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Is this a simple energy conservation problem?

  • Thread starter Vibhor
  • Start date
  • #1
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Homework Statement



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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
 

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  • #2
Titan97
Gold Member
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Why isn't it a simple energy conservation?

Initial energy is ##U_1+\frac{1}{2}mv^2##
Final energy is ##U_2##
 
  • #3
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Hi ,

What is U1 and U2 ?
 
  • #4
Titan97
Gold Member
451
18
Initial and final internal energy.
 
  • #5
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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 ?
 
  • #6
ehild
Homework Helper
15,394
1,802
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.
 
  • #7
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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 .
 
  • #8
ehild
Homework Helper
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1,802
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.
 
  • #9
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Ok .

So you are suggesting that accelerating/decelerating a gas container increases the temperature of gas . Right ??
 
  • #10
ehild
Homework Helper
15,394
1,802
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.
 
  • #11
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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) ??
 
  • #12
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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.
 
  • #13
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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 ??
 
  • #14
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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?"
 
  • #15
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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 ???
 
  • #16
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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.
 
  • #17
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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 ?
 
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  • #18
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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.
 
  • #19
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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 ??
 
  • #20
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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.
 
  • #21
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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
 
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  • #22
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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
:oldcry:
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 ?
 
  • #23
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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 .
 
  • #24
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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 :rolleyes:.
 
  • #25
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:oldcry:


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.
 

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