Problem 1:

Consider an ideal gas of N atoms following classical statistics.  Find the contribution to the average Helmholtz free energy and entropy from the internal electronic level structure considering the atoms as two-level systems with the ground state energy 0 and excited state energy E.  The system's temperature is T.  The internal electronic energy levels are non-degenerate.

The Helmholtz free energy is defined as F = U - TS.  F = -Nk_BT ln(Z).  

Solution:

• Concepts:
  Boltzmann statistics
• Reasoning:
  In thermodynamic equilibrium P(E_i) ∝ g_iexp(-E_i/(kT).
  Here the degeneracy g_i is 1 for all i.
• Details of the calculation:
  P(E) = Cexp(-E/(k_BT)).
  Σ_jP(E_j) = CΣ_jexp(-E_j/(k_BT)) = 1, C = 1/Σ_jexp(-E_j/(k_BT)) = 1/Z.
  Z is partition function for a single particle in the system.
  Let β= 1/k_BT.
  Here Z = 1 + exp(-βE).
  F = -Nk_BT ln(Z) = -Nk_BT ln(1 + exp(-βE) = -Nk_BT ln(1 + exp(-E/(k_BT)).
  S = -∂F/∂T = Nk ln(Z) + (NkT/Z) ε exp(-βε) (1/kT²)
  = Nk ln(1 + exp(-E/(k_BT)) + (N/Z) (E/T) exp(-E/(k_BT).

Problem 2:

A log of length L and radius r is weighted at one end with a lump of lead so it can float vertically in water (of density ρ_water).  The total mass is M.  The log is pushed in the water by a small amount compared to the equilibrium position. 
(a)  Find the equation of motion for the log and describe the motion.
(b)  If the motion is periodic, give an expression for the period.

Solution:

• Concepts:
  Buoyancy, harmonic motion
• Reasoning:
  The log is acted on by gravity and the buoyant force,
• Details of the calculation:
  (a)  This is a one-dimensional problem.  We choose the coordinate system from the figure and let the sign be the direction indicator.  When the log floats at rest in equilibrium, we have -B = Mg.  B is the magnitude of the buoyant force.
  B = ρ_water gπr²D, where D is the length of the log that is under water in equilibrium.
  When the log is pushed down a distance z, B increases by an amount ρ_water gπr²z, and the net force on the log becomes F = -ρ_water gπr²z.
  Equation of motion for the log:  d²z/dt² = -(ρ_water/M)gπr²z.  (Hooke's law)
  The log will execute harmonic motion, z = z₀cos(ωt).
  
  (b)  ω = 2π/T = (ρ_watergπr²/M)^½.
  The period is T =  2(πM)^½/(ρ_watergr²)^½.

Problem 3:

Consider that the Universe is spherical and an adiabatic expansion of a mono-atomic hydrogen gas
(γ = 5/3).  The current Universe's radius is at least 46 billion light years, and its temperature is 2.7 K.  Estimate the temperature when the Universe had the size of a cherry (1 cm radius).

Solution:

• Concepts:
  The ideal gas law, adiabatic processes
• Reasoning:
  For an ideal gas PV = NkT. 
  For an adiabatic process dU = -dW = -PdV.
• Details of the calculation:
  We also have U = N(3/2)kT for an ideal mono-atomic gas.
  Therefore we have U = (3/2)PV,
  dU = (3/2)(PdV + VdP). 
  Equating our two expressions for dU we have -PdV = (3/2)(PdV + VdP), 
  dP/P + (5/3)dV/V = 0.  PV^5/3 = C = constant,  (PV)V^2/3= constant,
  TV^2/3 = constant.
  Therefore the initial temperature of the gas was T_i = T_f(V_f/V_i)^2/3 = T_f(r_f/r_i)².
  Plugging in numbers:  T_i = (2.7 K)[(46*10⁹ ly)*(9.46 10¹⁵ m/ly)/(0.01 m)]² ~ 5*10⁵⁷ K

Problem 4:

Find the heat capacity C_V of:
(a) one mole of argon in the gas phase,
(b) a carbon dioxide (CO₂) molecule in the gas phase (a linear molecule),
(c)  an alcohol molecule (C₂H₆O) in the gas phase (NB, not a linear molecule),
(d)  0.1 mol of lead atoms in a lead crystal,
all in the high temperature limit.  Assume ideal gases in (a) through (c).

Solution:

• Concepts:
  Heat capacity:  C = dQ/dT
• Reasoning:
  dU/dT = dQ/dT at constant volume.
  Note:  For N-atom molecules, there are 3N degrees of freedom, including 3 translational DOF (center of mass motion), and 3N - 3 internal DOF (vibrations, rotations).  For linear molecules, rotations about the axis running down the length of the molecule don't count.  For non-linear molecules, there are always 3N - 6 vibrational DOF and 3 rotational DOF (associated with the 3 principal moments of inertia).  In the calculation of the thermal energy, vibrational contributions count double: each vibrational mode contributes ½K_BT or kinetic energy and ½K_BT for potential energy.
• Details of the calculation:
  (a)  f = 3,  U = (3/2)N_AK_BT,  C_V =  (3/2) N_AK_B = (3/2)R.  (There are only three translational DOF; no internal DOF).
  (b)  N = 9;  3N - 3 = 6 .  There are 2 rotational and 4*2 vibrational DOF. 
  Hence, f = 3 + 2 + 8 = 13 --> U = (13/2)k_BT,  C_V = (13/2)k_B.
  (c)  3N - 6 = 27 - 6 = 21 vibrational DOF, f = 3 + 3 + 2*21 = 48.
  U = 2 k_BT.  C_V = 24k_B.
  (d)  Here we can ignore the translational and rotational contributions of the entire crystal to the thermal energy.  There are 3N_A/10 *2 DOF where each DOF contributes ½k_BT to the thermal energy.  
  Hence U = 0.3N_Ak_BT = 0.3RT and C_V = 0.3R.