj]

I still have issues with this problem. Consider method 1. Assuming no friction between the piston and the walls, if you instantaneously remove 500 N of weight, the pistin will oscillate about the final point forever. However, if you remove 500 N and then using your finger allow the piston to move upwards slowly, the work done by the piston will be the total of the change in potential energy of the weight AND the work done on the finger that allows the piston to move slowly. Allowing the piston to move slowly, and keeping the temperature the same, will cause a PV curve like in method 2 and the result will be 60J.

 

[Chestermiller]

I still have issues with this problem. Consider method 1. Assuming no friction between the piston and the walls, if you instantaneously remove 500 N of weight, the pistin will oscillate about the final point forever.

This is not a correct assessment. Even if the piston is totally frictionless, the oscillation will still be damped, and will decay to zero over time, leaving the piston stationary at its final point. From your training as an engineer, do you remember learning about viscosity? Viscous damping/dissipation in the gas will cause the piston to finally come to rest. The total cumulative work done by the gas on the piston will then be the same as you calculated for method 1.

I am currently working out the details of how to present all this in a Physics Forums Insight article, so that members can understand it better.

Chet

 

[barryj]

Consider the following. Let's say that the weight is 1500N of sand. Then you gradually start removing the sand slowly until you only have 1000N left. I say that the PV curve will be as I have shown in method 2 and this will integrate to 60J. So, I say the 10J difference is due to the sand that is slowly being removed to be partially elevated. So at the final point, 10 J of work was used to partially elevate the slowly removed sand and 50J is the result of the final movement. ?

 

[Chestermiller]

Consider the following. Let's say that the weight is 1500N of sand. Then you gradually start removing the sand slowly until you only have 1000N left. I say that the PV curve will be as I have shown in method 2 and this will integrate to 60J. So, I say the 10J difference is due to the sand that is slowly being removed to be partially elevated. So at the final point, 10 J of work was used to partially elevate the slowly removed sand and 50J is the result of the final movement. ?

Yes. This is what would be the case if the process were carried out reversibly. This is a perfect description of a reversible expansion of the gas, and the associated work done. As the sand is gradually removed at different elevations, the removed sand achieves higher potential energy than it did to begin with.

But, in the case where the 500 N is removed suddenly, the process is irreversible, and part of the work is lost as a result of viscous dissipation of some of the mechanical energy. There is no increase in the potential energy of the sand that was removed all at one time at the initial elevation.

Chet

 

[barryj]

This may be the difference in the two methods. I was assuming the weight was removed gradually. If you removed the weight suddenly,then there would be an oscillation, and the viscosity of the gas would cause dampening to the amount of 10J? The viscosity would tend to heat the gas but since it is a isothermal process by defination, the generated heat of 10J would be transferred to the environment, yes??

 

[Chestermiller]

This may be the difference in the two methods. I was assuming the weight was removed gradually. If you removed the weight suddenly,then there would be an oscillation, and the viscosity of the gas would cause dampening to the amount of 10J?

You're on the right track now. The viscous damping would generate heat in the gas, and this would reduce the amount of heat that the gas could receive from the temperature bath, by 10 J (and still maintain the same final temperature). This would reduce the amount of work done by 10 J. Along with the viscous stresses present and acting at the piston face, the total force per unit area on the piston would be different.

The viscosity would tend to heat the gas but since it is a isothermal process by defination, the generated heat of 10J would be transferred to the environment, yes??

Not exactly. It would reduce the amount of heat transferred from the environment, and this would reduce the amount of work done.

Chet