# Tag Info

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 ...

10

Temperature is not the average kinetic energy of a particle it is the average energy per mode.1 In very simple models (i.e. the monoatomic ideal gas) the number of modes per particle is fixed and can not vary, so that the heat capacity of these simple models is indeed fixed. And in fact, good approximations to mono-atomic, ideal gases (noble gases, other ...

9

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 ...

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

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 ...

5

If equipartition holds, temperature corresponds to average energy per degree of freedom, which - besides kinetic ones - include internal ones like vibrational and rotational degrees of freedom. Even in cases of structurally similar molecules with the same degrees of freedom, because of energy quantization (in particular vibrational energy), heat capacity ...

4

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 ...

4

It seems to me that you're looking for the Boltzmann transport equation: $$\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}$$ with $f$ the distribution in phase-space, $\mathbf p$ the particle momentum, $Q$ some source term, and the RHS an interaction ...

4

The mechanism of EM radiation emission in dilute gases is different from solids, liquids and dense gases. In a solid the main source of the continuous emission, i.e. black body emission, is lattice vibrations causing local oscillations in electron density. The resulting transient dipoles emit EM just like any oscillating dipole. This isn't a resonant ...

4

Since the comment answered your question I'll just go ahead and set out a more generalised version. It's straightforward to simplify things back down to your case. Consider the following continuity equation: $$\dot{N}(x,t) = -\nabla\cdot\vec{\Gamma}(x,t) + S(x,t),$$ where $\vec{\Gamma}(x,t)$ is the flux (in your case $\vec{\Gamma}(x,t)= -D\nabla N(x,t)$, ...

3

As far as I know, Gallavotti proved the ergodicity of the Lorentz gas, while Sinai proved that of a system of $N \leq 5$ rigid spheres. Anyway, this is a minor detail. For certain aspects, a more suitable model for the Drude model is the Boltzmann gas. Lanford has shown (in 1970s, I think) that the entropy for this model is always increasing, but anyone has ...

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

The velocity of the system after collision is $$V=\frac{m_b}{m_b+m_s}v_b$$ The lost of kinetic energy can be assumed as the ejected heat in question, \begin{align} \Delta KE &= m_b v_b^2 /2 - (m_b+m_s)V^2 /2\\ &= m_b v_b^2 /2 - \frac{m_b^2}{m_b+m_s}v_b^2 /2\\ &= m_b v_b^2/2 \left(1-\frac{m_b}{m_b+m_s} \right)\\ ...

2

The expression you quote is for a ideal monatomic gas, and we get $C_v = 3/2$ for the three degrees of freedom. For ideal diatomic gases we do indeed have to count rotational degrees of freedom and we get $C_v = 5/2$. See the Wikipedia article on ideal gases for more info.

2

If you want to combine the two equation the easiest way is probably taking a "pseudo-derivative" of the equation of status: $2kV\Delta V = n R\Delta T$ And then substitute $\Delta T$. I don't find the resulting expression for $C_x$ very illuminating, however the initial hypothesis is pretty weird.

2

Alternate method to derive drift velocity: Consider a field $\vec{E}$ inside the conductor. Using equations of motion we can say that for every charge inside the conductor, $$\vec{v_1}=\vec{u_1}+\frac{\vec{E}e}{m}t_1$$ $$\vec{v_2}=\vec{u_2}+\frac{\vec{E}e}{m}t_2$$ $$.$$ $$.$$ $$.$$ $$\vec{v_n}=\vec{u_n}+\frac{\vec{E}e}{m}t_n$$ where $t_1,t_2,...t_n$ are the ...

2

In terms of the Hamiltonian formalism the MB distribution can be obtained for a set of particles without potential energy between them (it means free particles), under the assumption that in equilibrium they satisfy the MB statistics. It is worth noting that this is not an obvious fact, the MB statistics can be applied to any kind of object, in particular ...

2

Think of the volume as divided into many unit cells. The probability that a molecule of type Ai is present in the cell is proportional (in the leading term) to the number of molecules of type Ai. The combined probability of having n1 molecules of type A1, etc, is the product of the individual probabilities (assuming independent events). This boils down to ...

2

inelastic collisions are necessary in order to properly distribute energy among the internal molecular degrees of freedom (vibration, rotation). They are, but the ideal gas (nominally) does not have such internal degrees of freedom.1 Allowing them is a simple extension that is often made without announcing it2 and with that extension you get some ...

2

I think Maxwell-Boltzmann distribution should be valid for molecules in liquid too, at least according to classical statistical physics, because the factor $e ^{−\beta p^2/2m}$ in the Gibbs-Boltzmann probability density does not depend on potential energy and is the same whether the molecule is in gas, or a liquid. I do not know if there is a measurement ...

2

What makes water boil/evaporate is the thermodynamic concept derived from the first and second law of thermodynamics. You can read this article to find out the derivation from entropy to the Clapyeron equation. http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation#Derivation_from_state_postulate $$\frac{\mathrm{d} P}{\mathrm{d} T} = \frac ... 2 G.J.M. Hagelaar and L.C. Pitchford give an elegant derivation of fluid equations in the scope of two-term formulation of the Boltzmann equation. Yours equation above appears in (39) (see "Solving the Boltzmann equation to obtain electron transport coefﬁcients and rate coefﬁcients for ﬂuid models", http://dx.doi.org/10.1088/0963-0252/14/4/011) 2 Prahar is correct that generally we have an energy contribution of {1 \over 2} kT per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of {3 \over 2} kT. Often people talk about thermal energy being 'kT' because of the exponential expression in N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}} ... 2 Yes, the CF is a way of comparing RGs and IGs kept at same temperature and pressure. It is a comparison of volumes, as stated in the question. Hence the name "compressibility factor". The CF equation can be better stated as: P_{obs}V_{m,obs}=ZRT_{obs}, where the m stands for "molar", and "obs" means "observed". This is in contrast with ... 2 Collision theory will not help for calculating chemical rates in the liquid phase since the mass transport there is governed by diffusion. It is not perfectly clear from your question, so I will have to assume that you are interested in a two-phase gas/liquid system. First, liquid phase is much denser, molecules there are always in contact and don't have to ... 2 the discrepancy between the Boltzmann equation and the experiment may always be blamed upon the experiments. If one does things right, the Boltzmann equation is theoretically completely exact (issues will be discussed below). However, it's pretty hard to measure the distribution functions "directly", especially if you want the full dependence both on momenta ... 2 The easiest way would be to use statistical physics. Gases are effectively described with a one-particle distribution function:$$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 ...

2

I don't have any accurate measurements, but from my attempts at measuring my blankets with a ruler it looks to me as if two layers of blanket are more than twice as thick as one layer of blanket. I'd guess this is because some fibres of wool stand proud of the surface and they keep the two layers apart. Anyhow, this implies that two layers will be thicker ...

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