# 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

You have to take into account the differentials. The actual equation is $$f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z = n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z.$$ Changing to spherical coordinates, we get $$\text{d}v_x\text{d}v_y\text{d}v_z = ... 8 My understanding is that this question is being asked in the context of the kinetic theory of classical gases. In that context, here is the argument: If the system is rotationally invariant, then we should have \langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle. Thus \langle v^2 \rangle = \langle (v_x^2 + v_y^2 + v_z^2 )\rangle  ... 6 The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzman's constant, k_B = 1.3806503 × 10^{-23}\frac{m^2kg}{s^2K}  and is equal to the ideal gas constant divided by Avagadro's number, \frac{R}{N_A}. So where does the equation come from? The short answer: The equation above is ... 5 The kinetic theory of gases does in fact say that molecules in a gas move very rapidly (although some move quite slow, and others even faster). However, there is another crucial component to the theory. The idea of Mean Free Path. Here, a molecule is moving very fast but doesn't get very far before hitting another molecule. This is why things like odors ... 5 This follows from the equipartition theorem. The equipartition theorem states that in thermal equilibrium, the average energy of each degree of freedom (each independent way the system can move) is k_B T/2, where T is the temperature and k_B (or just k) is called the Boltzmann constant. There are three independent directions in which a gas particle ... 5 If equipartition holds, temperature corresponds to average energy per degree of freedom, which - besides kinetic ones - include internal ones like vibrational and rotational degrees of freedom. Even in cases of structurally similar molecules with the same degrees of freedom, because of energy quantization (in particular vibrational energy), heat capacity ... 5 It seems to me that you're looking for the Boltzmann transport equation:$$ \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} $$with f the distribution in phase-space, \mathbf p the particle momentum, Q some source term, and the RHS an interaction ... 4 The Boltzmann equation is an approximation. It is obtained by closing the so-called BBGKY hierarchy that links N-particle distribution functions to N+1 particle distribution functions. The closing is obtained by introducing an assumption, the 'molecular chaos' assumption. In short, molecular chaos asumes the absence of correlation in the incoming state for ... 4 Gravity makes molecules gradually accelerate downwards. Neglecting collisions, the molecules closer to the earth would thus be (on average) moving faster. You cannot neglect collisions, at least not in the part of the atmosphere where the atmosphere acts like a gas. Collisions remain important until you get to the exobase. Above the exobase, the ... 4 Since the comment answered your question I'll just go ahead and set out a more generalised version. It's straightforward to simplify things back down to your case. Consider the following continuity equation:$$ \dot{N}(x,t) = -\nabla\cdot\vec{\Gamma}(x,t) + S(x,t), $$where \vec{\Gamma}(x,t) is the flux (in your case \vec{\Gamma}(x,t)= -D\nabla N(x,t), ... 4 The mechanism of EM radiation emission in dilute gases is different from solids, liquids and dense gases. In a solid the main source of the continuous emission, i.e. black body emission, is lattice vibrations causing local oscillations in electron density. The resulting transient dipoles emit EM just like any oscillating dipole. This isn't a resonant ... 3 As far as I know, Gallavotti proved the ergodicity of the Lorentz gas, while Sinai proved that of a system of N \leq 5 rigid spheres. Anyway, this is a minor detail. For certain aspects, a more suitable model for the Drude model is the Boltzmann gas. Lanford has shown (in 1970s, I think) that the entropy for this model is always increasing, but anyone has ... 3 To understand the physics of inflated balloons you have to understand curved membranes under tension. If there's a membrane under tension T (for simplicity assumed to be isotropic), with fluids on both sides, it will be flat (planar) unless there is a pressure difference between the two fluids. If you notice that it's curved with radius of curvature R ... 3 Prahar is correct that generally we have an energy contribution of {1 \over 2} kT per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of {3 \over 2} kT. Often people talk about thermal energy being 'kT' because of the exponential expression in N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}} ... 3 When one says that "kinetic energy is conserved in an elastic collision" that means that the total kinetic energy of the system of particles involved in the collision doesn't change. It does not mean that the kinetic energy of each particle is unchanged. For a two particle system, the kinetic energy of each will change, but the sum won't. Also, your ... 3 The velocity of the system after collision is$$ 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)\\ ... 3 The internal energy of the gas should not include the GPE. A well-known example of how things work in this way is stars. Protostars have to heat up before they can produce energy through fusion. The initial temperature increase comes from the transfer of GPE to KE as the particles condense. As the gas loses GPE, it gains KE (and therefore ... 2 Ok, so David Hammen (rightly) debunked my previous answer, but I want to have another go because I feel that he hasn't provided a clear explanation of the molecular basis of hydrostatic pressure in a gas, which was the OP's original question: I agree with David that, in a steady state, increased kinetic energy of the molecules is not the answer, as that ... 2 And what is mean kinetic energy??? In statistical mechanics Temperature is identified as for gases , so it is called the average kinetic energy identified by the root mean square of the velocity. There must be inevitably deviation from their mean kinetic energy due to collisions; how can then there be same temperature at each & every point of ... 2 how can then there be same temperature at each & every point of the system? Temperature is a bulk property of a system. Once a system has reached thermal equilibrium, it will not be the case that each object in the system has the same KE. Instead, there is a well-defined distribution of energies for the objects in the system. For everyday ... 2 The molecular chaos assumption is not that all particle velocities are always uncorrelated. If you make two particles collide, of course their final velocities will be correlated. Rather, molecular chaos assumes that between collision events, each particle has "forgotten" that its velocity came from a prior collision. Since the particles have no way to ... 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

The difference is only in the properties of the material of a body. You can see in this video If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored in the lattices If it is not elastic the body will stay ...

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)

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