Difficulty in understanding the scenario for this Q

Question

Q:If a closed rigid wall container filled with 20 gm of He moving at a constant speed of 10 m/sec is stopped abruptly to stand still, find the increment in temperature of this He sample.

My question regarding this problem :

1. when kinetic energy increases , velocity increases and therefore , temp also increases . Now , in the above Q . If the container is put to a stop after some time , how can there be an increment in temperature since the molecules speed would decrease now. Therefore , their temp should decreases and so should kinetic energy ?

Theory 1:Does the Q mean to say the increment in temperature during the time it was travelling or after the time it was put to stop.

Theory 2:Also , I read online about this Q. It says that my thinking was wrong and what the Q means to say is that when the container is put to stop. There will be collision inside the container which causes the increase in temperature.

Below is how I solved:

Using first law of thermodynamics,

I found kinetic energy which has to be W or macroscopic work done. Then , we have q and $\delta$E left.

q must be 0 since it is a closed container. Therefore , in end . We just equate W and $\delta E$.

Main Q is that which theory is right and is my method of solving above correct ?

Answer 1

One basic notion behind this is that temperature of a substance is defined as the average kinetic energy of all the atoms or molecules of that substance.

Now assume gas in a balloon. Also assume that the balloon is not moving, so the average velocity of the gas inside the balloon is zero. Despite the average velocity (as a vector average) being zero, the individual molecules have non zero velocities, and therefore the kinetic energy is greater than zero.

So, the kinetic energy is non zero, even though the average velocity is zero (the balloon is not moving). If you heat up the gas (or if you reduce the volume of the balloon very fast - thus increase the pressure), and keep the balloon stationary, then the average kinetic energy will increase, despite the average velocity (as a vector sum) still remaining zero.

What I am trying to stress is that

The temperature is the average kinetic energy of the molecules .

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So, in the problem the idea is that the kinetic energy of the balloon ($E= \frac 1 2 m \cdot v^2$), is converted into heat energy, and increases the temperature of the molecules in the balloon. So basically you need to equate:

$$\frac 1 2 m \cdot v^2 = m \cdot C_p \Delta T$$

Answer 2

At the instantaneous moment of stopping the Container, there will be a rush of the molecules of the rear end to the front end. After some time, $\delta t$ this rush will cause extra pressure on the front of the container and less pressure on the back of the container until there is an equilibrium and the rush stops. Therefore at the time $t= \delta t$ there are stratified strata of higher temperature at the front end and lewer at the end assuming He as an ideal gas.

$$pV=n \bar{R}T$$

• p = pressure

• V = volume

• n = number of moles

• $\bar{R} = 8.31 J/ mol/ K \ universal$

• T = temperature in Kelvins.

This imbalance in the pressure will cause a rush back of the molecules of the He to the end and start a damped cyclical change of pressure and temperature at the two ends of the container.

The damping effect will give rise to an increase in the temperature of the container after some time.