Why don't we consider ordered motion as part of the internal energy of a gas?

In summary, the translational energy of gas particles in the internal energy is not considered due to the fact that it can always be transformed away by changing frames, unlike the internal energy which is present regardless of the frame. The internal energy is defined in the local rest frame and does not depend on volume, only on temperature. This is due to the fact that ideal gas particles interact only via short-ranged interactions, allowing for the gas to be compressed without any work due to particle interactions.
  • #1
Frigus
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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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  • #2
Hemant said:
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.
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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  • #3
Indeed, it's defined so by definition. All the quantities referring to the intrinsic properties of macroscopic matter are defined in the (local) rest frame of the matter, i.e., temperature, internal energy, pressure, entropy, etc. This is very important in the context of relativity theory since it tremendously simplifies things, using scalar quantities to describe the intrinsic properties of matter rather than non-covariant definitions of older treatments, which worked to some extent for special relativity but is pretty useless within the general theory.
 
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  • #4
Hemant said:
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.
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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  • #5
Nugatory said:
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)?
 
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  • #6
DrStupid said:
Wouldn't that include potential energy in an external field (which is not part of internal energy)?
Yes, but in thread we're talking about kinetic energy.
 
  • #7
Nugatory said:
Yes, but in thread we're talking about kinetic energy.

OK, I didn't considered "whatever" to be limited to this thread ;)
 
  • #8
DrStupid said:
Wouldn't that include potential energy in an external field (which is not part of internal energy)?
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 as
$$\epsilon=\frac{\rho}{2} \vec{v}^2 + \rho U_0(s,\rho) + \rho V(\vec{x}),$$
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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  • #9
Um, B-level, everyone.

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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  • #10
Can anyone please give the reason of the line written in the box.
 

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  • #11
It follows directly from the fact that for a non-interacting gas the equilibrium phase-space distribution is the Maxwell-Boltzmann distribution,
$$f(\vec{x},\vec{p})=\frac{\exp[\mu/(k_{\text{B}} T]}{(2 \pi \hbar)^3} \exp \left (-\frac{\vec{p}^2}{2m k_{\text{B}}T} \right).$$
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.
 
  • #12
Hemant said:
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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  • #13
Nugatory said:
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.
Thanks,
Understood today what you want to say and your reply helped me to get relief from this problem.😊
 
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  • #14
Hemant said:
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.
 
  • #15
Vanadium 50 said:
Um, B-level, everyone.

Why do we do not consider ordered motion as part of the internal energy of a gas?
Because wind and heat are different things.
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.
Thanks.
 
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Related to Why don't we consider ordered motion as part of the internal energy of a gas?

1. Why is ordered motion not considered part of the internal energy of a gas?

Ordered motion, also known as macroscopic motion, refers to the overall movement of a gas as a whole. This type of motion does not contribute to the internal energy of a gas because it does not involve the random movement of individual particles within the gas. The internal energy of a gas is determined by the kinetic energy of the particles and their interactions, not the overall motion of the gas.

2. How is the internal energy of a gas related to its temperature?

The internal energy of a gas is directly related to its temperature. As the temperature of a gas increases, the average kinetic energy of its particles also increases, resulting in a higher internal energy. This is because the particles are moving faster and colliding more frequently, leading to a greater amount of energy in the system.

3. Can ordered motion affect the pressure of a gas?

No, ordered motion does not contribute to the pressure of a gas. Pressure is determined by the force of the gas particles colliding with the walls of their container. Ordered motion does not involve these collisions and therefore does not impact the pressure of the gas.

4. How does the internal energy of a gas change during a phase transition?

During a phase transition, such as from a liquid to a gas, the internal energy of a gas remains constant. This is because the energy is used to overcome the attractive forces between the particles and break the bonds holding them together, rather than increasing the kinetic energy of the particles.

5. Is the internal energy of a gas affected by external factors?

Yes, the internal energy of a gas can be affected by external factors such as changes in temperature, pressure, and volume. These changes can alter the kinetic energy and interactions of the particles within the gas, resulting in a change in the internal energy of the system.

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