# Interpretation of kinetic energy and temperature in thermodynamics Consider the question above.

Till now I have interpreted kinetic energy as a consequence of temperature in thermodynamics, i.e., the kinetic energy of a gas is directly proportional to its temperature.

Provided this, in the above question, I have three arguments:

1. If the gases were to stop, shouldn't the temperature of the gas become zero degrees Kelvin?

2. If they were to stop, should the energy go into the walls of the container?

3. In the ideal case, the energy is stated to be purely kinetic in books. But shouldn't the energy of the system include nuclear energy, bond energy, etc? In this case, can the kinetic energy appear in these ways?

Which argument is correct?

Update:

I misinterpreted 'suddenly stopped' which lead to the formulation of argument 1 in my mind. I agree (as pointed in the @Bob D answer) that it does not apply. Anyway, I am not deleting it.

• The question must have been framed in ideal conditions ignoring all those energies – Anusha Aug 14 '20 at 5:02

Argument 1

The macroscopic kinetic energy of the 0.03 kg mass of the gas moving as a whole at 100 m/s is 150 J ($$1/2 mv^2$$). This is not the same as the internal microscopic kinetic energy which is due to the random velocities of the molecules and which determines the temperature. So your argument 1 does not apply.

Argument 2

The author appears to be assuming the energy is absorbed by the gas as discussed below

Argument 3

Only the molecular kinetic energy is involved

As I said above, based on the author’s answer the author appears to be making the (perhaps questionable) assumption that all of the macroscopic kinetic energy when the gas stops is absorbed by the gas increasing its internal energy and temperature, as follows:

For an ideal gas the change in it’s internal energy depends only on It’s change in temperature. For one mole

$$\Delta U =C_{v}\Delta T$$.

For a diatomic gas

$$C_{v}=\frac{5}{2}R$$

Therefore

$$\Delta U =\frac {5}{2}R\Delta T$$.

Setting that equal to 150 J

$$\Delta U =\frac {5}{2}R\Delta T=150J$$

$$\Delta T=\frac{60}{R}$$

Hope this helps

First, let me state that if the box has stopped the particles in the box won't have stopped moving. This would be the case if they all had zero velocity in the non-moving box, meaning that the velocity of the box would be imparted to all of them (as is the case of particles in a box with non-zero temperature). You have to calculate the kinetic energy of the collection of particles in the box, due to the motion of the box, which seems not too difficult, I guess.
When the box stops moving this extra energy (on top of the kinetic energies of all particles) is gone. Of course, it's absorbed by the gas of diatomic particles (as stated in the cited question), and later on, depending on the temperature outside the box, it's absorbed or not (energy coming in). That's why it's best in this question it's best to state that the box is a perfect insulator.
As you have calculated the kinetic energy of all particles due to the moving box (which is the same as in gas at zero Kelvin temperature contained in a moving box), you can calculate the temperature increase of the gas.
All the other forms of energy you mentioned don't contribute. These only contribute to the final mass of the particles and thus to the final kinetic energy. The final mass (and thus kinetic energy), after all the interactions you stated, is the one used in the molecular weight.