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"Thermal energy" is a bit of a misnomer because "thermal" really refers to a method of energy transfer, not energy storage. When energy moves from one system to another, it can do so via a thermal process (e.g. conduction, convection, radiation) or a mechanical process (something pushes on something else). So technically, I wouldn't call $\frac{3}{2}kT$ the ...

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Hopefully, I will not be "bad mouthing" those who have a background similar to my own, but here goes. In the world of chemical engineering, there are quite a few equations that are much more empirical than "first principle" based. This is particularly true for heat transfer problems, which usually have to deal with turbulent fluids. From a practical ...

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When the chemical potential is 0 the extra free energy needed to add or remove a particle to the system is 0(i.e $\mu=\frac{dA}{dN}=0$. So particles can leave and enter the system without changing the (free) energy. In A BEC all particles have condensed to the ground state of the system. Particles entering or leaving the system will be added to the ground ...

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Yes, the temperature of the gas would decrease quite fast, given that the molecules in the container are still, which implies zero temperature for container. However, if the container's temperature is non-zero, it sometimes happens that gas molecules will instead gain energy because the molecule it collides with is moving fast enough in the opposite ...

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If you assume that the gas is ideal then each collision of a molecule of gas with the wall conserves the kinetic energy. Hence the temperature will stay the same.

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The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.

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Why it is colder in mountains, at high altitudes? One answer is that mountains on Earth aren't all that tall. An impossibly tall mountain would see temperatures vary with altitude as depicted below. Tall as it is, even Mount Everest doesn't extend into the stratosphere. This is a question about the lowest layer of the atmosphere, the troposphere. ...

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An atmosphere in absolute equilibrium in fact is isothermal (see below for more detailed analysis of your cannonball). However, if the atmosphere is mixed by wind, gas expands and contracts adiabatically. If the mixing is fast enough, it obeys relatively well the adiabatic invariant, which multiplied by suitable form of ideal gas law ($(T/(pV))^\gamma = ... 1 The air becomes colder because of the ideal gas law,$PV=nRT$. where$P$is pressure,$V$is volume,$n$is the number of moles of the gas,$R$is the ideal gas constant, and$T$is the temperature of the gas in Kelvin. If we rearrange$PV=nRT$, we can solve for$T$. By looking at$T=\frac{PV}{nR}$you can see that reducing pressure will reduce the ... 2 Imagine wind blowing along a plane with the air by the ground all a nice and steady temperature. Now this wind encounters a mountain range, so is forced upwards. The pressure is lower at higher altitude since there is less remaining atmosphere above it. The temperature of gas decreases when the pressure is lowered, which is why this same air gets ... 1 as molecules jump higher, they loose energy/speed due to gravity. This results in molecules slower at heights and therefore you have lower temperatures at the heights, boy. Molecules do not jump up, they scatter off each other every which way. The difference in gravitational energy within the nanometers of the molecule's path before a scatter on ... -5 I guess we all agree that a cannonball does not get cold when flying upwards, it loses kinetic energy, not heat energy. Why would ascending air not do the same thing? So equivalently to an ascending cannonball we get: When one molecule in a rising column of air bounces upwards, it loses kinetic energy while moving upwards. That loss of kinetic energy is a ... 1 Not sure the calculation was done correctly (on first glance). However, that is not important here, just think about what you are doing: You have N particles all of which are mutually interacting via a super-long-ranged potential (Coulomb-interaction$\sim r^{-1}$would be considered long-ranged, you are using a parabolic$\sim r^2$potential). So, what you ... 6 The potential energy for a diatomic molecule is not $$U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2$$ but is instead $$U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2,$$ where$r_0$is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector ... 0 Are you interested in the relative pressure difference between two points or the absolute pressure? In a container the pressure on the walls due to a gas can be calculated using the ideal gas law (pV=NRT). However, for a column of water, or the pressure at earth's surface due to a column of air, the pressure can be calculated as P=F/A=ρgh. The first ... 2 A degenerate gas is one where more than one electron (in fact, two) occupies each possible low-energy state up to the Fermi energy. I suppose the term "degenerate" comes from the multiple occupancy of each energy level. 0 A normal gas consists of particles that do not interact much except for elastic collisions. Often, describing a gas in a simplified way ignoring the other interactions completely is good enough. This simplified model is the "ideal gas". This simplified description fits for the electron gas also, as long as the pauli exclusion principle does not become ... 0 The ideal gas law is derived without consideration of the gravitational field. The kinetic behavior of a gas is, by far, more complex than the very simple example of the ideal gas law. So, answering to your two questions: Yes, the range of application of each equation is different. Use the hydrostatic equation and not the ideal gas one. Anyway, kinetic ... 1 Here it is an incomplete answer, but just to think about some issues. An ideal gas is assumed to have particles non-interacting and with no extension (point-like particles). In particular for an ideal gas you discard gravity. While the Stevin's law (the relation$\rho g h$) is a direct consequence of the fact that you have a fluid in a gravitational field. 5 Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because$E=\frac ...

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In slow expansion the gas is always in equilibrium state. Rapid expansions are basically rapid so there is no time for energy transfer so they are adiabatic.

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