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22

I think you are right. A perhaps more precise relation between temperature and velocity is the Maxwell–Boltzmann distribution: \begin{equation*} P(\textbf{v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \text{exp} \left[-\frac{m ( \textbf{v} - \textbf{v}_0)^2}{2 k_B T} \right]. \end{equation*} where you see that the mean velocity $\textbf{v}_0$ and the ...


13

I think your view is correct, and you can think about the following real word example. In labs here on earth, we can use laser cooling techniques to cool atoms to $\mu$K scales in the lab frame. But the lab is on earth, and the earth is moving very fast around the sun, and the sun is moving very fast around the galactic center and so on. We don't take ...


11

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

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 ...


10

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 = ...


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 ...


8

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 ...


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 $ ...


8

Introduction Let us define the density of particles of species $s$ in a volume element, $d\mathbf{x} \ d\mathbf{v}$, at a fixed time, $t$, centered at $(\mathbf{x}, \mathbf{v})$ as the quantity $f_{s}(\mathbf{x},\mathbf{v},t)$. I assume this function is non-negative, contains a finite amount of matter, and it exists in the space of positive times and ...


7

According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ...


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 ...


6

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 ...


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

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 $N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}}$ ...


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

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 ...


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 ...


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

Background Generally, all phase transitions require some input energy in order for the transition to occur. For instance, the transition from solid-to-liquid or vice versa requires what is called the enthalpy of fusion or latent heat of fusion. This is the amount of energy needed to change the total interal energy (i.e., enthalpy) of a substance in order ...


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 ...


4

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

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 ...


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

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

The assumptions that are being made here are perfectly elastic collisions with the walls where the atom will rebound with the same speed as it came it with. Its trying to calculate the average force based on the number of interactions per unit time with the wall, and is therefore fundamentally related to how long it takes for successive collisions between ...


3

Since question is common to both chemistry and physics, answer by jheindel can be given here too.If there is any objection feel free to delete it. While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically. In deriving the ...


3

1) There are more carefully derived values assuming a distribution of particles (I do not have the reference here), but instead of considering all particles as distributed between different initial positions and velocities, lets us take them all equal to the same "typical" particle: the one with the typical velocity, let us say the median velocity $u_x$; ...


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 ...


3

The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ...



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