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Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer. Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(...


12

Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing). Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic ...


11

What you say is a good idea, but is not quite correct because we must maximize the distribute subject to two constraints. I will reiterate the derivation following along with McQuarrie's Statistical Mechanics. Constraints in the Ensemble: The constraints I mention are actually the properties that determine the ensemble we are working with: $$ \sum_ja_j=A $$ $...


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Temperature vs kinetic energy [OP:] I've read at many places that temperature is the average kinetic energy of particles present in an object. Temperature has to do with the average kinetic energy of particles, but to say the two concepts are the same is incorrect. What is correct is that if the particles in two mono-atomic gas samples have the same ...


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The partition function $q=\sum_i\exp(-E_i/k_BT)$ in your question can be regarded as the effective number of levels accessible to the molecule at a given temperature. It also means that in the equilibrium distribution the partition functions tells us how the systems are partitioned or divided up among the different energy levels. In determining the form of ...


8

In the solution there are two types of molecules $N_1$ and $N_2$. Assume that they do not interact with one another but simply occupy particular 'lattice' sites by blocking them. The total number occupied sites is $N=N_1 + N_2$. The first molecule can be placed at any of the $N$ sites, the second at $N-1$ empty sites and so on. The total number of ...


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Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$. Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation $$\begin{align} Q_\mathrm{tr,...


8

A simple validation The result you quoted is the average translational kinetic energy for an ideal gas. First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics. The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each ...


7

Here follows a complete mathematical derivation of the expressions for the internal energy and isochoric heat capacity. I am not sure why taking the high-temperature limit is unphysical, but maybe someone else knows. The potential is $$ V(x) = \frac{k_0x^2}{2} + \alpha x^4 $$ for one dimensional, localized oscillators. The Hamiltonian then becomes $$ H(p,...


7

I'm confused why you're interpreting the partition function as a count of states. It can't be a count; it's continuous. The zero point energy doesn't actually matter because you can just shift the energy scale so that it starts at zero. The main point of zero point energy is that the ground state of the harmonic oscillator is such that there is energy, and ...


7

Quantum mechanics is about the physics of very small things, molecules and smaller. Classical mechanics is about macroscopic things. Quantum mechanics covers the whole of classical mechanics as well, but in the macroscopic limit both become equivalent. For example discretized energy states become so close, that you can thing of them as a continuum of states....


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Within statistical mechanics (SM) a molecular property $X$ is computed by $$\left<{X}\right>_{SM}=\sum^{states}_i X_i p_i$$where $X_i$ is the value of $X$ for energy state $i$ and $p_i$ is the probability of being in energy state $i$ with energy $E_i$:$$p_i=\frac{e^{-E_i/kT}}{\sum_i e^{-E_i/kT}}$$Within molecular dynamics (MD) the corresponding ...


7

Temperature is related to kinetic energy, but it can't be simply equated to the average kinetic energy of the system. As I wrote in response to another answer, different systems can have different average kinetic energies/particle, but the same temperature. E.g., at the same temperature the avg. kinetic/energy particle of a diatomic gas is greater than ...


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In principle you can, assuming you are given $V(\{\bf{r_i}\})$ where $\{\bf {r_i} \}$ is the set of variables that define a configuration in the system - this is typically the coordinates of the atoms, but it might include things like external fields. The problem is in practice it will be at best horribly, horribly inefficient, and in fact almost certainly ...


6

The answer to this is linked to how you define the “volume” of your system, and the boundary conditions. The simplest case is simply to consider an isolated systems, like a molecule in gas phase. This corresponds to $N$ atoms (or molecules or particles) with no restrictions of their coordinates and no boundary conditions. This is essentially an infinite box. ...


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Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that $kT$ has to have the same units as $E_i-E_j$, so the terms in the exponent cancel. If you're measuring temperature in $\pu{K}$ and Boltzmann's constant in $\pu{J/K}$, then the units of energy you have to use are joules, $\pu{J}$.


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Given the accuracy it sounds like you're looking for, and the systems you're studying, this indeed appears to be a Very Hard Problem. I think your intuitions are well founded. Briefly, to your questions: 1) How to handle the low-frequency modes? A lot of manual exploration of the PES, doing things like running normal mode scans and optimizing intermediate ...


6

From plain Wikipedia, Statistical mechanics is a branch of theoretical physics that uses probability theory to study the average behaviour of a mechanical system, where the state of the system is uncertain. (Thus, very useful for things like thermodynamics, where everything we measure is an average in some sense.) Statistical mechanics per se doesn't ...


6

To answer the question of the title. In Adam-Gibbs theory the activation enegy is defined as [1]: $$E_\mathrm a=-R\left[ \frac{\partial \ln k}{\partial (1/T)} \right]$$ Now clearly this is a first order linear differential equation, as you were looking for. This can easily be solved by seperation of variable: $$\int E_\mathrm a\, \mathrm d(1/T) = -R\int \...


5

In principle, you can use the integrated Clapeyron and Clausius–Clapeyron equations to calculate $P$ vs. $T$ co-existence lines and find the crossing point which is at the triple point. You need the $\Delta H$ for fusion and evaporation and the densities of the solid and liquid phases. The equations are $$ p_\text{liq-vap}=\exp\left(\frac{\Delta H_\text{vap}}...


5

You really have to calculate out things exactly, and particularly carefully if electrons are in the same orbital, e.g. $p^2$, when the Pauli principle has to be checked. Its a bit tedious but it can be very quick with a bit of practice especially if electrons are in different orbitals. You start by calculating the spin, orbital and total of all angular ...


5

You're on the right track. Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet. The sum in the denominator is called the partition ...


5

(1) You have to maximize $\Omega$ subject to the constraints that the total number of particles and the total energy is constant, which is more complicated than just taking the derivative and setting it equal to zero. You can do it using Lagrange multipliers or (which is more typical in an undergraduate course) move to the canonical ensemble where the ...


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Ignoring multiplicative constants, the required integral is (in general form) $$ \int \sqrt{x} e^{ax} \mathrm{d} x = \frac{1}{a}\sqrt{x}e^{ax}+\frac{i\sqrt{\pi}}{2a^{3/2}} \mathrm{erf}(i\sqrt{ax})$$ where $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\mathrm{d}t$$ Setting $x=E$, $a=\beta$, and the integration limits as $0$ and $\infty$, ...


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$E_n$ represents the energy of the $n$th electronic state relative to the ground state (i.e. the ground-state electronic energy $E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies $E_1, E_2, \cdots$ are far greater than $k_\mathrm{B}T$, such that $\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$. Under these conditions, the ...


5

The actual partition function is unimaginably formidable. For just $N$ point particles in a 3D box, it's already got $3N$ dimensions. If the box is length $L$, GROMACS would probably divide the box into a machine-precision grid and calculate the energy from $\sim L \times 10^{10}$ values in each dimension. The partition function would incorporate all of that ...


4

We first come across the canonical partition function $Q(\beta) := \sum_i e^{-\beta E_i}$ as a normalization constant for the probability that a given microstate is occupied, as given by the Boltzmann distribution. It is the Boltzmann distribution that allows for all the remarkable properties of the normalization constant $Q$. We mention the two most ...


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In the simplest terms and most convenient definitions, it represents the total amount of states that the energy can be in. The probability expression you wrote emphasizes this. A probability is a part over a whole. So in the case of Boltzmann's distribution, it is one state over the sum of all the states.


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I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion means ...


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I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently. This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes. That is, if we ...


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