Hot answers tagged kinetic-theory
10
The analogy is a very good one, because heat transfer is in fact modelled by phonons, which you could also use to describe sound waves.
The crucial difference is that sound waves have a much longer wavelength (at least in the range of some millimetres) than thermal phonons (not more than a few orders of magnitude bigger than the atomic lattice scale). These ...
8
My understanding is that this question is being asked in the context of the kinetic theory of classical gases. In that context, here is the argument:
If the system is rotationally invariant, then we should have $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$. Thus $\langle v^2 \rangle = \langle (v_x^2 + v_y^2 + v_z^2 )\rangle $ ...
6
The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzman's constant, $k_B = 1.3806503 × 10^{-23}\frac{m^2kg}{s^2K} $ and is equal to the ideal gas constant divided by Avagadro's number, $\frac{R}{N_A}$. So where does the equation come from? The short answer: The equation above is ...
5
Sure, you can generalize the mean free path to a different number of dimensions. But first, let's understand the derivation in 3D.
A particle will collide with any other particle that it comes within a distance $d$ of. So if it moves a length $\ell$, it will collide if there is another particle in a volume $\pi d^2 \ell$. Call this the volume swept out by ...
5
The kinetic theory of gases does in fact say that molecules in a gas move very rapidly (although some move quite slow, and others even faster).
However, there is another crucial component to the theory. The idea of Mean Free Path. Here, a molecule is moving very fast but doesn't get very far before hitting another molecule. This is why things like odors ...
5
This follows from the equipartition theorem. The equipartition theorem states that in thermal equilibrium, the average energy of each degree of freedom (each independent way the system can move) is $k_B T/2$, where $T$ is the temperature and $k_B$ (or just $k$) is called the Boltzmann constant. There are three independent directions in which a gas particle ...
3
To understand the physics of inflated balloons you have to understand curved membranes under tension.
If there's a membrane under tension $T$ (for simplicity assumed to be isotropic), with fluids on both sides, it will be flat (planar) unless there is a pressure difference between the two fluids. If you notice that it's curved with radius of curvature $R$ ...
3
Just to add some explicitness to the above answers: take an isolated particle at rest; it's KE is zero. Now switch to a reference frame with relative velocity $\beta$ wrt the particle. In this frame, it's KE is $$KE' = E - mc^2 = (\gamma - 1)mc^2 = \frac{1}{2}mv^2 + O(\beta^4).$$ We see that $KE' \neq KE,$ thus it's not a scalar.
3
The Boltzmann equation is an approximation. It is obtained by closing the so-called BBGKY hierarchy that links N-particle distribution functions to N+1 particle distribution functions. The closing is obtained by introducing an assumption, the 'molecular chaos' assumption. In short, molecular chaos asumes the absence of correlation in the incoming state for ...
2
the discrepancy between the Boltzmann equation and the experiment may always be blamed upon the experiments. If one does things right, the Boltzmann equation is theoretically completely exact (issues will be discussed below). However, it's pretty hard to measure the distribution functions "directly", especially if you want the full dependence both on momenta ...
2
Yes, the CF is a way of comparing RGs and IGs kept at same temperature and pressure. It is a comparison of volumes, as stated in the question. Hence the name "compressibility factor".
The CF equation can be better stated as: $P_{obs}V_{m,obs}=ZRT_{obs}$, where the m stands for "molar", and "obs" means "observed". This is in contrast with ...
2
You're basically assuming an infinite mean free path for the air molecules, whereas
people normally would use the Navier-Stokes equations which assume an
infinitesimal mean free path. You will therefore underestimate the pressure
difference.
Further, instead of solving the full fluid flow problem, people normally simply model
a hole as "an impedance to ...
2
The easiest way would be to use statistical physics. Gases are effectively described with a one-particle distribution function:
$$f \sim \exp \left[-\frac{E}{kT} \right]$$
which leads to a spatial part of $f$ (concentration) in a uniform gravity field to be
$$n_i \sim \exp \left[-\frac{m_i g z}{kT} \right]$$
That's the easiest way to show that "the ratio ...
1
As for you distribution, I think it should be correct, because you can note $p^0=\frac{m}{\sqrt{1-v^2}}$, so your distribution is actually
$$
f_k=\frac{1}{m}n'_k\delta(p-p_k),
$$
where $n'_k$ is the particle density in your lab frame, which is (up to your normalization $m$ which is unclear to me) really the phase-space distribution of dust.
I think that a ...
1
I think you have an error in assumption 2. If $N$ is the number of molecules, then the mass of the sample would be $N$ multiplied by the mass per molecule, not $N$ multiplied by the total mass of the sample. You are kind of "overcounting" mass. If you take $m$ to be the mass per molecule (molecular mass), then I believe it works out.
1
I do not understand the relation which you try to put up with the hydrogen atom. In the hyrdogen atom the energy is due to the Coulomb attraction. As far as I understand your question, you are interested in the rotational energy instead.
The rotational energy of a rigid rotor is given by
$$E_\text{rot} = \frac{L^2}{2I}$$
with $L$ the angular momentum and ...
1
First, the derivation. The condition says that it's very unlikely for the particle number to be greater than one – the number of filled one-particle states is a small fraction of the number of all one-particle states and the doubly/multiply filled ones may be considered a higher-order correction that may be neglected. This approximate inequality is ...
1
I've looked at the notes. There are several things going on here, some of them not stated. And we have to make some assumptions. One is that the average kinetic energy of a particle moving in a direction $v$ is $mv^2/2$, where $m$ is the mass of the particle. This isn't too hard to prove, but doing so would take us far afield.
Another assumption is that ...
1
Nonlocal in kinetic theory means you substitute local variables like the heat flux: q(r,t)=ec(r,t)*v by a non local variable wich usually involves an integral kernel. One needs to do so if the particle distribution function has a long tail in the high energy domain. Particles traveling at hight velocity do have a non negligible effect on the heat flux, even ...
1
I'm going to try to accomplish something the other comments don't quite get.
The challenge here is that we're really talking about a fluids problem, really a hydrostatic problem. You can't just write the ideal gas equation and be done with it, although no one has really written that equation yet, so I'll write it here. But I'm not just going to write it. ...
1
The statement that the entropy increases because of collisions is incorrect. The conservation of phase space volume is a theorem of Hamiltonian mechanics, and therefore applies to all known physical systems, regardless of whether they contain nonlinear forces, collisions or anything else.
What actually happens is that although the phase space volume ...
1
The entropy increase comes from the assumption that you can close the system on the kinetic level, thereby (i) making the dynamics tractable and getting a transport equation, and (ii) disregarding extremely high frequency contributions and paying for this with an entropy increase.
Any interaction leads to collision terms; the details only matter for the ...
1
The formula is $$\mu_s \geq \frac{g\tan\theta}{1+\frac{k^2}{r^2}}$$ where $\mu_s$ is static friction coefficient for the ball-incline interface. $\theta$ is the angle of the incline, and $k$ is the radius of gyration of the ball (for a solid uniform spherical ball, $k=R\sqrt{\frac{2}{5}}$). R is the radius of the ball. If you have a more complicated body, R ...
1
Collision theory will not help for calculating chemical rates in the liquid phase since the mass transport there is governed by diffusion. It is not perfectly clear from your question, so I will have to assume that you are interested in a two-phase gas/liquid system.
First, liquid phase is much denser, molecules there are always in contact and don't have to ...
1
If one goes to the wiki article on evaporation one sees that
For molecules of a liquid to evaporate, they must be located near the surface, be moving in the proper direction, and have sufficient kinetic energy to overcome liquid-phase intermolecular forces. Only a small proportion of the molecules meet these criteria, so the rate of evaporation is ...
1
leftaroundabout gave an excellent explanation for the thermal conduction
of insulators. However, in the case of metals, a significant amount of
energy is carried by the excitations of electrons (the width of their
Fermi-Dirac distribution). The thermal conductivity is then related to
how far an excited electron can travel before being scattered, and is
...
1
A scalar is something that doesn't transform under coordinate transformations. A vector is something that transforms "like a vector," in other words, its coordinate transforms are realized as a local multiplication by a linear operator (e.g., a matrix).
In Newtonian mechanics, it's simple to see that kinetic energy, being proportional to the square of a ...
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