Thursday, May 19, 2011

Basic Physics: From Mechanics to Heat (Pt. 10)

We're now at the point to make the transition from Mechanics to Heat and Thermodynamics by applying the laws of Newton we've seen to the case of a gas and specifically the behavior of gas molecules to arrive at a form of the "ideal gas law" and thence to the kinetic theory of gases. We start using the diagram shown and focusing on the motion of one gas molecule in the direction shown. We supposed 'N' such gas molecules are in a box of dimensions: a x b x c.

Suppose the molecule has a velocity v with components: v_x, v_y and v_z e.g. such that: v^2 = v_x^2 + v_y^2 + v_z^2. When the molecule indicated hits the shaded face of the box the change in its momentum is:

delta p = (-mv_x) - (-mv_x) = - 2mv_x to the right

The impulse Ft, applied to the shaded face is then: 2mv_x to the right.

The time interval dt (d for delta) before a 2nd molecule makes a 2nd collision is:

dt = 2c/ v_x (i.e. v_x dt = 2c )

Hence, the number of collisions on the shaded face in unit time is:

Frequency = 1/dt = v_x/ 2c

Then the force exerted by a molecule on the face is:

delta (mv)/ dt= (2 mv_x) 1/dt = (2mv_x)(v_x)/2c= mv_x^2/c

The Pressure at the shaded face = (total force exerted by N molecules) divided by the area of the shaded face)

P = m/c[v_x^2)1 + (v_x^2)2 + ....(v_z^2)N]/ ab

Now substitute n = N/abc and gas density rho = nm, i.e. abc = N/m:

-> N(mv_x)^2/ abc = N(mv_x)^2/ (N/n) = nmv_x^2

= rho(v_x^2)

->P = rho(v_x)^2 but v_x^2 = v_x^2 = v^2/3 (since v_x^2 = v_y^2 = v_z^2)

so: P = rho v_^2/ 3

This represents the derivation of the average translational kinetic energy of a monatomic gas from Newtonian mechanics. By comparison, the gas equation of state is:

PV = N kT and thus: P = Nk T/ V

where T is the absolute temperatue (in Kelvin deg.) and k is Boltzmann's constant, or k = 1.38 x 10^-23 J/K). ON combining the two forms:

N kT/ V = rho v_^2/ 3 = nmv^2/3

The average translational KE = E(k) = ½ mv^2, but N/V = n

Therefore: kT = mv^2/3 and mv^2 = 3kT

so: ½ mv^2 = 3kT/2

Which is one of the more important relations in thermodynamics.

Next: Measuring the Mechanical Equivalent of Heat

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