Intro to Statistical Thermodynamics: Quantum Mechanics, Radiation, and Equilibrium - Prof., Study notes of Physical Chemistry

Download Intro to Statistical Thermodynamics: Quantum Mechanics, Radiation, and Equilibrium - Prof. and more Study notes Physical Chemistry in PDF only on Docsity! Chemistry 431 Lecture 1 Introduction Statistical Averaging Electromagnetic Spectrum Black body Radiation NC State University Overview Quantum Mechanics Failure of classical physics Wave equation Rotational, vibrational and translational motion Electronic structure Spectroscopy Statistical Mechanics Partition function Average energy Entropy Boltzmann’s constant Note that we k instead of R in the exponent. K is called Boltzmann’s constant and it is related to R by: where NA is Avagadro’s number. kT has units of Joules and is a measure of the thermal energy of a molecule RT has units of Joules/mole and is a measure of the thermal energy of a mole k = RNA Microscopic probability When we look at the world in terms of energy levels we can see that the probability of being in a given level depends on temperature. We define the probability of being in level j as or Pl = number of molecules in state j total number of molecules Pl = Nj NX + NY Microscopic probability We can calculate the probability of levels A and B Of course, PX + PY = 1 PX = NX NX + NY = 11 + NY/NX = 1 1 + e– ΔE/kT PY = NY NX + NY = NY/NX1 + NY/NX = e – ΔE/kT 1 + e– ΔE/kT Partition function for ladder of energy levels The denominator gives the average number of levels that are accessible at a given temperature. We can simplify it as follows: Q = 1 + e– ΔE/kT + e– 2ΔE/kT + e– 3ΔE/kT + ... Let x = e– ΔE/kT, then Q = 1 + x + x2 + x3 + ... Q – 1 = x + x2 + x3 + x3 +... and xQ = x + x2 + x3 + x3 +... since Q – 1 = xQ Q = 11 – x = 1 1 – e– ΔE/kT Calculating the energy It is not obvious how to obtain the energy for this system. We can consider the general formula and make the substitution β = 1/kT. For the energy ladder we have < E > = PjEjΣ i = 1 ∞ = e– βΔEΔE + 2e– 2βΔEΔE + 2e– 3βΔEΔE + ... 1 + e– βΔE + + e– 2βΔE + + e– 3βΔE + ... = – ∂Q/∂βQ < E > = – 1 – e– βΔE – e– βΔE –ΔE – 1 – e– βΔE 2 = ΔE e– βΔE – 1 High temperature limit As the temperature approaches infinity all of the levels become equally populated. The average energy can be calculated assuming β = 1/kT << 1. Thus, kT is a classical energy for an averaged system at high temperature. In units of joules/mole The energy of a gas is related to its PVm product (Vm is the molar volume) thus, < E > = ΔE e– βΔE – 1 = ΔE1 – βΔE+ .... – 1 = 1 β = kT < E > = RT PVm = RT Black body Radiation • An ideal emitter of radiation is called a black body. • Observation: that peak of the energy of emission shifts to shorter wavelengths as the temperature is increased. • Wien displacement law: λmaxT = 2.88 x 106 nm-K. Dilemma for Classical Physics • The maximum in energy for the black body spectrum is not explained by classical physics. • The cavity modes of the black body are predicted to be Where ρ is the radiant energy density. This function increases without bound as λ → 0. This law is known as the Rayleigh-Jeans law. ρ = 8 π k B T λ 4 p (x10° Wm") 40 30 20 10 UV Catastrophe T T —— Rayleigh-Jean 82k pT p ! | l 4 1000 2000 3000 4000 Wavelength (nm) 5000 Planck’s Innovation Classical radiation All frequencies are possible Quantized radiation Only frequencies nhν are allowed Mathematical Form of the Planck Law • The energy levels will be populated according to a thermal weighting. The higher levels will be less populated than the lower levels. • In the Planck theory the energy density becomes: ρ = 8 π hc λ 5 1 e hc /λ k B T – 1 p (x10° Wm") 40 30 20 10 UV Catastrophe T T —— Rayleigh-Jean 82k pT p ! | l 4 1000 2000 3000 4000 Wavelength (nm) 5000 Consistency with Experiment • The temperature behavior of the Rayleigh- Jeans law is recovered because ehc/λkT –1 ≈ hc/λkT as T → ∞ • The integral of the total energy is proportional to T4 which gives the Stefan-Boltzmann law, W = σT4. W is the flux or energy/area. • The Wien displacement law is recovered from differentiation of ρ. Setting dρ/dλ = 0 gives the maximum in the distribution law. Wien displacement law 0 = ∂ ρ∂ λ = ∂ ∂ λ 8 π hc λ 5 1 e hc /λ k B T – 1 Stefan-Boltzmann law W = ρ d λ 0 ∞ = 8 π hc λ 5 1 e hc /λ k B T – 1 d λ 0 ∞ = σ T 4 Use the Stefan-Boltzman law W = σ T4 (W is the flux or power per unit area) σ = 5.6704 × 10-8 kg s-3 K-4 (Watts/m2/K4) Assuming that the temperature at the surface is 5500 K and the diameter is 1.4 x 106 km. What is the radiant power of the sun? The way that the Sun's diameter is measured is by taking angular diameter measurements and then translating them to linear diameter measurements. The angular diameter of the Sun can be measured using a telescope during a total solar eclipse or by timing Mercury when it is in transit in front of the Sun. The first series of measurements were taken in the early 1700's by Jean Picard in Paris, France. First calculate the area and then the flux (power). The area is A = 4πR2 A = 4(3.1416)(7 x 108)2 A = 6.16 x 1018 m2 The flux is W = σ T4 W = 5.6704 × 10-8 (5500)4 W = 5.19 x 107 Watts/m2 The total power is P = WA= (5.19 x 107 Watts/m2)(6.16 x 1018 m2 ) P = 3.2 x 1026 Watts What is the radiant power of the sun? What is the radiant power at the surface of the earth? We use the distance from the earth to the sun to obtain the flux at the earth. The earth is Re = 1.5 x 108 km from the sun. The area irradiated is Ae = 4πRe2 Ae = 2.83 x 1023 m2 Pabs = Pemit = 6 x 1017 Watts Pemit = σ Tearth4 Aearth [Aearth = 4π Rearth2 = 2.1 x 1015 m2] Tearth = (Pemit/σ Aearth )1/4 Tearth = (6 x 1017/5.6704 × 10-8/2.1 x 1015)1/4 Tearth = 266 K This is close, but it is a little frosty. Why is this? P.S. What is 266 K in oC? What is 266 K in oF? What is the temperature at the surface of the earth? Pabs = Pemit = 6 x 1017 Watts Pemit = σ Tearth4 Aearth [Aearth = 4π Rearth2 = 2.1 x 1015 m2] Tearth = (Pemit/σ Aearth )1/4 Tearth = (6 x 1017/5.6704 × 10-8/2.1 x 1015)1/4 Tearth = 266 K This is close, but it is a little frosty. Why is this? We ignored the fact that the earth has an atmosphere! The atmosphere reflects back some of the radiated heat. What is the temperature at the surface of the earth? 1. Some molecules in the atmosphere absorb incident light. Ozone absorbs UV light and prevents harmful radiation from reaching the surface of the earth. 2. Molecules can also absorb emitted or radiated light. What is the wavelength of such light? It can be obtained from the Wien displacement law. λmaxT = 2.88 x 106 nm-K Thus, for the sun with T = 5500 K, λmax = 523 nm For the earth with T = 266 K, λmax = 10,800 nm = 10.8 μm The sun’s emission is peaked in the visible region of the Electromagnetic spectrum and the earth emits in the infrared. What is the role of the atmosphere?