# If temperature is related to average kinetic energy in an ideal gas, then does speeding up the gas container affect its temperature? [duplicate]

This question is related to the following textbook question:

The temperature of an ideal gas is directly proportional to the average kinetic energy of its molecules. If a container of ideal gas is moving past you at $$2000\ \frac{\mathrm{m}}{\mathrm{s}} ,$$ is the temperature of the gas higher than if the container was at rest? Explain your reasoning.

My approach to this problem was to refer back to the derivation of the kinetic energy- temperature relation. The equation $$E_\mathrm{kin} =\frac{3}{2}nRT$$ was derived in the containers reference frame, with the momentum transfer from particles impacting the wall playing an important role. So even if the container is moving 2000 m/s, the particles inside will be moving relative to this speed as well, so the momentum transfer would be the same in the box's frame of reference, and we'ed get the same temperature.

I guess first I'd like to know if this reasoning is accurate and/or if I've missed any pieces?

Moreover, as an extension, I'm wondering what would happen if we suddenly stopped the container while letting the interior gas retain its original speed. Going off my original argument w/ the equation being derived in heavy part from the momentum transfer and reference frame considerations, for a short period of time the particles inside should still be going much faster w.r.t. the container, and so we'ed expect the temperature to jump up upon stopping the box due to an increased rate of collisions. In the physical world, im guessing this heat would probably be lost pretty quickly though, assuming the container can take the stress without breaking.

TL;DR box and particles move at same speed, but if box stops particles fast, so temp increase. Am I on the right track?

• Temperature is proportional to average kinetic energy in the rest frame of the gas. – Bert Barrois Jun 22 '19 at 11:31
• This is a surprizingly common question. See physics.stackexchange.com/questions/96327/…, physics.stackexchange.com/questions/336386/…, physics.stackexchange.com/questions/90343/…, and links therein. – dmckee --- ex-moderator kitten Jun 22 '19 at 14:06
• If you define temperature as what a thermometer measures, it is easy. If the thermometer is in the rest fram of the gas, it obviously gives the same temperature. If it is in your rest frame, it either gets destroyed quickly or if it survives then it shows a higher temperatiure. – lalala Jun 22 '19 at 14:33
• I think it would not be in thermal equilibrium if you give it a boost. Thermal equilibrium is quite an important condition for temperature to make sense due to the Zeroth Law of Thermodynamics – Jepsilon Jun 22 '19 at 23:36

In addition to what has already been said, if you were able to measure a change in the temperature of the gas moving at constant velocity, it would violate the special theory of relativity which states that the laws of physics are the same in all inertial frames. Which is the same as saying the results of any experiment should be the same in all inertial frames, including the measurement of temperature.

If you were able to detect a change in temperature for the gas moving at constant velocity, you would have a means to detect absolute motion.

Hope this helps.

• I fail to see how relativity is relevant here as in the question clearly the container is moving relative to a stationary observer. If the observer is moving with the container itself there is no velocity for the container as observed by the observer. So no change in kinetic energy. – Kolandiolaka Jun 22 '19 at 13:49
• @Kolandiolska It certainly is relevant. If the temperature were different when measured in the moving frame then when measured in the stationary frame you would have a means to detect absolute motion – Bob D Jun 22 '19 at 14:11
• Yes but that doesn't fit the premise of the question. The question is an external observer observing a moving container. In this case the temperature can vary much like the observed length and time and I believe at relativistic velocities it does matter as the container volume and the collision frequency and such is observed differently by an external observer. – Kolandiolaka Jun 22 '19 at 14:16
• I beg to differ. The question is "If a container of ideal gas is moving past you at 2000m/s, is the temperature of the gas higher than if the container was at rest?" It doesn't say who is making the measurement. My answer is based on the premise that the measurement is being made by an observer in the reference frame of the container. I'm in a car at rest with respect to the road. I measure the temperature of the gas. The car is now moving at 2000 m/s with respect to the road. I measure the temperature again. It does not change. If it did, I would have a means of detecting absolute motion. – Bob D Jun 22 '19 at 15:02
• @Kolandiolaka BTW Obviously a car can't move at 2000 m/s. The example is for constant velocity (an inertial frame of reference) – Bob D Jun 22 '19 at 18:07

Temperature is a frame invariant. Different observers moving at different relative speeds shouldn't observe different temperatures. But Kinetic energy (and therefore its average) is frame dependent (the kinetic energy of a bullet measured from the gun's frame of reference is crazy huge while the same as measured from the bullet frame of reference is zero). Your reasoning is quite on point in my opinion.

Temperature is related to the avarege kinetic energy of the particles as measured from the frame of reference of the container or the center of mass of the particle system. It is the component of the kinetic energy associated with the random motion so it has to be done from that reference frame as to avoid adding the kinetic energy of the overall consistent motion of the particles (a.k.a. the cointainer moving all together).

As per your second question. Yes! if you suddently stop your container the abrupt change in motion of the solid container would be seen in the frame of reference of the gas particles (of their common center of mass) as a sudden accelleration towards one of the walls of the container. This means that particles would gain kinetic energy quickly (since an accelleration is a change in speed and a change in speed is translated as a change in kinetic energy). The thing is that since the particles are not gaining a common speed component but are gaining a common change in speed component the argument is different from the one applied before. The particles would collide randomly trnsporting this new kinetic energy rising the overall average kinetic energy as measured from the frame of reference of the gas center of mass. Thus temperature should increase.

In fact this is the same reason why in any real collision part of the energy is transfered as heat, rising the temperature of the colliding bodies.

In school I learned that temperature corresponds to the unordered kinetic energy of the constituent particles of the gas. As Bert Barrois pointed out, this means that a net momentum of the center of mass of a gas under consideration is typically not considered to contribute to the temperature, i.e. the mean kinetic energy per degree of freedom is really evaluated in the center of mass frame of the gas, so you were completely right.

I also agree with you that once you would more or less abruptly stop the container, the kinetic energy of the gas' center of mass motion would start to equilibrate within the box and heat it up (from the perspective of the gas, the container back wall then does something similar in flavour to how you pushing the piston in when pumping up your bicycle tyre makes the gas hot, and you can often feel it (if you ever pumped up a bicycle tyre manually)).

As an example: In a 22.4 litre box, there'd be about 1 mol of nitrogen molecules weighing 28 grams at room temperature and atmospheric pressure. Their inner energy would be $$E = \frac{5}{2} n R T = \frac{5}{2}\ \text{mol} R T$$ at $$n = 1\ \mathrm{mol}$$, and 3 translational and 2 rotational degrees of freedom per molecule. With $$R = 8.31\ \mathrm{J/(K\ mol)}$$ that makes unordered kinetic energy of $$6090 \ \text{J}$$. The center of mass motion of the gas has $$E_\text{kin} = 1/2 m v^2 = 1/2 \cdot 0.028 \ \text{kg}\ (2000\ \text{m/s})^2 = 56000 \ \text{J}$$.

So the gas would get pretty hot: $$T_\text{final} \approx \frac{62090}{6090} \cdot 293 \ \text{K} \approx 3000 \ \text{K}$$.