Problem 1:
A hollow ball with a volume V is held in place in a tank under water by a
wire under a sloped plank as shown in the figure. The water density is ρ and average ball density is ρ/5.
The plank makes an angle α with the horizontal, with tan(α) = 1/3. What is the tension in the wire if
the whole system is accelerating horizontally with acceleration a = g/6.
Solution:

Concepts:
Freebody diagrams
 Reasoning:
Let the xaxis point towards the right and the yaxis point
up. For the net force F on the ball we have F_{y} = 0,
F_{x} = ma.
 Details of the calculation:
The buoyant force acting on the ball is B_{t} = ρVg
j  ρVa i.
F_{y} = mg + B_{y}  Ncosα = ρVg/5 + ρVg  Ncosα = 4ρVg/5  Ncosα = 0.
N = 4ρVg/(5cosα).
F_{x} = Nsinα + T + B_{x} = ma. T = 4ρVg/15  ρVg/30
+ ρVg/6 = 2ρVg/5.
Problem 2:
Particles evaporating in the zdirection from a hole in a container with an
ideal gas at temperature T.
(a) Compute the speed distribution, f*(v), of the evaporating particles.
(b) Compute the mean zcomponent of the velocity, <v_{z}>* of the
escaping particles.
(c) Compute the mean energy of the particles leaving the container.
Assume that the equilibrium in the interior of the container is not disturbed by
the evaporating particles.
MaxwellBoltzmann speed distribution: f(v) = (m/(2πkT))^{3/2 }4πv^{2}exp(mv^{2}/(2kT)
Solution:
 Concepts:
The MaxwellBoltzmann speed distribution, geometry
 Reasoning:
The atoms escape through a small hole of area A. We need to find the rate at
which atoms with different v or v_{z} escape from the hole.
 Details of the calculation:
Consider an area dA and gas atoms with speeds between v and v + dv in a ring
at a distance between r and r + dr from dA. Assume that the gas atoms travel in
straight lines. The probability that gas atoms from the ring will reach
the area dA is dA cosθ/(4πr^{2}). This is the solid angle subtended by
dA at any dV in the ring.
The number of atoms from the ring reaching dA with speeds between v and v +
dv
is
nf(v)dv2πr^{2}sinθ dr dθ dA cosθ/(4πr^{2}) = ½nf(v)dv sinθ
dr dθ
dA cosθ.
Here n = N/V is the density of the gas.
The total number of atoms with speeds between v and v +
dv reaching dA in a time
interval dt comes from rings with radii between r = 0 and r = vdt. Per
unit area per unit time it is therefore given by
∫_{0}^{π/2}
dθ nf(v)dv v sinθ cosθ/2 = ¼ n v f(v) dv.
The escape rate per unit area per unit time is
R_{esc} = ½n∫_{0}^{∞}dv ∫_{0}^{π/2}dθ
sinθ cosθ v f(v) = ¼ n ∫_{0}^{∞} vf(v) dv = n (kT/(2mπ))^{½}.
We therefore have f*(v) = ¼ n v f(v)/R_{esc} = ½(m^{2}/(kT)^{2})^{
}v^{3}exp(mv^{2}/(2kT).
(b) Atoms from a ring reaching dA with speed v have a velocity component v_{z}
= vcosθ.
The average velocity component <v_{z}> for the particles escaping from
the hole is given by
[½n∫_{0}^{∞}dv ∫_{0}^{π/2}dθ sinθ cos^{2}θ
v^{2} f(v)]/R_{esc}
= (3/4)(2πkT/m)^{½} [∫_{0}^{π/2}dθ sinθ cos^{2}θ]/[
∫_{0}^{π/2}dθ sinθ cosθ] = ½(2πkT/m)^{½}.
(c) The mean energy of the particles is given by <E> = ½m∫_{0}^{∞}v^{2}f*(v)
dv.
<E> = ¼(m^{3}/(kT)^{2})∫_{0}^{∞}v^{5}exp(mv^{2}/(2kT)
dv = 2kT.
Problem 3:
(a) One mole of ideal gas with constant heat capacity C_{V} is
placed inside a cylinder. Inside the cylinder there is a piston which can move
without friction along the vertical axis. Pressure P_{1} is applied to
the piston and the gas temperature is T_{1}.
At some point, P_{1}
is abruptly changed to P_{2} (e.g. by adding or removing a weight
from the piston). As a result, the gas volume changes adiabatically. Find the
temperature T_{2} and the volume V_{2} after the thermodynamic
equilibrium has been reached in terms of C_{V}, P_{1}, T_{1},
and P_{2}.
Use the relation between heat capacities C_{V} and C_{P} to
simplify the formulas.
Definition of C_{V}: dU = C_{V}dT, C_{P} = C_{V}
+ R
(b) After the thermodynamic equilibrium has been established in part (a),
the pressure is abruptly reset to its original value P_{1}. Compute
final values of the temperature T_{f }and the volume V_{f}
after the thermodynamic equilibrium has been reached again.
Compute the difference in temperatures (T_{f}  T_{1}) and show
that it is quadratic in (P_{2}  P_{1}).
Comment on the sign of the temperature difference.
Solution:
 Concepts:
The ideal gas law, adiabatic processes, the first law of thermodynamics
 Reasoning:
ΔQ = 0 for adiabatic processes, and thus the first law of thermodynamics
becomes
ΔU + ΔW = 0, where ΔW is the work done by gas, and U is its internal energy.
 Details of the calculation:
(a)
Using ΔW = P_{2}ΔV and ΔU = C_{V}ΔT we obtain:
C_{V}(T_{2}  T_{1}) + P_{2}(V_{2} 
V_{1}) = 0.
Using the ideal gas PV = RT we get
T_{2} = (C_{V}T_{1} + P_{2}RT_{1}/P_{1})/(C_{V}
+ R) = T_{1}(C_{V} + RP_{2}/P_{1})/(C_{V}
+ R),
V_{2} = RT_{2}/P_{2} = (C_{V}T_{1}
R/P_{2} + R^{2}T_{1}/P_{1})/(C_{V} + R).
Using the relation between heat capacities C_{P}  C_{V} = R
we get
T_{2} = T_{1}(C_{V} + RP_{2}/P_{1})/C_{p},
V_{2} = RT_{2}/P_{2} = (RT_{1}/P_{1})(C_{V}P_{1}/P_{2}
+ R)/C_{p} = V_{1}(C_{V}P_{1}/P_{2} +
R)/C_{p}.
(b) Using the result from part (a) we immediately obtain
T_{f} = T_{2}(C_{V} + RP_{1}/P_{2})/C_{p},
V_{f} = (RT_{2}/P_{2})(C_{V}P_{2}/P_{1}
+ R)/C_{p} = V_{2}(C_{V}P_{2}/P_{1} +
R)/C_{p}.
T_{f}  T_{1} = T_{1}(C_{V} + RP_{2}/P_{1})(C_{V}
+ RP_{1}/P_{2})/C_{p}^{2}  T_{1}= T_{1}(C_{V}^{2} + R^{2} +C_{V}RP_{2}/P_{1}
+ C_{V}RP_{1}/P_{2})/C_{p}^{2}  T_{1}
= RT_{1}C_{V} (2 +P_{2}/P_{1} + P_{1}/P_{2})/C_{p}^{2}
= (V_{1}/P_{2})(C_{V}/C_{p}^{2})(2P_{1}
P_{2} + P_{2}^{2} + P_{1}^{2})
= (V_{1}/P_{2})(C_{V}/C_{p}^{2})(P_{2}
 P_{1})^{2}The change in temperature is is quadratic in (P_{2}  P_{1}). It is always positive, as expected from the second law of thermodynamics.
Problem 4:
The ground state of the neutral lithium atom is doubly
degenerate. The first excited state is 6fold degenerate, and it is at an
energy 1.2 eV above the ground level.
(a) In the outer atmosphere of the sun, which is at a temperature of 6000 K,
what fraction of the neutral lithium is in the first excited level?
Since all
the other levels of Li are at much higher energy, it is safe to assume that they
are not significantly occupied.
(b) Find the average energy of a lithium atom at temperature T. (Again,
consider only the ground and first excited level.)
(c) Find the contribution of these levels to the specific heat per mole, C_{V},
and sketch C_{V} as a function of 1/T. Discuss the curve.
Solution:
 Concepts:
Boltzmann statistics
 Reasoning:
The probability that a system is in a
particular state s with energy E_{s} is proportional to exp(E_{s}/(k_{B}T)).
 Details of the calculation:
(a) Let ∆E = 1.2 eV, β = 1/(k_{B}T), k_{B} = Boltzmann
constant.
P(E_{0 }+ ∆E) ≡ P(∆E) = 6*exp(β(E_{0}
+ ∆E))/[2*exp(βE_{0})
+ 6*exp(β(E_{0 }+ ∆E))]
= 6*exp(β∆E)/[2 + 6*exp(β∆E)].
We can safely choose our energy scale so that E_{0} = 0.
∆E = 1.2 eV, k_{B} = 8.62*10^{5} eV/K,
T = 6000 K, β∆E = 2.32.
P(E) = 6*exp(2.32)/(2 + 6*exp(2.32)) = 0.23.
(b) <E> = ∆E*6*exp(β∆E)/[2 + 6*exp(β∆E)].
= ∆E*3*exp(β∆E)/[1 + 3*exp(β∆E)]
=
3∆E/[exp(β∆E) + 3] = 0.27 eV higher than the ground state energy.
(c) C_{V} = N_{A}∂<E>/∂T_{V}
= N_{A}3∆E^{2}exp(β∆E)(∂β/∂T)[exp(β∆E)
+ 3]^{2}_{ }
= 3N_{A}k_{B}β^{2}∆E^{2}exp(β∆E) [exp(β∆E) + 3]^{2}.
N_{A} = Avogadro's number.
β∆E is proportional to 1/T.
As T > ∞,
β∆E > 0, C_{V} > 0. All atoms are in the excited state.
As T
> 0,
β∆E > ∞, C_{V} > 0.
The thermal energy available is not
enough to overcome the energy gap ∆E.
The function has a maximum between
these limits.
Problem 5:
A glowing incandescent lamp
filament may be regarded as a black body. If the filament is heated to 2400 K
with a power of 100 W, find the surface area of the filament in cm^{2}.
Solution:
 Concepts:
The StefanBoltzmann Law
 Reasoning:
The StefanBoltzmann Law gives the total energy emitted at all wavelengths by a black body.
 Details of the calculation:
Radiated power = σ * T^{4} * Area .
Here σ is the StefanBoltzmann constant, σ = 5.67*10^{8}W/(m^{2}K^{4}).
Area = 100W/(5.67*10^{8}W/(m^{2}K^{4)}*(2400 K)^{4})
= 5.32*10^{5} m^{2} = 0.532 cm^{2}.