Change in temperature when a gas is in a moving container

[Krushnaraj Pandya]

Homework Statement

Some gas at 300K is enclosed in a container. Now the container is placed on a fast moving train.is the change in temperature of the gas the same as the change dT observed when train suddenly stops?

Homework Equations

1/2mv^2=dU?? or dU=0??

The Attempt at a Solution

intuitively, there is no change in relative velocity of molecules anyhwere so dT=0 everywhere. But according to my high school textbook, when train suddenly stops- ordered kinetic energy changes to disordered kinetic energy therefore raising internal energy by 1/2mv^2

 

[haruspex]

intuitively, there is no change in relative velocity of molecules anyhwere so dT=0 everywhere

Consider what happens to the molecules when the container moves relative to them. They will hit one side, on average, faster than against the other. The rebounds will be faster, then they will strike other molecules, and pretty soon the extra KE will be in every direction.

 

[Krushnaraj Pandya]

So the temperature and internal energy changes when the train moves with a velocity or suddenly stops?? What's got me confused is that in a lot of places including the book 'Concepts of Physics by HC Verma' it says that since it is bulk motion of gas, there is no change in internal energy and therefore temperature change is zero. I read a similar conclusion on a previous forum answered by 'phinds' and 'haruspex' (I can't find the link right now). so which one to consider? Also, if the starting or stopping of train was gradual-would the temperature change be negligible as the inertial forces would be very small??

 

[haruspex]

So the temperature and internal energy changes when the train moves with a velocity or suddenly stops?? What's got me confused is that in a lot of places including the book 'Concepts of Physics by HC Verma' it says that since it is bulk motion of gas, there is no change in internal energy and therefore temperature change is zero. I read a similar conclusion on a previous forum answered by 'phinds' and 'haruspex' (I can't find the link right now). so which one to consider? Also, if the starting or stopping of train was gradual-would the temperature change be negligible as the inertial forces would be very small??

I suspect it depends on how sudden the change in speed is. If a moving box is gradually brought to rest then the gas exerts a greater pressure against the leading face than against the trailing, so does work on the box.
In the case of a sudden stop from velocity v it is clear that ½mv² gets added to the internal energy, where m is the mass of the gas.
We can run this in reverse, e.g. by using a frame of reference moving at velocity v. In this frame, the box is suddenly jerked from rest to moving at velocity v. The internal energy should be the same in both frames, therefore accelerating the box to velocity v added ½mv² to the internal energy and another ½mv² to the bulk KE. Thus, it takes more work to accelerate a box of gas than to accelerate the same mass of a solid.
This should not surprise. It is akin to an inelastic collision. We can simplify it to a single molecule bouncing back and forth parallel to the added velocity. If the box has mass M and the molecule mass m and we set the box moving at speed v then by conservation of momentum the bulk speed attained by the system is v' where (M+m)v'=Mv. The new KE is less than the original KE (½Mv²) by ##\frac{Mmv^2}{2(M+m)}##. For M>>m this reduces to ½mv². Therefore the work done to accelerate the molecule was mv².

@Chestermiller , does that sound ok to you?

 

[256bits]

@haruspex Is this akin to a vertical column of gas in a gravitational field.
In essence the gas is continuously accelerating, and AFAIK the gas does not increase in temperature overall over time.
There is a pressure differential and I suppose a dT due to that from end to end, but constant.
A box of gas changing velocity should experience a cooler end and a hotter end during the process.

 

[haruspex]

@haruspex Is this akin to a vertical column of gas in a gravitational field.
In essence the gas is continuously accelerating, and AFAIK the gas does not increase in temperature overall over time.
There is a pressure differential and I suppose a dT due to that from end to end, but constant.
A box of gas changing velocity should experience a cooler end and a hotter end during the process.

That's uniform acceleration. The steady state for that would be a pressure gradient but a uniform temperature. In a sudden stop or start it is not steady state.

 

[Krushnaraj Pandya]

That seems about right. I think I can solve this question now as per the context given in any future tests, thanks a lot :)