Is this a simple energy conservation problem?

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

The answer to this problem can be obtained by equating $\frac{1}{2}mv^2 = ΔU$ .

But I am not sure why this is to be done . In fact I think this is not quite right .

1) The Kinetic energy $\frac{1}{2}mv^2$ is the energy due to bulk motion whereas ΔU relates the microscopic motion .Can we really equate the two ?

2) Is this an application of 1st law of thermodynamics . I do not think so .

3) Is this a simple energy conservation problem . Again I doubt that is the case .

Any help is appreciated .

Thanks

 

[Titan97]

Why isn't it a simple energy conservation?

Initial energy is $U_1+\frac{1}{2}mv^2$
Final energy is $U_2$

 

[Vibhor]

Hi ,

What is U1 and U2 ?

 

[Titan97]

Initial and final internal energy.

 

[Vibhor]

Ok .

And what is the mechanism by which this energy is transferred from macroscopic kinetic energy to microscopic internal energy , since work done on the gas is zero ?

Do you agree work done on the gas is zero ?

 

[ehild]

Ok .

And what is the mechanism by which this energy is transferred from macroscopic kinetic energy to microscopic internal energy , since work done on the gas is zero ?

Do you agree work done on the gas is zero ?

The gas does not gain energy. Initially, all molecules had a velocity component equal to the velocity of the vessel and a random component corresponding to the velocity distribution of the initial thermodynamic state of the gas. When the vessel stops, that regular component is randomized through collisions between the molecules. The internal energy is the energy of random motions. It is increased when the macroscopic KE, the KE of the regular motion, transforms to the KE of the random motions of the molecules.

 

[Vibhor]

Hello ,

Initially, all molecules had a velocity component equal to the velocity of the vessel and a random component corresponding to the velocity distribution of the initial thermodynamic state of the gas. When the vessel stops, that regular component is randomized through collisions between the molecules. The internal energy is the energy of random motions. It is increased when the macroscopic KE, the KE of the regular motion, transforms to the KE of the random motions of the molecules.

An increase in internal energy means an increase in temperature of the gas . But I have studied that the bulk motion of the vessel does not affect the random motion of the gas . Irrespective of whether the vessel accelerates /decelerates/moves with uniform speed , the temperature does not change . But this is not the case here .An increase in internal energy means an increase in the temperature .

I am seeing a contradiction in the concepts .

 

[ehild]

Hello ,
An increase in internal energy means an increase in temperature of the gas . But I have studied that the bulk motion of the vessel does not affect the random motion of the gas . Irrespective of whether the vessel accelerates /decelerates/moves with uniform speed , the temperature does not change . But this is not the case here .An increase in internal energy means an increase in the temperature .

I am seeing a contradiction in the concepts .

Initially, the average velocity of the molecules was equal to Vm, the velocity of the vessel, as the random component of the velocities (Vr) summed up to zero. When colliding with the wall, only the random component of the molecule changed. Assume one dimensional motion: after the vessel stopped, a molecule colliding with a wall with velocity Vm+Vr will rebounce with velocity -(vm+vr). At the end there will be equal amount of molecules moving with -(vm+vr) as those moving with vm+vr so the average velocity becomes zero. And the velocities randomize further by the collisions between molecules, and even other degrees of freedom (mainly rotational) are excited during the collisions. At the end all initial macroscopic energy transforms into the energy of random motion.

 

[Vibhor]

Ok .

So you are suggesting that accelerating/decelerating a gas container increases the temperature of gas . Right ??

 

[ehild]

Ok .

So you are suggesting that accelerating/decelerating a gas container increases the temperature of gas . Right ??

Not quite. You can speak of temperature of the gas when it is in steady state. Till the molecules interact with an accelerating wall, the gas is not in a state which is called a "state" by Thermodynamics.

 

[Vibhor]

Not quite. You can speak of temperature of the gas when it is in steady state. Till the molecules interact with an accelerating wall, the gas is not in a state which is called a "state" by Thermodynamics.

But after the gas container stops accelerating/decelerating temperature of gas does increase . Right ??

Isn't this exactly what is happening in the original problem (the temperature increases by ΔT when the vessel stops) ??

 

[Chestermiller]

But after the gas container stops accelerating/decelerating temperature of gas does increase . Right ??

Isn't this exactly what is happening in the original problem (the temperature increases by ΔT when the vessel stops) ??

For this closed system, the 1st law of thermodynamics tells us that:
$$\Delta U+\Delta(KE)+\Delta(PE)=Q-W$$
For this particular problem, the equation reduces to:$$\Delta U+\Delta(KE)=0$$
But, mechanistically, how can the kinetic energy of the gas get converted to internal energy. The gas within the container has inertia, and when the container is brought to a stop, a flow and compression/expansion action takes place within the gas. This dynamic movement and deformation of the gas is damped out by viscous stresses. In the end, all the non-random kinetic energy KE is converted to random kinetic energy U (i.e., internal energy) by viscous dissipation.

 

[Vibhor]

For this closed system, the 1st law of thermodynamics tells us that:
$$\Delta U+\Delta(KE)+\Delta(PE)=Q-W$$
For this particular problem, the equation reduces to:$$\Delta U+\Delta(KE)=0$$
But, mechanistically, how can the kinetic energy of the gas get converted to internal energy. The gas within the container has inertia, and when the container is brought to a stop, a flow and compression/expansion action takes place within the gas. This dynamic movement and deformation of the gas is damped out by viscous stresses. In the end, all the non-random kinetic energy KE is converted to random kinetic energy U (i.e., internal energy) by viscous dissipation.

Would you agree that irrespective of whether the vessel(thermally insulated) is stopped suddenly or gradually ( decelerated ) , the temperature of gas would increase ??

 

[Chestermiller]

Would you agree that irrespective of whether the vessel(thermally insulated) is stopped suddenly or gradually ( decelerated ) , the temperature of gas would increase ??

No.

So what are your thoughts on why I answer this question with a "no?"

 

[Vibhor]

No.

So what are your thoughts on why I answer this question with a "no?"

I think if gas container is stopped gradually ,the container would be doing work on the external agent ( whatever it may be ) . So by doing work , the non random translational kinetic energy of the gas transfers to the external agent , leaving the random kinetic energy unchanged . Hence temperature of gas remains same as before .

Makes sense ?

 

[Chestermiller]

I think if gas container is stopped gradually ,the container would be doing work on the external agent ( whatever it may be ) . So by doing work , the non random translational kinetic energy of the gas transfers to the external agent , leaving the random kinetic energy unchanged . Hence temperature of gas remains same as before .

Makes sense ?

Extremely well-analyzed!

So, in this case, even though the volume of the gas does not change, work is still done on the gas by its surroundings (the container) by unbalanced forces applied to the gas through a displacement. This is combined with the fact that there are negligible viscous stresses and viscous dissipation, since these depend on the rate at which the gas is deforming. In this case, the 1st law energy balance approaches:
$$\Delta (KE)=-W$$
which is basically the work energy theorem. The gas in this case behaves analogous to a rigid body.

 

[Vibhor]

So, in this case, even though the volume of the gas does not change, work is still done on the gas by its surroundings (the container) by unbalanced forces applied to the gas through a displacement.

I am having a little trouble understanding the work term"W" in the 1st law of thermodynamics as applied in setup where we are gradually stopping the vessel .

Is "pdV work " (due to volume change) and "bulk work" ( work done by the force applied by the container ) two different types of work we need to sum in the "W" term while applying the 1st law of thermodynamics ?

 

[Chestermiller]

I am having a little trouble understanding the work term"W" in the 1st law of thermodynamics as applied in setup where we are gradually stopping the vessel .

Is "pdV work " (due to volume change) and "bulk work" ( work done by the force applied by the container ) two different types of work we need to sum in the "W" term while applying the 1st law of thermodynamics ?

Basically, yes. But recognize that both of these involve a force at the boundary of the gas integrated over a displacement.

 

[Vibhor]

Ok .

Suppose the vessel is initially at rest . Now it is accelerated till it reaches a speed 'v' .

Would you agree that same reasoning applies here as well i.e when the vessel is suddenly moved ,the temperature of gas rises whereas if it is accelerated gradually the temperature remains unchanged ??

 

[Chestermiller]

Ok .

Suppose the vessel is initially at rest . Now it is accelerated till it reaches a speed 'v' .

Would you agree that same reasoning applies here as well i.e when the vessel is suddenly moved ,the temperature of gas rises whereas if it is accelerated gradually the temperature remains unchanged ??

Yes.

 

[Chestermiller]

I've thought about this situation some more, and I'm afraid I'm going to have to take back what I said in posts running from #12 to #20. Your questions about gradual acceleration and deceleration prompted me to think more deeply into this problem, particularly post #19, and allowed me to gain better perspective.

Here's what I arrived at.

If I very gradually decelerate the container with gas in it, then the gas will not be deforming rapidly and viscous effects will be negligible. The container will exert a greater force on the leading edge of the gas than on the trailing edge, and the net work of these forces will result in the decrease in kinetic energy. There will be no dissipation of mechanical energy, and the internal energy of the gas will not change. It will be the same as if the gas were a rigid body.

Now, if I decelerate the container more rapidly, the kinetic energy change will be the same, but I will have to do more net work on the gas because of the viscous stresses that are now becoming more important. So, there will be an increase in the internal energy of the gas equal to the additional work needed to overcome viscous stresses (which is basically equal to the new larger net amount of work minus the change in kinetic energy). During the deceleration, the net force on the gas will be larger, but the distance will be smaller. However, the larger force will win out, and the new amount of work will exceed the integral of the force times distance present in the gradual acceleration case. Also, in this more rapid deceleration case, if the the deceleration only increases modestly, the change in internal energy will still be relatively small, and not anything near that approaching the change in kinetic energy. Without doing a detailed gas dynamics analysis of the gas in the container, it would not be possible to quantify the increase in internal energy.

Finally, if we go to the limit of a very rapid deceleration (virtually instantaneous), the same situation would prevail, but to a greater extreme. The net amount of work to stop the gas would be even higher because of viscous stresses. The net work would be higher than the change in kinetic energy, but by an unknown amount. And this unknown amount would determine the increase in internal energy. So, in a very rapid stop, the increase in internal energy is not going to be equal to the change in kinetic energy. It will probably be less.

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.

Sorry for any confusion I may have caused.

Chet

 

[Vibhor]

I've thought about this situation some more, and I'm afraid I'm going to have to take back what I said in posts running from #12 to #20

Here's what I arrived at.

...and the net work of these forces will result in the increase in kinetic energy.

Increase or decrease ??

By the way you are referring to non random(macroscopic) kinetic energy due to bulk motion. Right ?

 

[Vibhor]

Now, if I decelerate the container more rapidly, the kinetic energy change will be the same, but I will have to do more net work on the gas because of the viscous stresses that are now becoming more important. So, there will be an increase in the internal energy of the gas equal to the additional work needed to overcome viscous stresses (which is basically equal to the new larger net amount of work minus the change in kinetic energy). During the deceleration, the net force on the gas will be larger, but the distance will be smaller. However, the larger force will win out, and the new amount of work will exceed the integral of the force times distance present in the gradual acceleration case. Also, in this more rapid deceleration case, if the the deceleration only increases modestly, the change in internal energy will still be relatively small, and not anything near that approaching the change in kinetic energy.

Could you please explain this using mathematical notations just like you did in post#12 .

 

[Vibhor]

Finally, if we go to the limit of a very rapid deceleration (virtually instantaneous), the same situation would prevail, but to a greater extreme. The net amount of work to stop the gas would be even higher because of viscous stresses. The net work would be higher than the change in kinetic energy, but by an unknown amount. And this unknown amount would determine the increase in internal energy. So, in a very rapid stop, the increase in internal energy is not going to be equal to the change in kinetic energy. It will probably be less.

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 .

 

[Chestermiller]

Increase or decrease ??

Decrease. I went back and changed it.

By the way you are referring to non random(macroscopic) kinetic energy due to bulk motion. Right ?

Yes.

 

[Chestermiller]

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]

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.

 

[Vibhor]

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 .

 

[Vibhor]

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 ??

Please share your views .

Many Thanks

 

[Chestermiller]

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 ??

Please share your views .

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