How to deduce $E=(3/2)kT$?

Question

It says in my course notes for undergraduate environmental physics that a particle has so-called "kinetic energy"

$$E=\frac{3}{2}kT=\frac{1}{2}mv^{2}$$

Where does this formula come from? What is $k$?

Answer 1

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 can move (three independent components of velocity), so the total kinetic energy is $3\times k_B T/2$. It is important to realise that this is a statistical formula that is only true on average for a large number of particles in equilibrium: each individual particle may in fact have a kinetic energy different from this. The Boltzmann constant provides a link between the microscopic and macroscopic worlds, by relating the typical average energies of microscopic particles to the energies required to change the temperature of a macroscopic mass by a measurable amount.

Answer 2

The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzmann's constant, $k_B = 1.3806503 \times 10^{-23}~\mathrm{\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 derived from the ideal gas law as well as the experimentally verified fact that 1 mole of any gas at STP occupies a constant volume (measured to be 22.4L). We can use this relation with the mass of the given particle to prove that average kinetic energy is proportional only to temperature of the gas.

The long answer: This page provides an in-depth derivation of the formulas above.

Hope this helps!

Answer 3

This comes from Statistical Mechanics. I'm not sure of your background, but I'll post the derivation I know. If you need to expand, pick one Statistical Mechanics book like Fundamentals of Statistical and Thermal Physics by Reif.

In order to be precise, the whole idea is: considering the classical phase space $M$ endow it with a probability density function $\rho : M\to \mathbb{R}$. This probability density is undestood exactly as you might think:

$$P(A)=\int_A \rho(x)dx$$

is the probability that the system is in a microstate contained in the region of position and momentum $A$.

The question one tries to answer then is: considering the system is in thermal equilibrium at temperature $T$, what is $\rho$?

The answer, whose derivation is found in Reif's book, is (considering $M = \mathbb{R}^{6N}$ being $N$ the number of particles)

$$\rho(p,q)=\dfrac{1}{Z}e^{-\beta H(p,q)}$$

with $p = (p_1,\dots,p_N)$ the momenta of the particles, $q = (q_1,\dots,q_N)$ their coordinates, $H$ the Hamiltonian of the system, $\beta = (k_B T)^{-1}$ and $Z$ being the normalization factor called the "partition function" given by

$$Z = \int e^{-\beta H(p,q)} d^{3N}p d^{3N}q$$

Notice also that $p_i = (p_{ix},p_{iy},p_{iz})$ and $q_i=(q_{ix},q_{iy},q_{iz})$.

Now what is the energy? The energy is the mean value of $H$, considering the probability density $\rho$, in other words:

$$E = \langle H\rangle=\int \rho(p,q)H(p,q) d^{3N}p d^{3N}q=\dfrac{1}{Z}\int H(p,q)e^{-\beta H(p,q)}d^{3N}p d^{3N}q$$

but it is clear that

$$H(p,q)e^{-\beta H(p,q)}=-\dfrac{\partial}{\partial \beta}e^{-\beta H(p,q)}$$

thus

$$E = -\dfrac{1}{Z} \dfrac{\partial}{\partial \beta} \int e^{-\beta H(p,q)}d^{3N}p d^{3N}q=-\dfrac{1}{Z}\dfrac{\partial Z}{\partial \beta}$$

Now think of $N$ free particles of same mass - i.e., a gas. In that case the Hamiltonian is given by $H(p,q) = \sum \frac{p_i^2}{2m}$ - i.e, the sum of their kinetic energies. Now in this case finding $Z$ is simple because the integrals will factor:

$$Z=\int e^{-\beta \sum \frac{p_i^2}{2m}}d^{3N}pd^{3N}q = \prod_{i=1}^{N}\int e^{-\beta \frac{p_i^2}{2m}}d^3p_i \int d^3q_i,$$

but each integral over the coordinates just gives the volume $V$ of the box where the particles are, so we have

$$Z = V^N \prod_{i=1}^N \int e^{-\beta \frac{p_{ix}^2}{2m}} d^3 p_{ix} \int e^{-\beta \frac{p_{iy}^2}{2m}} d^3 p_{iy} \int e^{-\beta \frac{p_{iz}^2}{2m}} d^3 p_{iz}$$

the three integrals are identical, so we compute just once using the well known Gaussian Integral

$$Z = V^N \left(\int e^{-\beta \frac{p^2}{2m}} dp\right)^{3N}=V^N \left(\sqrt{\dfrac{2m\pi}{\beta}}\right)^{3N}=V^N \left(\dfrac{2m\pi}{\beta}\right)^{3N/2}$$

Now compute $E$

$$E = -\dfrac{1}{Z}\left(V^N (2m\pi)^{3N/2}\right)\left(-\dfrac{3N}{2} \beta^{-3N/2-1}\right)=\dfrac{3N}{2}\dfrac{\beta^{3N/2}}{V^N(2m\pi)^{3N/2}}\left(V^N (2m\pi)^{3N/2}\right)\beta^{-3N/2-1}$$

This is exactly $E = \dfrac{3N}{2} \beta^{-1}$ or else

$$E = \dfrac{3N}{2} k_{B}T$$

which for a single particle gives your result.

Answer 4

I've looked at the notes. There are several things going on here, some of them not stated. We also 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 there is an average energy per particle in a gas and that this average depends on temperature. By a rather complicated argument one can show that the average energy per particle is $kT/2$, where $T$ is the absolute temperature in kelvins and $k$ is a constant known as Boltzmann's constant.

The argument then goes as follows: The particle can move in three independent directions. That is, its motion will change its $x$, $y$, and $z$ components independently. So we have to multiply our formulas by three, which almost give us the formula you have above.

The difference is the $v^2$ term. That's defined as the sum the average energies in the three independent directions, $v^2 = v_x^2 + v_y^2 + v_z^2$. Since all directions are equivalent and independent, we assume that each contributes equally to the sum. Thus $v^2 = 3v_x^2$, where I've picked $v_x$ for convenience, I could have used any of the $v$'s.

Putting all the bits together gives us the formula above.

Answer 5

We can use the ideal gas equation and the equation for adiabatic process to work this out. (We use adiabatic since the work we do on the gas is not transfered out thermally in an adiabatic process, so it's all going to the internal energy).

We find the work generally for gases with $$dW=PdV$$ In the case of an adiabatic process, we have $$PV^{\gamma}=const$$ for a monatomic gas, $\gamma=\frac{f+2}{f}=\frac{5}{3}$, where f is the degrees of freedom (one for each of the 3 spacial dimensions), so $$PV^{\frac{5}{3}}=const$$ Solving for P, we get $$P=\frac{const}{V^{\frac{5}{3}}} $$ which we plug into our first equation, $$dW=\frac{const}{V^{\frac{5}{3}}} dV$$ Integrating, we get $$\int dW=\int \frac{const}{V^{\frac{5}{3}}} dV$$ $$W=\frac{3}{2} \frac{const}{V^\frac{2}{3}}$$

We now bring in the ideal gas equation law, $$PV=NkT$$ (Note that $nR=Nk$.) Plugging our fourth equation into the ideal gas equation law, we get, $$\frac{const}{V^\frac{5}{3}} V=NkT$$ $$\frac{const}{V^\frac{2}{3}}=NkT $$

We now plug this into our energy equation, $$W=\frac{3}{2} \frac{const}{V^\frac{2}{3}}=\frac{3}{2} NkT$$ So the energy we put into the system per atom is $$U=\frac{3}{2} kT$$

Answer 6

In Molecular study of a closed container, firstly by using first Equation of motion we get "a=Vf/t" as initially Vi=0, Then we can put in Force formula to get F=m(vf/t) Now using second equ, of motion we get "d=1/2at^2 " as Vi=0

Now put in Work done formula to have "W=1/2mVf^2 " as work done is same as K.E so E(k,E) = 1/2mVf^2 now using the relation of centre mass (in terms of velocity ecpression) m1V1+m2V2=(m1+m2)Vcm ,where Vcm is the velocity of Centrre of Mass of two body systems m1 and m2 ... As we know that Momentum(P)= mV2 then we have P1+P2=(m1+m2)Vcm as m1=m2=m (as molecules have same mass) so According to law of conservation of momentum P=P1+P2 P=2mVcm we consider only X axis motion so, we can write P=2mVx AS force is ratio of momentum and time
so from above relation we can say F=2mVx/t....(alpha) we can get t=2L/Vx ...... (Beta) by considering Molecule covers 2L distance in time t with velocity Vx.... Putting (Beta) in (alpha) F=mVx/2L/Vx .....=> F=mVx^2/L Now total average velocity expression would be V^2=Vx^2+Vy^2+Vz^2 Let particle has same Velocity in all axis so V^2=Vx^2+Vx^2+Vx^2 Or we can say V^2=3Vx^2, from this equ we can get Vx^2=1/3V^2 and this equation for N number of Particles may be written as F=mNV^2/3AL .... (GaMMA) Where A is the force area of the gas container For Volume of closed chamber in which molecules are located (since it is a Cube type rectangular chamber) so its VOlume ought to be Area*Length so Vp=(mNV^2)/3 Where we got Vp from AL of equation (GaMAA) Now Multiplying and Dividing By 2 we get PV=(2/3)(N)(1/2mV^2) .'. As K.E = 1/2mV^2 so PV=(2/3)(N)K.E We know from ideal gas equation .'. PV=nRT nRT=(2/3)(N)K.E From This we can get K.E=3/2nRT/N For unit mole of the gas n=1 and according to BOLTZMAN R/N=K so K.E= (3/2) KT

Answer 7

The "3" comes from entropy (and thereby $k$) depending on 3 different possible directions for the momentum that holds the energy.
$$ \begin{aligned} E&=dQ=TdS\\ S&=k \log (\text{states in 3D}) = 3 k \log(\text{states in 1D})\\ E&= 3 k T \log(\text{states in 1D}) \end{aligned} $$

Maybe the $1/2$ has something to do with the average of entropy when raising the $v$ from $0$ to $v$. Or trying to get a momentum $mv$ from energy via $mv (v/2)$.

$k$ is a conversion factor from temperature to energy. Specifically it is the slope of temp over heat energy removed as you go to from $Q$ energy to $0$ (or $T$ to $0$).