suddenly stop a moving moving vessel filled with gasses. What will change in temp?

vkash

1 g mole of oxygen at 27 degree Celsius and 1 atmospheric pressure is enclosed in a vessel. The vessel is thermally insulated. This vessel is moving with speed of v₀. Suddenly it is stopped. what will change in temperature of the gas.

How i did it.
first of all i find the root mean speed of the gas in position when there is no motion of vessel.
[tex]\sqrt[2]{\frac{3kT}{m}}[/tex]
after putting values it comes out to be 483.3 ms^-1
(till here it is done similarly in book)
after this i apply this formula
[tex]\frac{T1}{T2}=(\frac{v1}{v2})^2[/tex]
Here v represent the root mean velocity of the gas molecules. I put T₂=300,T₁ is 301. v₁=483.3 and v₂=483.3+v'. v' is velocity of the container. I assume that when we suddenly stop the container. velocity v' became the velocity of gas molecules,
but after applying second formula i got answer around ~10ms^-1 but answer is 36ms^-1.

can you guys please point out my mistake.

ehild

It is not the velocity, but the translational kinetic energy of the vessel & gas that transforms into the kinetic energy of the molecules - both translational and rotational. Your formula does not take into account that the oxygen (O₂) molecules have both translational and rotational energy.
The gas in the moving container has its internal energy U. What is the internal energy for a two-atomic gas? Initially, the whole gas moves with speed v₀, so it has 1/2 mv₀² energy in addition to its internal energy. When the container stops, it has only internal energy. Use conservation of energy between the initial and final states.

ehild

Andrew Mason

Why would the temperature change at all?

Assuming the container is rigid, the work done in stopping the container and gas is done not by the internal energy of the gas but by the external applied force. Why would the force applied to one side to stop the container (and thereby press in on the gas inside) not be completely offset by the force of the same magnitude that is applied to the other side in the container?

AM

Dick

Because the gas in the container doesn't suddenly stop moving. It will pile up against the forward side of the container and form an acoustic wave that will eventually dissipate. Adding to the internal energy of the gas. I agree with ehild's analysis.

vkash

Does the second formula i write is incorrect?
thanks for answering ,This method is similar to that in book, it gives me correct answer.

ehild

The formula is correct, but it gives the root-mean square speed of the molecules. You do not get the initial translational velocity just by subtracting the lower RMS value from the higher one.
The velocity of an individual molecule is the sum of its random thermal velocity v_r and the velocity of the container Vc. The velocities are vectors. The rms speed is the sum of the velocity-squares for all molecules, divided by the number of molecules. The new rms speed is v'=(1/N)∑(v_r(i)+Vc)².

ehild

vkash

Little tough to understand(??v_r??).The main thing i understand is that, that V_rms can't be used as velocity of the gas molecules.
I will discuss it with my physics teacher.
thanks ehild, for your valuable answers.

ehild

Vrms is speed, not velocity. The velocity of the molecules can have any direction.

Imagine a one-dimensional world where the molecules can move either to the right or to the left, and all move with the same speed. In the tank in rest, the velocity of one half of the molecules is [itex]\vec{v}[/itex] and the velocity of the other half is [itex]\vec{-v}[/itex]. The rms speed is
[tex]V_{rms}=\sqrt{1/2 (\vec v)^2+1/2 (-\vec v)^2}=\sqrt{v^2}=v[/tex]
the magnitude of the velocity of the individual molecules. If the tank moves to the right with velocity [itex]\vec{V}[/itex] , half of the molecules will move with velocity [itex]\vec{V}+\vec{v}[/itex], the other half move with velocity [itex]\vec{V}-\vec{v}[/itex]. The rms speed is

[tex]V_{rms}=\sqrt{1/2 (\vec V+\vec v)^2+1/2 (\vec V-\vec v)^2}=\sqrt{V^2+v^2}[/tex]

ehild

Andrew Mason

Ok. So let's suppose the stop is not sudden but very gradual. Would that make a difference in the energy of the compression wave? I don't see how you can determine the energy of any compression wave created without knowing something about the dimensions of the container and knowing more detail about how it stops.

AM

Dick

It's as ehild described. The translational kinetic energy of the gas has to go someplace. It's can't escape as heat since the vessel is insulated so it goes into heating up the gas. Yes, I think if the slowdown is very gradual you've got a different problem on your hands, But it's certainly true if the stop is abrupt. Anyway, I think that's the intent of the problem.

ehild

It is a Introductory Physics problem. The solution has to be based on a simple model. "Stops suddenly" and "thermally insulated" means adiabatic process, that all work done on the gas transforms into the random motion of the molecules, that is change of internal energy, as the time is not enough for any kind of energy transfer between the gas and surroundings. If the process is slow it is not adiabatic any more. The initial translational energy of the gas as whole will be transformed to some kind of vibration energy of the walls of the container and to the ambient and it will increase the temperature of the Universe at the end.


ehild

Dick

If the gas is decelerating uniformly it will develop a pressure difference between the forward facing wall of the vessel and the backward wall making the vessel harder to decelerate. So whatever is providing the work (well, it's actually absorbing the energy, I guess) to decelerate it will need to put in the extra work to decelerate the gas in addition to the vessel.

vkash

great example.


You are discussing about the case when the object is stopped with gradual deceleration. I think if wall are fully adiabatic(insulated, it's an ideal case) then temperature of gas will increase by same amount. since the extra kinetic energy of box will be transferred to atomic particles after the box comes to rest. Am i correct???

ehild

The KE of the box will be "used up" in the interaction with the external agent which stops it. It is the extra translational KE of the gas which transforms into the thermal energy, energy of the random motion of the molecules.
The temperature of an object is connected to the random motion of its constituent particles. The temperature of a ball thrown is not higher as that in rest. A flow of gas does not have higher temperature because of its speed of flow.

You are correct, the temperature of the gas will increase by the same amount either the container stops suddenly or gradually, if the walls are completely adiabatic, that is the box is perfectly insulating and is completely unaffected by the molecules of the gas. But there is no such box in reality.

Imagine you have marbles in a box, which moves with some velocity and stopped suddenly. Because of their inertia, the marbles move further and the front ones hit the wall violently, and reflect from it with high speed. After reflection, the marbles move backwards and collide with the other marbles moving forward. The collisions are not central, so the velocities become random, that corresponds to higher temperature.
If the container stops gradually, the marbles collide with the wall with small relative velocity. The collisions with the yet unaffected marbles is not that violent, it does not change the direction of velocities too much, so the other marbles collide also with the wall, and the marbles moving backwards will reach the back wall, and the whole set of marbles will oscillate between the walls. The marbles will move in an ordered way even after the box is stopped, and the KE of an ordered motion does not adds to the thermal energy. Of course the motion of the marbles is randomised sooner or later, and all KE of their ordered motion changes into thermal energy. But the marbles interact with the walls many times and no collisions are perfectly elastic. Some fraction of the KE is transferred to the wall. This fraction is negligible for the short time of interaction when the box stops suddenly, but gets considerable in a long time.

ehild

vkash

what would happen if
there is seen something like this. that there is a box of mass m having adiabatic walls filled with ideal gas of *** M . IT is moving with velocity v toward a spring of spring constant k(very large spring constant). What will be maximum compression in the spring.
For this condition i think. following equation will work
[tex]\frac{mv^2}{2}=\frac{kx^2}{2}[/tex] NOT [tex]\frac{(m+M)v^2}{2}=\frac{kx^2}{2}[/tex]
Am i correct sir?

What if spring constant is small so that box does not stop instantly??? can you please help me in figure out this condition.

ehild

I think you meant the equation of conservation of energy for an ideal spring, with a point mass or rigid body connected to it, and v is the maximum speed when the spring is relaxed and x is the maximum change of length of the spring. If that is the case, the box oscillates forever. When the velocity is zero, the elastic energy is maximum, and the spring force is maximum. Of course the total mass counts in that case, the mass of the box + the mass of filling.

If the content of the box can move with respect to it the motion of the box is not a simple harmonic motion any more. The CM will perform a SHM, with angular frequency sqrt(k/m+M) but both the box itself and the object inside will do some other motion, depending on the force between the object and the box. The "x" in your equation will be the maximum deviation of the CM and v will be the maximum speed of the CM. The KE of the whole system consist of the translational KE of the CM and the KE belonging to the inner degree(s) of freedom. In case of a gas, there are about 10²³ particles inside, all interacting with the walls of the container and with each other, moving according to all interactions they take part in, and each of them having its own kinetic energy.