Thermodynamics Ch 18 – The Kinetic Theory of Gases

Molecular Model of an Ideal Gas

Let’s look at a molecule striking a wall of a container. The a particle of mass, m, and a velocity, vx, strikes a wall elastically we know that the impulse (F Dt) (F Dt) is Dp = (-mvx – mvx) = -2mvx (it also could be +2mvx) F Dt = -2mvx x = Dx/Dt; (time for the incoming particle also is the same for the elastic Dt = Dx/ outgoing particle, so total time is 2x/v ) F (2x/vx) = -2mvx F = -mvx^2/x According to Newton’s 3^rd law, the wall will exert an equal and opposite force on the particle. Fwall = mvx^2/x Fwall = (m/x) vx^2 by one particle And if all same type of particle (homogeneous) the average m/x for each particle will be equal, and the average velocity of each particle will be x^2 = S(vx1^2 + vx2^2 + vx3^2 + …) / N (N = number of particles)

So the total force by all the particles is F = N (m/x) x^2 v^2 = vx^2 + vy^2 + vz^2 ; since completely random x = y = z v^2 = 3vx^2 F = N /3 (m^2 / d) and since P = F / A

Which means temperature is a direct measure of the average molecular kinetic energy. And since x = 1/3 ½kBT = ½mx^2 ½kBT = ½my^2 ½kBT = ½mz^2 The theorem of equipartition energy states that each degree of freedom (x, y, z) contributes ½kbT to the energy of the system Etranslational = N (½m^2) Etranslational = 3N(½kBT) Etranslational = 3/2 NkBT Etranslational = 3/2 nRT vrms = Ö^2 rms à root mean square vrms = Ö(3kBT/m) 3/2 kBT = ½m^2 vrms = Ö(3RT/M) M = mNA Remember kB = R / NA kB = 8.314 / 6.022 x 10^23 kB = 1.38 x 10^-23 J/K

P = F / A P = F / d^2 P = N/3 (m^2 / d) / d^2 Eq 21.2 in book P = 2/3 N/V (½m vave^2)

PV = nRT kB = R/NA PV = NkBT N = n NA

PV = PV 2/3 N (½m^2) = NkbT T = 2/3kB (½m^2) 3/2 kBT = ½m^2 à K = 3/2 kBT

Example: A spherical balloon of volume 4000 cm^3 contains helium at an (inside) pressure of 1.20 ´ 10^5 Pa. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is 3.60 ´ 10^–22 J? P = 2/3 N/V (½mvave^2) Eq 21.2 in book (derived above) N = 3/2 P V / KE N = 3/2 (1.2x10^5)(0.004) / 3.6x10^-22 N = 2 x 10^24 molecules n = N/NA = 2 x 10^24/ 6.02 x 10^23 n = 3.32 moles

Demo: Crooke’s Radiometer: TH-D-CR

Molar Specific Heat of an Ideal Gas

Q = n CVΔT The energy added to the gas by heat at a constant volume only the internal energy, Eint, of the gas increases. (No work is done; Area under curve is zero since a vertical line.)

W = -∫PdV Work = 0 J ΔEint = n CVΔT + W ΔEint = Q + W (but W = 0) ΔEint = Q

Constant Volume We know Eint = Ktrans = 3/2 nRT Eint = n CVT 3/2 nRT = n CVT CV = 3/2 R

This result is if the only internal energy is translation …thus monatomic atoms. Thus we predict CV = 3/2 R for all monatomic atoms, CV = 3/2 (8.314 J/ K*mole) CV = 12.5 J/ K*mole We find that this is true experimentally (Table 12.2)

Q = n CPΔT The energy added to the gas by heat for constant pressure not only increases the internal energy, Eint, but also must account for the energy transferred out of the system by the increase volume, Work (area under the curve.) We see that T+ΔT is the same for both processes (constant pressure and constant volume), but the constant pressure path has a change of volume which represents the negative amount of work that is also done on the system such that

ΔEint = Q + W ΔEint = nCPΔT + -P ΔV n CVT = nCPΔT + -nRΔT CP - CV = R CP – 3R/2 = R CP = 5R/2

We know ΔEint for both constant pressure and constant volume is the same, n CVΔT, due to both go to the same isotherm.

A ratio of these two specific heats is useful in solving problems, thus γ = CP/CV = 5R/2 / 3R/2 γ = 1.67

Once again, the result is seen experimentally to be true.

Example: Calculate the change in internal energy of 3.00 mol of helium gas when its temperature is increased by 2.00 K.

Eint = 3/2 n R ΔT Eint = 3/2 (3)(8.314)(2K) Eint = 74.8 Joules

Adiabatic Processes for an Ideal Gas

Adiabatic Process is a process where no energy is transferred to surrounding environment. These processes are either quick processes or insulated processes and is represented by PV^γ = constant

ΔEint = Q + W ΔEint = 0 + nCVΔT -PdV = nCV dT

but for adiabatic processes (insulated where Q = 0), the internal energy only depends on Temperature

PdV + VdP = n R dT

The total differentiation of the ideal gas state, PV = nRT is

PdV + VdP = nR(-PdV/nCV) PdV + VdP = -(R/CV) P dV

Solve -PdV = nCV dT for dT = -PdV/nCV and substitute in

dV/V + dP/P = (CV – CP)/CV dV/V dV/V + dP/P = (1 – γ) dV/V γ ∫dV/V + ∫dP/P = 0

R = CP – CV and divide by PV γ = CP/CV

γ ln (V) + ln (P) = 0 which is equivalent to PV^γ = constant

Example Air in a thundercloud expands as it rises. If its initial temperature is 300 K, and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has doubled?

PV^γ = constant PV^γ = PfVf^γ T V^γ^-1 = Tf Vf^γ^-1 300 V^γ^-1 = Tf (2V)^γ^-1 (V/2V)^1.40-1 = Tf /300 Must use γ for diatomic molecules, 1.40 Tf = 227 °C

eq. 21.20 P V^γ = Pf Vf^γ (nRT/V)V^γ = (nRTf/Vf)Vf^γ (T/V) V^γ = (Tf/Vf) Vf^γ T V^γ^-1 = Tf Vf^γ^-1

The Equipartition of Energy

As we’ve seen the model for molar specific heat agrees only with monatomic gases as expected. Also our value of CP – CV = R does prove to be accurate for all gases. We need to include (a) translational KE AND (b) rotational (c) vibrational.

As we can see the distance the diatomic atom is from the y-axis so the Inertial, KR = ½Iyω^2 = 0 Joules, but Inertia is far from zero for rotation about the x & z-axes Thus the total inertia for any diatomic molecule will just be two of the three axes. (We can rotate the axis so that Inertia will always be zero about one of the axes.)

To write Eint with both (a) translational and (b) rotational energy yields Eint = ETrans + ERot (each degree of freedom contributes Eint = 3N(½kbT) + 2N(½kbT) on average ½kbT per molecule, for an Eint = 5N(½kbT) additional two degrees of freedom Eint = 5/2 nRT for rotation)

CV = 1/n (dEint/dT) CV = 5/2 R CP – CV = R CP – 5/2 R = R CP = 7/2 R γ = CP/CV = 7R/2 / 5R/2 γ = 1.40

We know from the previous section monatomic atoms γmono = 1.67 And for diatomic atoms γdi = 1.40 The results agree well with experimental data.

To write Eint with both (a) translational and (b) rotational energy (c) and vibrational yields Eint = ETrans + ERot + EVib (2 more degrees of freedom for Eint = 3N(½kBT) + 2N(½kBT) + 2N(½kBT) vibration; each atom is vibrating) Eint = 7/2 nRT Eint = 7N(½kBT)

CV = 1/n (dEint/dT) = 7/2 R CP – CV = R CP – 7/2 R = R CP = 9/2 R γ = CP / CV γ = 9R/2 / 7R/2 γ = 1.29

We know from the previous section monatomic atoms, γmono = 1.67 diatomic atoms, γdi = 1.40 And for polyatomic atoms, γpoly = 1.29 The results agree well with experimental data.

To explain molar specific heat of a solid at high temps, we use small displacements of an atom from its equilibrium position which is approximated by simple harmonic motion. Ex = ½mvx^2 + ½kx^2 (review from Ch 7 & 8) Similarly along the y & z-axes, thus for a total of 6 degrees of freedom. Eint = Ex + Ey + Ez (Remember…energy isn’t a vector) Eint = 2N(½kBT) + 2N(½kBT) + 2N(½kBT) Eint = 3NkBT = 3nRT CV = 1/n (dEint/dT) CV = 3 R (at lower temperatures classical physics just isn’t a good enough approximation; this is only good at high temps)

Example A certain molecule has f degrees of freedom. Show that an ideal gas consisting of such molecules has the following properties:

(1) total internal energy is fnRT/2; Eint = N(½kBT) for each degree of freedom theorem of equipartition energy So if, f, molecules…then Eint = fN(½kBT); where n = N/kB From Ch 19: kB = R / NA ; n = N / NA; N = nR / kB Eint = fn(½RT) (2) molar specific heat at constant vol. is fR/2; CV = 1/n ( dEint / dT) CV = 1/n (d [fn(½RT)] / dT) CV = ½fR

(3) molar specific heat at constant pressure; is (f + 2)R/2; CP – CV = R CP = ½fR + R CP = fR/2 + 2R/2 CP = ½(f + 2)R (4) specific heat ratio is g = CP/CV = (f + 2)/f γ = CP / CV γ = ½(f + 2)R / ½fR γ = (fR + 2R) / fR γ = (f + 2) / f

The Boltzmann Distribution Law

The number density from statistical mechanics is nV(E) = n0e^-E/k^B^T Thermal Excitation of Atomic Levels, p. 655 Demo: Bromine Diffusion: TH-D-BD

Distribution of Molecular Speeds

To the right is the observed speed distribution of gas molecules in thermal equilibrium. NV is called the Maxwell-Boltzmann speed distribution. The number of molecules having speeds in the range of v to v + dv is equal to the area of the shaded rectangle, NV dv. As NV approaches zero as v approaches infinity (to the right à ).

NV = 4π N(m / 2πkBT)^3/2 v^2e^-KE/kBT NV = 4π N(m / 2πkBT)^3/2 v^2e^-½mv^2 / kBT This is the curve in the active figures

Example: From the Maxwell-Boltzmann speed distribution, show that the most probable speed of a gas molecule is given vmp = (2kBT/m)^1/2. Note that the most probable speed corresponds to the point at which the slope of the speed distribution curve dNv /dv is zero.

vrms = (3kBT / m)^1/2 vrms = 1.73 (kBT / m)^1/2 vave = (8kBT / π m)^1/2 vave = 1.60 (kBT / m)^1/2 vmp = (2kBT / m)^1/2 vmp = 1.41 (kBT / m)^1/2 vrms > vave > vmp

Solution In the Maxwell Boltzmann speed distribution function take dNV / dv = 0 to find 4πN (m/2πkBT)^3/2 exp(-½mv^2/kBT) (2v – 2mv^3/2kBT) = 0 exp(-½mv^2/kBT) (2v – 2mv^3/2kBT) = 0 2exp(-mv/2kBT) 2v (1 – mv^2/2kBT) = 0 exp(-mv/2kBT) v (1 – mv^2/2kBT) = 0 and solve for v to find the most probable speed Reject as solutions v = 0 and v = ∞ Retain only (1 – mv^2/2kBT) = 0 Then vmp = (2kBT/m)^1/2

Demo: Diffusion in Air: TH-D-DA

Mean Free Path

The mean free path, L, is the average distance, d = vavet, divided by the number of collisions that occur in that time interval

L = vavt / (Acircle d)nV If motion of L = vavt / (πd^2 vavt)nV molecules are included L = 1 / ( πd^2 )nV L = 1 / √2 πd^2nV

With a collision frequency of

f = πd^2vavenV

f = √2 πd^2vavenV f = vave (√2 πd^2nV) f = vave / L

Where the inverse of the frequency is the mean free time (period)

Tmeanfree = 1 / πd^2vavenV

In an ultrahigh vacuum system, the pressure is measured to be 133 ´ 10^–10 Pa. Assuming the molecular diameter is 3.00 ´ 10^–10 m, the average molecular speed is 500 m/s, and the temperature is 300 K, find the number of molecules in a volume of 1.00 m^3 and the mean free path of the molecules

N = P V NA / R T N = 133x10^–10(1)6.02x10^–23/8.314(300) N = 3.21 x 10^12 L = 1 / √2 πd^2nV L = V / N √2 πd^2 L = 1 / 3.21x10^12 √2 π (3.00x10^–10)^2 L = 7.79 x 10^5 meters

PV = n R T PV = N/NA RT