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There is some ideal gas in a container moving with some velocity on a smooth surface and you suddenly stop it( say by using your hands) , will the temperature of the gas increase? It seems to me that since you're suddenly stopping it there is no work done (because of no displacement ) so the kinetic energy must change to internal and hence temperature of the gas will increase. Is this right?

Now consider the same container but it is now moving on a rough surface . Assume there is a constant frictional force which stops it after travelling a certain distance. Now , will the temperature of the gas increase ? Now here I think since friction does work , the work will be equal to change in kinetic energy so the temperature will not increase . This seems strange. I think I'm not right.

What is the difference between these two cases ? (Or , is there a difference?) Will the temperature change be same in both these cases ? (If at all it takes place).

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    $\begingroup$ When the box is stopped it changes momentum. Let's say the momentum is $p$, then the change of momentum is equal to force times time $p = Ft$. The faster you stop the box, i.e. the smaller $t$ is, the greater the force $F$ will be. If you could stop the box in zero time (which you can't) the force $F \rightarrow \infty$. So your first and second cases are the same. They differ only in how large the force is and how long it is applied for. $\endgroup$ – John Rennie Jan 20 '15 at 10:48
  • $\begingroup$ @JohnRennie i see , so there is work done in the first case also? Still then , the work will be equal to change in kinetic energy so the temperature will not increase . This seems strange, because temperature does increase due to friction. $\endgroup$ – A Googler Jan 20 '15 at 10:53
  • $\begingroup$ To convince yourself that indeed the temperature of a fluid within a container will increase you can perform a simple real world experiment. Obtain two pint size water bottles and fill each half full of water at the same (room) temperature. measure and record the temperature. Take one bottle and shake it vigorously for 5 minutes. measure and record the temperature of both bottles. You will see that the bottle that was shaken will have increased its temperature over the bottle that remained at rest. Gas molecules are much harder to agitate in this same manner but nevertheless affected. (cont) $\endgroup$ – docscience Jan 20 '15 at 15:07
  • $\begingroup$ The temperature changes would just be much smaller and more difficult to measure. And I agree with John Rennie, in terms of work done on the fluid - it's the same. Just two different paths to get there. $\endgroup$ – docscience Jan 20 '15 at 15:10
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Hockey Puck

First lets take a look at a simpler example like a hockey puck (a rigid body). A hockey puck gliding along the ice that's suddenly stopped by a hockey stick looses its kinetic energy in the impact. In this case the energy turns into heat caused by the deformation and restitution of materials at the impact site.

If instead the hockey puck started sliding over a rough patch of ice it would loose its kinetic energy to friction, some of the heat produced would be transferred to the puck and some to the ice.

In both scenarios the same amount of energy was turned into heat however the temperature of the puck at the end would be different and very complicated to predict as it would depend on all of the surface properties of the puck the ice and the hockey stick. These complications would exist for a container carrying a gas as well, and would probably dominate the temperature change of the gas. If this is what you're interested in you would need to look into the tribology of the surfaces in question.

Simplify

If instead you're interested in how a container of gas would differ from a rigid body, let us simplify away those complications. If we claim that the container holding the gas is perfectly insulative then the heat generated from friction or the impact will stay on the outside of the container and we only have to consider heat generated within the gas itself. In this case the only way heat could be generated is due to the acceleration applied to the container, and the two cases can be modeled as a slow process (a small constant acceleration over a long duration) or a fast process (a strong acceleration over a short duration)

Slow

Now if a constant acceleration heated gasses we'd be in trouble, as gravity can be treated like a constant acceleration. So in the case of the slow constant acceleration, heat could only be generated at the beginning and end of the acceleration. At the beginning of the acceleration, the gas would redistribute in the container to create a pressure gradient $a\rho$ and then and the end the gas would redistribute to a neutral pressure gradient. At the beginning of the deceleration the pressure gradient would not be fully developed so the force required to sustain the acceleration would ramp up from zero to $F=ma$ once the pressure distribution had equilibrated. Then at the end, when the container came to a stop and the acceleration returned to zero, the pressure would still be unequal so keeping the container at zero acceleration would still require some force. These differences would balance exactly in terms of momentum. However due to the velocity being zero at the end, the extra force would not transfer any energy to the container and thus less overall energy would be transferred to the container. Of course in our example all of the energy transferred to the container is absorbed in either the impact or as friction.

Fast

If the process above is sped up, a larger and larger percentage of the energy is lost due to the starting and stopping effects being greater in magnitude. Eventually, the starting and stopping effects may even overlap as the pressure gradient doesn't have time to form before the acceleration is over. This can be taken to an extreme case where the pressure gradient doesn't even have a chance to form at all, before the deceleration is over. In this case nearly all of the energy would be converted to heat inside the gas. However, I should note that in such cases the impacting surface would have to be sticky as otherwise the gas would cause the container to bounce.

Really Fast

If the container decelerated from mach 2 instantaneously, you might expect a pretty different picture. A shock wave would be created at the forward wall and a vacuum at the rear. The shock wave would pass rearward through the container at Mach 0.65 eventually hitting the vacuum (whose edge would have been moving forward at slightly less than mach 2) and once the shock wave hit the far edge then the system would balance everything out with sounds waves. This would actually be pretty much the same as the previous scenario except if rebounding was allowed in which case this would have a much lower coefficient of restitution.

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i say there is no difference between the two cases.

In both cases, when the container has some acceleration w.r.to ground, psuedo-force is applied on the gas particles relative to the container. as gravity shows no effect on the temerature, even this horizontal psuedo-force also shows no effect.

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I think the First Law of Thermodynamics is applicable here:

$Q = \delta{U} + W$

Consider all the gas particles collectively in a free body diagram. The deceleration would have the effect of increasing the pressure on the gas particles much as you feel when you are accelerating.

Since no heat was added to the system by say radiation for example, we have:

$ -\delta{U} = W$

Where U is the internal energy of the gas. The gas was compressed, therefore work was done on the gas and its internal energy increased as manifested by an increase in temperature as that pressure was applied. There will be a difference in the temperatures between the two cases because the decelerations and therefore applied pressures will be different.

The full-blown analysis would be more complicated.

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Stopping a bowl of water or a container of gas (think of the analogy)?

The situation you described is very analogous to stopping a bowl of water either suddenly or very gradually. Quite obviously the sudden stop of water will generate lots of waves/plops, that will eventually dissipate to thermal energy, while stopping it slowly doesn't.

So where did the difference come from? In the first situation, if the box is stopped really suddenly ($\Delta t \ll c_sl$, where $\Delta t$ is the time it takes to stop, $c_s$ is the speed of sound in the gas and $l$ is the length of the box in the direction of movement), the process is not anymore adiabatic, and the entropy of gas increases. This can be seen as follows.

Stopping the container suddenly makes you only absorb the kinetic energy of the box, but not the gas (in case of water this is not entirely true, as water is incompressible, as opposed to a gas). This happens, because the waves generated by the rapid acceleration of the walls will not have enough time to travel and most of the gas retains it's momentum just after the impact. Thus the kinetic energy absorbed by your hand (must be a very strong one though) is $m_{cont}v^2/2$, not $(m_{cont} + m_{gas})v^2/2$, which leaves $m_{gas}v^2/2$ into the gas. If time approximately $c_sl$ passes, the waves generated will reflect and the momentum is passed to the container. However, no work can be done anymore, as the velocity of the container is already $0$.

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