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Expansion of an Ideal Gas Uri Lachish, guma science Contents:
1. Introduction 2. Elastic collisions
The velocities of the particles before the collisions are vm and
vM, and after the collision they are vm' and
vM'. m vm + M vM = m vm' + M vM ' (1) and the law of conservation of energy states that: m vm2+ M vM2 = m vm'2 + M vM'2 (2) Division of equation (2) by equation (1) yields: vm - vM = -(vm' - vM') (3) That is, the relative velocity between the particles is conserved during the collision, and the direction is
reversed. (1 + m/M) vM' - (1 - m/M) vM = (m/M) 2vm (4)(1 + m/M) vm' + (1 - m/M) vm = 2vM (5) Assume that m is a microscopic particle and M is a macroscopic body. In the limit (m/M) --> 0: vM' = vM (6)vm' = -(vm - 2vM) (7)The direction of the microscopic particle is reversed and its velocity changes by 2vM, two times the velocity of the macroscopic body. The velocity of the macroscopic body is almost not affected by the collision. If the velocity of the macroscopic body is extremely slower than that of the microscopic particle, vM << vm, then the kinetic energy, DEk, transferred from body to body will be, by equations (6) - (7): D Ek = (m/2) (vM'2 - vM2) = 2m vm vM (8)The energy is proportional to the velocity of the macroscopic body and is independent of its mass. 3. The gas pressure
It will collide with the wall during a time Dt, if its distance from the wall is less than |v| Cosq Dt. The number DNcol of particle collisions with a unit area of the wall during Dt is: D Ncol = [n ∫f(v) |v| Cosq d3v] Dt (9)n is the particle density, f(v) is the probability that the particle velocity is v, normalized by ∫4p v2 f(v) dv = 1, and d3v = 4p v2 dv Sinq dq . Substituting in equation (9): DNcol = n (∫4p v2 f(v) v dv) (∫Cosq Sinq dq) Dt (10) The three-dimensional average velocity <v> of particles that collide with the wall is <v> = ∫4p v2 f(v) v dv. <v> is not zero since the integration is only over positive values. The trigonometric integral over half the sphere is ∫Cosq Sinq dq = 1/2, ( 0 < q < p/2), therefore: DNcol = n (<v>/2) Dt (11) Upon a collision with the wall the particle will transfer a momentum 2mv Cosq to the wall. The overall momentum DP, delivered to a unit area of the wall during Dt, is the sum of momenta delivered by all the colliding particles that come from all directions of the half sphere. Similarly to equations (9) - (10): DP = n 2m (∫4p v2 f(v) v2 dv) (∫Cos2q Sinq dq) Dt (12) The three-dimensional average is
<v2> = ∫4p
v2 f(v) v2 dv, p = n m <v2>/3 (13) The thermal average is calculated by applying Maxwell-Boltzmann distribution: f(v) = (m / 2pkT)3 / 2 exp(-mv2 / 2kT) (14) to the calculation of <v> and <v2> (see appendix): <v> = 4p ∫f(v) v3dv = 2 (2kT / pm)1/2 (15) <v2> = 4p ∫f(v) v4 dv = 3kT / m (16) Substituting <v2> in equation (13) yields the pressure: p = n kT (17) Equation (17) is the state equation of an ideal gas. 4. Drag within an ideal gas Gas particles that collide with the disc's front surface will leave it with a higher velocity according to the energy conservation law, equation (2). Particles that collide with the disc's back surface will leave it with a lower velocity according to the same law. Particles that collide with the front surface gain a kinetic energy: D+ = (1/2)m ((vm + 2vM)2 - vm2 ) (18) and particles that collide with the back surface lose a kinetic energy: D- = (1/2)m (vm2 - (vm – 2vM)2) (19) So that the average energy given to two particles by the disc is: D+ + D- = (1/2)m 8vM2 (20) According to equation (11) the number of particles that collide with the disc's front surface during a time interval Dt is DNcol = (n <v>/2) Dt, and a similar number of particles collide with the back surface. Therefore, by equation (20), the energy transformed to the particles during this period is: DE = (1/2)m 8vM 2(n <v>/2) Dt (21) The work of pushing the disc a distance Dx through the gas is pexDx = pex vM Dt, where pex is the force (per unit area) that pushes the disc. This work is equal to the energy given to the particles, equation (21), therefore: pexvM D t = 2m vM2 (n <v>) Dt (22) and by inserting <v> = 2 (2kT / pm)1/2 from equation (15): pex = g vM = 4n (m/2) (2kT/pm) 1/2 vM (23) So that the drag coefficient is: g = n 4(mkT/2p)1/2 (24) The faster particles, after colliding with the disc's front surface, and the slower particles, after colliding with its back surface, will thermalize with the other particles in the gas, and the drag energy will be transformed into heat. Thus, the movement of any surface within a gas, including pistons, is an irreversible thermodynamic process. It becomes reversible only in the limit of slow motion. 5. Expansion DE = vM n 2m (∫4p v2 f(v) v2 dv) (∫Cos2q Sinq dq) Dt (25) By using equation (12): DE = p vM Dt = p DV (26) vM Dt is the distance traveled by the partition and it is also the change of the volume DV of the box per unit area of the moving partition.The calculation demonstrates that elastic particle collisions with a moving partition is the transformation mechanism of microscopic thermal energy of the particles into macroscopic mechanical work of the partition. 6. Isothermal expansion D E = N kT (DV/V) (27)n = N/V , V is the box volume and N is the number of particles in it. The energy is integrated in an isothermal system:∫dE = N kT ∫dV/V (28) 7. Adiabatic expansion of a mono-atomic gas A single particle collides DNcol / N = (v/2) Dt / V times, equation (11), with the partition during Dt. During each collision the particle loses (or gains) a velocity 2vM, equation (7), twice the partition velocity in the direction of its movement. A partition that moves in the x-direction will change the vx component of the particle velocity. However, the velocity change will eventually spread to the vy and vz components. Therefore, the change of a particle velocity during Dt will be: dv = -(1/3) 2vM (v/2) Dt / V = -(v/3) dV/V (29)since vMDt = dV. The equation of adiabatic expansion of a mono-atomic gas, without internal degrees of freedom of gas particles, is calculated by integration of equation (29): v/v0 = (V0/V)1/3 (30)and by equation (13): p/p0 = (V0/V)5/3 (31)or, by the state equation (16), T/T0 = (V0/V)2/3, or, p/p0 = (T/T0)5/2. 8. Expansion of a quantum well y (x, t) = (i/2L) (e-i(kx + wt) - e+i(kx - wt)) (32)where k = p/L and w = hp2 / 2mL2 . If the boundary at L is moving slowly with a velocity vM , then the waves reflected from it will be Doppler shifted: k' = k (1 - 2vM/c) (33)where c = w/k is the velocity of the waves. During a time Dt each wave is reflected c Dt / 2L times by the moving boundary, so the overall shift of the wave number will be: D k = - k vM Dt / L = - p DL/L2 (34)When the well expands from L1 to L2 the wave number will shift, by integration of equation (34), from p/L1 to p/L2. Therefore, the ground state wave function related to L1 will be shifted to the ground state function related to L2. See: Appendix: integrals 0∫ ∞xn exp(-a2x2) dx = G((n+1)/2) / 2an+1 G is the gamma function: G(n) = (n – 1)! G(x + 1) = x G(x) G(x) G(x + 1/2) = G(2 x) p1/2 / 22 x - 1 G(1/2) = p1/2 References On the net: December 2002, section on drag added July 2006. Download as pdf: http://urila.tripod.com/expand.pdf By the author:
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