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Why we do not considered the translational energy of the gases particles in the internal energy due to ordered motion but only due to random motion.

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- Thread starter Frigus
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Because we can always find an inertial frame in which the translational energy is zero; it's the center of mass frame. The internal energy is whatever cannot be transformed away just by changing frames, the energy that is there even when the center of mass is not moving.

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In the general form of the first law of thermodynamics, the energy of the gas is expressed as the sum of three parts: the kinetic energy of the ordered motion KE plus the gravitational potential energy PE, plus the combined energy of random motion and molecular interaction (the internal energy U). So we write:

$$\Delta U+\Delta (KE) + \Delta (PE)=Q-W$$

So the ordered kinetic energy change is definitely accounted for.

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The internal energy is whatever cannot be transformed away just by changing frames, the energy that is there even when the center of mass is not moving.

Wouldn't that include potential energy in an external field (which is not part of internal energy)?

- #6

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Yes, but in thread we're talking about kinetic energy.Wouldn't that include potential energy in an external field (which is not part of internal energy)?

- #7

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Yes, but in thread we're talking about kinetic energy.

OK, I didn't considered "whatever" to be limited to this thread ;)

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Sure, in non-relativistic physics it's usually customary to express everything in "quantities per mass", i.e., the energy density of an ideal fluid with velocity ##\vec{v}## may be written asWouldn't that include potential energy in an external field (which is not part of internal energy)?

$$\epsilon=\frac{\rho}{2} \vec{v}^2 + \rho U_0(s,\rho) + \rho V(\vec{x}),$$

where ##\rho## is the mass density, ##U_0## the internal energy of the gas per unit mass as measured in its rest frame, and ##V## is the potential of the external forces per mass. E.g., for the gravitational force on Earth you may write $$V(\vec{x})=-\vec{g} \cdot \vec{x}.$$

For an ideal fluid you have no friction, heat conduction or any other dissipative transport processes by assumption, and thus the equations of motion follow under the constraint ##\delta s=0##, i.e., the entropy per mass stays constant.

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Why do we do not consider ordered motion as part of the internal energy of a gas?

Because wind and heat are different things.

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$$f(\vec{x},\vec{p})=\frac{\exp[\mu/(k_{\text{B}} T]}{(2 \pi \hbar)^3} \exp \left (-\frac{\vec{p}^2}{2m k_{\text{B}}T} \right).$$

From this the total number of particles and internal energy follows

$$N=V \exp[\mu/(k_{\text{B}} T)] \int_{\mathbb{R}^3} \mathrm{d}^3 p f(\vec{x},\vec{p}), \\

U=V \exp[\mu/(k_{\text{B}} T)] \int_{\mathbb{R}^3} \mathrm{d}^3 p f(\vec{x},\vec{p}) \frac{\vec{p}^2}{2m} = \frac{3 N}{2} k_{\text{B}} T.$$

For a fixed number of gas particles the internal energy only depends on temperature, not on volume. The reason is that an ideal gas is defined as particles that interact only via short-ranged interactions with the average inter-particle distance being large compared to the range of these interactions, i.e., most of the time the gas molecules are free particles and thus you can squeeze them to a certain extent without any work due to particle interactions. The only work you have to do is against the gas pressure.

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Can anyone please give the reason of the line written in the box.

You mean why you reposted your initial question in photographic form? No idea.

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Thanks,Because we can always find an inertial frame in which the translational energy is zero; it's the center of mass frame. The internal energy is whatever cannot be transformed away just by changing frames, the energy that is there even when the center of mass is not moving.

Understood today what you want to say and your reply helped me to get relief from this problem.

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Can anyone please give the reason of the line written in the box.

It is usually a very good approximation. But is doesn't work for astronomic scales.

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I was just seeing my old posts that I didn't understand quite well and after seeing your this answer I was just blown away because earlier in that time I was just thinking that it's answer will be out of the world but its answer is so simple that now I think that how one can even ask question like that.

Why do we do not consider ordered motion as part of the internal energy of a gas?

Because wind and heat are different things.

Thanks.

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