# List the postulates of kinetic theory of gases. Derive one kinetic gas equation?

##### 1 Answer

The **main postulates** or assumptions are (for ideal gases):

- Gases are constantly in motion, colliding
**elastically**in their container. - They
**do not interact**(i.e. no intermolecular forces are considered), and are assumed to be point masses with**negligible volume**compared to the size of their container. - Ensembles of gases produce a
**distribution of speeds**, and the average kinetic energy of this ensemble is*proportional*to the translational kinetic energy of the sample.

As for "one kinetic gas equation", there is not just one. But I can derive the Maxwell-Boltzmann distribution, and from there one could derive many other equations... some of them being:

- collision frequency
- average speed
- RMS speed
- most probable speed

The latter three are derived in more detail here.

**The following derivation is adapted from Statistical Mechanics by Norman Davidson (1969). Fairly old, but I kinda like it.**

Consider the **classical Hamiltonian** for the free particle in three dimensions:

#H = (p_x^2)/(2m) + (p_y^2)/(2m) + (p_z^2)/(2m)# ,where

#p = mv# is the momentum and#m# is the mass of the particle.

The **distribution function** will be denoted as

From *Statistical Mechanics* by Norman Davidson (pg. 150), we have that

#(dN)/N = (e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n)/zeta# ,where:

#beta = 1//k_BT# is a constant, with#k_B# as the Boltzmann constant and#T# as the temperature in#"K"# .#H(p_1, . . . , p_n; q_1, . . . , q_n)# is the Hamiltonian for a system with#3N# momentum coordinates and#3N# position coordinates, containing#N# particles in 3 dimensions.#zeta = int cdots int e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n# is the classical phase integral.

This seems like a lot, but fortunately we can look at three coordinates for simplicity.

A **phase integral** is an integral over *phase space*, a coordinate system where two conjugate variables are orthogonal to each other. In this case, position

With

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(int_(-oo)^(oo) int_(-oo)^(oo) int_(-oo)^(oo) "exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)#

Now, let's take the bottom integral and separate it out:

#zeta = int_(-oo)^(oo) e^(-p_x^2//2mk_BT)dp_x cdot int_(-oo)^(oo) e^(-p_y^2//2mk_BT)dp_y cdot int_(-oo)^(oo) e^(-p_z^2//2mk_BT)dp_z#

In phase space, each of these integrals are identical, except for the particular notation. These all are of this form, which is tabulated:

#2int_(0)^(oo) e^(-alphax^2)dx = cancel(2 cdot 1/2) (pi/alpha)^(1//2)#

Notice how the answer has no

#zeta = (2pimk_BT)^(3//2)#

So far, we then have:

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(2pimk_BT)^(3//2)#

Next, it is convenient to transform into velocity coordinates, so let

#(dN(v_x,v_y,v_z; x,y,z))/(N) = (e^(-m^2(v_x^2 + v_y^2 + v_z^2)//2mk_BT)m^3dv_xdv_ydv_z)/(2pimk_BT)^(3//2)#

#= (m/(2pik_BT))^(3//2)e^(-m(v_x^2 + v_x^2 + v_z^2)//2k_BT)dv_xdv_ydv_z#

*Now, we assume that the gases are isotropic, meaning that no direction is any different from any other.*

To enforce that, suppose we enter **velocity space**, a coordinate system analogous to spherical coordinates, but with components of

#v_x = vsintheta_vcosphi_v#

#v_y = vsintheta_vsinphi_v#

#v_z = vcostheta_v#

With the differential volume element being

#(dN(v, theta_v, phi_v))/(N)#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v cos^2phi_v + v^2 sin^2theta_v sin^2phi_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

Fortunately, using

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v(cos^2phi_v + sin^2phi_v) + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#=> ul((dN(v, theta_v, phi_v))/(N) = (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv sintheta_v d theta_v d phi_v)#

We're almost there! Now, the integral over allspace for

#int_(0)^(pi) sintheta_vd theta_v = 2#

And the integral over allspace for

#int_(0)^(2pi) dphi_v = 2pi#

So, we integrate both sides to get:

#color(blue)(barul(|stackrel(" ")(" "(dN(v))/(N) -= f(v)dv = 4pi (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv" ")|))#

And this is the **Maxwell-Boltzmann distribution**, the one you see here:

**DERIVATION OF SOME OTHER EQUATIONS**

This is only a summary.

**Gas Speed Equations**

#<< v >> = int_(0)^(oo) vf(v)dv = sqrt((8k_BT)/(pim)#

#v_(RMS) = << v^2 >>^(1//2) = (int_(0)^(oo) v^2f(v)dv)^(1//2) = sqrt((3k_BT)/m)#

#v_(mp)# : Take the derivative of#f(v)# , set the result equal to#0# , and solve for#v# to get

#v_(mp) = sqrt((2k_BT)/m)#