# Is this a simple energy conservation problem?

Chestermiller
Mentor
Could you please explain this using mathematical notations just like you did in post#12 .
FOR DECELERATION
$$\Delta U + \Delta (KE)=-W$$where W is the positive work that the gas does on its surroundings (the container) and $\Delta (KE)$ is the change in kinetic energy (negative). So,
$$\Delta U=-(\Delta (KE)+W)\geq0$$with the = sign applying to very gradual deceleration and the $\geq$ sign applying to very rapid deceleration. Alternate and equivalently, $$W\leq-\Delta(KE)$$This equation says that, because of viscous dissipation of mechanical energy, the amount of work that the gas does on its surroundings (the container) is less than its decrease in kinetic energy. (Not all the kinetic energy can be converted to work).

FOR ACCELERATION
$$\Delta U + \Delta (KE)=-W$$
where, in this case, W is negative, since the surroundings are doing work on the system, and $\Delta (KE)$ is positive. So,
$$\Delta U=-(\Delta (KE)+W)\geq0$$with the = sign applying to very gradual acceleration and the $\geq$ sign applying to very rapid acceleration. Alternate and equivalently, $$-W\geq\Delta (KE)$$This equation says that, because of viscous dissipation of mechanical energy, the amount of work that the surroundings needs to do to accelerate the gas (-W) is greater than its increase in kinetic energy. (Additional work is required to increase the kinetic energy).

So we see that, both in rapid acceleration and rapid deceleration, the internal energy increases (actually, as we shall see shortly, by exactly the same amount). If we were to use the same rationale that we had used in the original analysis of the problem to determine the change in internal energy for acceleration, we would have written that $\Delta U=-\Delta (KE)$ and we would have concluded that, for acceleration ($\Delta (KE)>0$), $\Delta U$ is negative.

Actually, we will now show that for the same magnitude of the change in kinetic energy, the change in internal energy is exactly the same for extremely rapid acceleration as for extremely rapid deceleration. We know from Galileo that the laws of physics do not change when a system is observed from two different inertial frames of reference. So, consider deceleration as observed from the following two inertial frames of reference:

Frame 1. The laboratory frame that has already been used to analyze the deceleration problem (where the observer is stationary on the laboratory floor)

Frame 2. An inertial frame in which the observer is moving with the original velocity of the container (in the positive x direction) forever (relative to to the laboratory frame).

As reckoned by an observer in inertial frame 2, the container is standing still, and the barrier that is going to stop the motion of the container when it hits appears to be moving toward the container in the negative x direction with velocity v. Once the container hits the barrier, both the container and the barrier are moving with the velocity v in the negative x direction. So, as far as the observer in frame 2 is concerned, the container and gas start out standing still, and, after the barrier hits them, they are instantly accelerated in the negative x direction to a velocity of magnitude v. So, as reckoned from frame 2, the container has experienced an acceleration exactly equal to the deceleration reckoned by an observer in the laboratory frame (frame 1). This shows that the change in internal energy for the acceleration and deceleration situations must be exactly the same for the same magnitude of the change in kinetic energy. The two situations are indistinguishable physically (since neither inertial observer can prove that it is his frame that is stationary and it is the other frame that is moving).

Chestermiller
Mentor
Well , then how do we solve the original numerical problem in the OP ?

$\Delta U+\Delta(KE)=0$ was the basis on which the problem was to be solved .
Well we now know that this was not correct. Otherwise, it could not predict a different result for acceleration than for deceleration. You're the one who asked about acceleration, which led to this realization. You should be proud of yourself. I must admit that I was unaware of all this.

As I mentioned in post # 21, for some problems, one can not obtain the desired solution without analyzing the transient details of the flows and deformations occurring within the system. (Not every problem can be solved with equilibrium thermodynamics). To attack this problem, I would use gas dynamics to solve the problem for various imposed constant decelerations of the container from very low values to very large values. But it's not a simple problem.

The gas does not gain energy.
If we think of this problem as a simple introductory physics problem and apply energy conservation $U_1+\frac{1}{2}mv^2 = U_2$ ,then by the same logic if a stationary gas vessel is suddenly moved ,then its internal energy should decrease ( temperature falls) .

Please pardon me if I have misunderstood you . Kindly share your thoughts on this problem .

This problem has appeared in our national level engineering entrance exam and didn't evoke much doubts/reactions from students/teachers . Solutions available everywhere use energy conservation as mentioned in the OP .

This makes me wonder if I am missing some simplified assumptions .

As I mentioned in the previous post If we think of this problem as a simple introductory physics problem and apply energy conservation ,then by the same logic if a stationary gas vessel is suddenly moved ,then its internal energy should decrease .

@ehild , @TSny , @haruspex , @Chestermiller . Are all my favourite members on the same page that the original problem cannot be solved ??

Many Thanks

Chestermiller
Mentor
This problem has appeared in our national level engineering entrance exam and didn't evoke much doubts/reactions from students/teachers . Solutions available everywhere use energy conservation as mentioned in the OP .

This makes me wonder if I am missing some simplified assumptions .

As I mentioned in the previous post If we think of this problem as a simple introductory physics problem and apply energy conservation ,then by the same logic if a stationary gas vessel is suddenly moved ,then its internal energy should decrease .

@ehild , @TSny , @haruspex , @Chestermiller . Are all my favourite members on the same page that the original problem cannot be solved ??

Many Thanks
My analysis is still screwed up. I'm going to start over from scratch tomorrow using a model that I have developed and which I hereby guarantee will resolve all the issues and uncertainties. Please bear with me. It's bedtime here now (11 pm). I'll be back tomorrow morning.

Chet

ehild
Homework Helper
Thermodynamics that you study at high school really means "Thermostatics". All concepts refer to a gas in equilibrium with itself and with its surrounding. If the gas is not in equilibrium, you can not speak about pressure or about internal energy. The internal energy is defined as the sum of energies of all molecules, and the molecules are supposed to move randomly, so as the average of their translational velocities is zero, also the average of the angular momenta is zero.

You have a vessel filled with gas and the system is moving with velocity V. That means the average velocity of the gas molecules is also V.
There is an external force that stops the vessel. You can suppose that that force acts on the vessel, not on the gas molecules. For example, the vessel collides with a wall and sticks to it. The gas molecules interact with the walls of the vessel when they collide with them. It was said that the vessel stopped suddenly, in time shorter than the average collision time. The external force did some work, but you should not worry about that work. You should consider the result only that the vessel stopped in such a short time that the velocity distribution of the molecules did not change.
Such thing can happen, you certainly made experiment with row egg and hard-boiled egg. You can know which egg is row and which is boiled, if you spin them and suddenly stop the spinning eggs. When you release them, the boiled egg does not move, but the row one starts to spin a bit, as the inside kept spinning when you stopped the shell.
Imagine that you have a ball in a train and the train suddenly stops. The ball keeps its original velocity, rolls on the floor and collides with the front wall of the compartment and bounces back. Assuming ideally elastic collision, it moves back with the same speed and rebounds from the opposite wall. In very long time, the time average of the ball becomes zero. Imagine you have a lot of balls - they do the same. But they collide in different times. After long time, half of them will move forward, the other half backward. If the collision with the wall is ideally elastic, the energy does not change. Only the direction of motion of the balls became random. The KE of a ball does not depend on the direction of motion. It is the same when all balls move with the same speed - in any direction.

When the vessel changes velocity gradually, so there are lots of collisions with the wall during the acceleration/deceleration, the gas is always in equilibrium with itself. As Chest said, it behaves as a rigid body, and the force applied by the wall changes the bulk motion, not the random one.
In general case, the velocity distribution of the molecules will differ from the equilibrium distribution during the acceleration process. The average velocity differs from zero. The gas is not in equilibrium. But all ordered motions cease with time and the energy transforms to heat.

haruspex
Homework Helper
Gold Member
For the case of very gradual acceleration, we will have the analogous situation. The net work will be virtually equal to the increase in kinetic energy, and the change in internal energy will be essentially zero. Then, as we go to a more rapid acceleration, we will have to do more work to overcome viscous stresses, but the increase in internal energy will be determined by the net work minus the increase in kinetic energy. This quantity would be unknown without a gas dynamics analysis. Finally, at very rapid (virtually instantaneous) acceleration, we would experience the same situation. The net work would be still higher than the change in kinetic energy, and the internal energy would increase.

The reason I was driven to these conclusions was that, if one applies the same approach to the acceleration that we had previously used for the deceleration, we would determine that, for an instantaneous acceleration, since the kinetic energy increases, the internal energy must decrease. This of course does not make any sense. The internal energy must increase in both cases because of viscous dissipation.
For the sudden case, I believe your earlier analysis was better.
We can imagine that the transition from a stationary container to one moving at a steady velocity, or vice versa, is so sudden that no gas molecules collide with it in the interval. Thus, either way, the gas simply finds itself having a bulk velocity relative to the container. If we take the container walls as completely rigid and far more massive than the gas, and a perfect insulator, no energy is transferred to the container.

As I mentioned in the previous post If we think of this problem as a simple introductory physics problem and apply energy conservation ,then by the same logic if a stationary gas vessel is suddenly moved ,then its internal energy should decrease .
This I don't understand. If given a sudden acceleration from rest, the wall of the vessel (this time the "rear" wall) still impacts the gas molecules and therefore imparts energy to them. I think about it like this: If we have a perfectly insulated vessel of gas and we shake it vigorously for some time, we would expect the gas to heat up, due ultimately to viscous dissipation, right? But isn't shaking the vessel equivalent to a series of accelerations and decelerations? If your reasoning were correct, the cooling due to each acceleration would balance the heating due to each deceleration during the shaking and the gas would remain at the same temperature overall. Does that make sense?

I tend to think this is a deceptively tricky first law problem.

Chestermiller
Mentor
This I don't understand. If given a sudden acceleration from rest, the wall of the vessel (this time the "rear" wall) still impacts the gas molecules and therefore imparts energy to them. I think about it like this: If we have a perfectly insulated vessel of gas and we shake it vigorously for some time, we would expect the gas to heat up, due ultimately to viscous dissipation, right? But isn't shaking the vessel equivalent to a series of accelerations and decelerations? If your reasoning were correct, the cooling due to each acceleration would balance the heating due to each deceleration during the shaking and the gas would remain at the same temperature overall. Does that make sense?

I tend to think this is a deceptively tricky first law problem.
Yes. One of the things I neglected was that, when the container is suddenly started from rest (acceleration case), the walls of the container are moving, and thus do work on the gas. I will be introducing my model that will help us quantitatively resolve all our issues and uncertainties about this in my next post.

Chet

Chestermiller
Mentor

Consider the mechanical system shown in the Figure. This model system exhibits all the qualitative features of the actual gas behavior inside the container, and can be considered a surrogate for our system. It will help us to get a quantitative understanding of what is happening in our system when the container is suddenly stopped (deceleration) or is suddenly started (acceleration case).

The mass M in the figure is a surrogate for the mass of the gas, and can slosh back and forth within the container (as the actual gas does). The springs (with combined spring constant k) are a surrogate for the non-uniform compressive/expansion behavior of the gas; when one of the springs is compressed, the other is extended; this is a feature of how the gas behaves in the container, with non-uniform expansion and compression. The dampers (with combined damper constant C) are a surrogate for the viscous behavior of the gas; these cause the irreversible conversion of mechanical energy to internal energy U; The total energy dissipated in compressing and expanding the damper can be taken as a surrogate for the change in internal energy of the gas.

@Vibhor: Using Newton's laws, I intend to analyze the mechanical response of this system when the container is suddenly stopped and/or started. (In fact I have already done this, but I just wanted to be sure you were comfortable with the model). What the analysis has shown is that

1. The original analysis of the problem for a sudden stop was correct
2. For the case in which the container is suddenly started, the change in internal energy is exactly the same as for the case in which the container is suddenly stopped.
3. For the case in which the container is suddenly started, the work done by the container on its contents is exactly 2x the increase in kinetic energy, so that half the work goes into accelerating the mass and half is dissipated to produce an increase in internal energy. This is why the increase in internal energy is the same for a sudden start as for a sudden stop.

So, if are you comfortable with the model, I will proceed with the analysis.

I might mention that I have used this same spring-damper analog approach in a Physics Forums Insights article to analyze the fundamental difference between reversible and irreversible expansions and compressions of gases: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

Chet

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Chestermiller
Mentor
Actually, we don't even need the model I described in post #35 to get the results the we desire for the case of the container imposing a sudden velocity on the gas. At any time after the container is started in motion with velocity V, the momentum of the gas in the container is $$\int{\rho v d\phi}$$where $\rho$ is the density of the gas at a given location in the container, v is the local velocity of the gas at that location, and $d\phi$ is the differential volume. The integral is taken over the volume of the container. A momentum balance on the gas in the container then yields:
$$\frac{d[\int{\rho v d\phi}]}{dt}=F(t)$$
where F(t) is the net force that the container is exerting on the gas at time t. If we multiply this momentum balance equation by the container velocity V and integrate between time = 0 and infinite time (the final equilibrium state), we obtain:
$$V\left[\int{\rho v d\phi}\right]_{t=\infty}-V\left[\int{\rho v d\phi}\right]_{t=0}=\int_0^{\infty}{F(t)Vdt}$$
Since V is constant, the term on the right hand side of this equation can be recognized as the total work done by the container on the gas W*. The term $V\left[\int{\rho v d\phi}\right]_{t=0}$ on the left hand side is equal to zero, because the velocity of the gas is zero at time zero. Since at infinite time, the velocity v of the gas is uniform and equal to that of the container, the term $V\left[\int{\rho v d\phi}\right]_{t=\infty}$ on the right hand side of the equation becomes:
$$V\left[\int{\rho v d\phi}\right]_{t=\infty}=V^2\left[\int{\rho d\phi}\right]_{t=\infty}=MV^2$$
where M is just the total mass of the gas in the container. Therefore, we have:
$$W^*=MV^2$$
This shows that the work done by the container on the gas is exactly equal to twice the final kinetic energy of the gas. If we combine this with the 1st law of thermodynamics applied to the gas in the container, we obtain:
$$\Delta U+\frac{1}{2}MV^2=MV^2$$or
$$\Delta U=\frac{1}{2}MV^2$$
This is identical to the change in internal energy for suddenly stopping the container.

So the three conclusions I listed in post # 35 apply (not just to the surrogate mechanical model in post \$35 but) to the case of an actual gas:

1. The original analysis of the problem for a sudden stop was correct
2. For the case in which the container is suddenly started, the change in internal energy is exactly the same as for the case in which the container is suddenly stopped.
3. For the case in which the container is suddenly started, the work done by the container on its contents is exactly 2x the increase in kinetic energy, so that half the work goes into accelerating the mass and half is dissipated to produce an increase in internal energy. This is why the increase in internal energy is the same for a sudden start as for a sudden stop.

Thank you very much for sparing your invaluable time in doing the analysis and writing the above detailed posts .

What the analysis has shown is that

1. The original analysis of the problem for a sudden stop was correct
You mean $W=0$ and $\Delta U = -\Delta(KE)$ ??

Chestermiller
Mentor
Thank you very much for sparing your invaluable time in doing the analysis and writing the above detailed posts .

You mean $W=0$ and $\Delta U = -\Delta(KE)$ ??
Yes.

How is a sudden stop different from a sudden start in terms of work done , such that $W=0$ in the former case and $W≠0$ in the latter ??

Intuitively ,it should be same (atleast it is in case of a rigid body ) .

Pardon me if I am being stupid .

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Chestermiller
Mentor
How is a sudden stop different from a sudden start in terms of work done , such that $W=0$ in the former case and $W≠0$ in the latter ??

Intuitively ,it should be same (atleast it is in case of a rigid body ) .

Pardon me if I am being stupid .
You're not being stupid at all. The stupid one around here has been me.

In the case of a sudden start, there is displacement of the boundary during the time that the boundary force is being applied. It's the same as if the gas were inside a horizontal cylinder, and there were pistons on either end moving forward. There is no change in volume, but there are (unequal) forces being applied by the pistons as they are advancing. So there is a net force being applied to the gas through a displacement.

In the case of a sudden start, there is displacement of the boundary during the time that the boundary force is being applied. It's the same as if the gas were inside a horizontal cylinder, and there were pistons on either end moving forward. There is no change in volume, but there are (unequal) forces being applied by the pistons as they are advancing. So there is a net force being applied to the gas through a displacement.
But this is exactly what is happening when the vessel is stopped suddenly ( in reverse manner).

How are container walls not exerting force in the first case while they exert force ( and hence do work ) in the second case ??

Chestermiller
Mentor
But this is exactly what is happening when the vessel is stopped suddenly ( in reverse manner).

How are container walls not exerting force in the first case while they exert force ( and hence do work ) in the second case ??
When the container is suddenly stopped, the container walls do exert a force on the gas, but, during that time, the container walls are not moving, so there is no displacement and no work is being done.

When the container is started suddenly, the container walls also exert the a force on the gas, but, during that time, the container walls are moving, so there is a displacement and work is being done.

Look at the figure I presented in post #35. For the case of a sudden stop, both the container and the spring-mass-damper assembly are moving with a velocity V initially, but, at time zero, the container velocity suddenly drops to zero. However, the mass keeps moving, and the springs and dampers exert forces on the leading and trailing faces of the container. But these faces are no longer moving, so no work is done at the container faces by the spring-mass-damper assembly. What happens is that the mass just experiences a damped harmonic oscillation until it finally stops moving.

Ok . Thanks a lot .

What happens to the gas temperature when there is a gradual acceleration or deceleration of the gas vessel (thermally insulated) ? Is there a definite answer or does it require some additional information/details ?

Chestermiller
Mentor
Ok . Thanks a lot .

What happens to the gas temperature when there is a gradual acceleration or deceleration of the gas vessel (thermally insulated) ? Is there a definite answer or does it require some additional information/details ?
For very gradual acceleration or deceleration of the container, the internal energy does not change (no viscous dissipation) and the temperature does not change.

If you are interested in exploring the effect of the acceleration rate in more detail quantitatively, we can use the model I described in post #35. Any interest?

For very gradual acceleration or deceleration of the container, the internal energy does not change (no viscous dissipation) and the temperature does not change.
Do you believe that post#15 and #16 are correct and is the valid reasoning for the gradual acceleration/deceleration case ??

Chestermiller
Mentor
Do you believe that post#15 and #16 are correct and is the valid reasoning for the gradual acceleration/deceleration case ??
Yes, #16 a little more than #15. The focus should be on the gas, and part of the work done on the container by the outside agent is transferred to the gas.

Several responses have likened the gradual acceleration/deceleration of the gas to the acceleration/deceleration of a rigid body (since viscous stresses are going to be negligible in the gas). For the acceleration of a rigid body, the work-energy theorem applies.