Problem 1:
A planetesimal orbiting the Sun absorbs solar energy at a rate equal to the
solar flux at the orbital radius multiplied by the cross-sectional area of the
planetesimal multiplied by (1 - the albedo). The temperature of the
planetesimal can be estimated by setting the rate of solar energy absorption
equal to the rate at which the planetesimal is radiating energy.
(a) If we assume the planetesimal radiates as a black body, and the solar
luminosity is 3.8*1026 Watts, calculate the temperatures of
planetesimals as a function of distance from the Sun and albedo.
(b) For water ice, with an albedo of 0.35 and sublimation temperature of 200 K,
find the minimum distance from the Sun that a water ice planetesimal can remain
solid.
For reference, the Earth orbital radius is 1.5*1011 m.
Solution:
- Concepts:
The Stefan-Boltzmann Law
- Reasoning:
Radiated power = emissivity * σ * T4 * Area, σ = 5.67*10-8W/(m2K4).
- Details of the calculation:
(a) In steady state the planetesimal's surface radiates as much energy as it
absorbs per second.
The total power radiated by the sun is P = 3.8*1026 W.
At a distance R from the sun, the intensity of the sunlight is P/(4πR2).
If the planetesimal has a radius r, the absorbed power is
Pabs = Pπr2(1 - albedo)/(4πR2).
In equilibrium this equals the radiated power Prad = σ*T4
*4πr2.
Pπr2(1 - albedo)/(4πR2) = σ*T4 *4πr2.
P(1 - albedo)/(16 σ πR2) = T4.
T = [P(1 - albedo)/(16 σ πR2)]
¼.
(b) Given P, T, and albedo, we can solve for R. T must be less than 200 K
for the ice to exist, so
R2min = 0.1625*3.8*1026 W/(4π*5.67*10-8
Wm-2K-4*1.6*109 K4).
R = 2.33*1011 m.
Problem 2:
(a) Find an expression for the variation in atmospheric pressure P with
elevation h above sea level assuming that the temperature of the atmospheric air
T and the acceleration due to gravity g are both constant with elevation and
that the atmospheric air is an ideal gas with molar mass M.
(b) Now if the pressure at sea level is P0 = 1 atm, calculate the
atmospheric pressure at the top of Clingmans Dome, which at 2025 m is the
highest point in the Great Smoky Mountains National Park. Assume that the
temperature remains constant at T = 20oC,
and the molar mass of air is M = 28.8*10−3 kg/mol. Give your
result in atm.
(c) Melting and boiling points depend on pressure. A good approximation for
the rate of change of the melting or boiling temperature with pressure is given by the Clausius-Clapeyron relation dT/dP
= T ∆V/L, where L is the latent heat of the substance (per kg) and T is the
melting or boiling temperature at standard pressure, and ∆V is the change in
volume (per kg).
To first order estimate the change in the boiling point of water on top of
Clingmans Dome. (For small changes of the melting point temperature
expand ∆T = (dT/dP)∆P.)
The density of steam at 1 atm and 100 oC is 0.59 kg/m3,
and the latent heat of vaporization for water (per kg) is 2264.7 kJ.
Solution:
- Concepts:
The ideal gas law
- Reasoning:
Let us consider a column of gas containing ρparticle particles of
mass m per unit volume near the Earth's surface.
The pressure times the area (i.e. the force) at height y must exceed that at
height y + dy by the weight of the intervening gas.
We need Py+dyA - PyA = -mρparticleVg = -mρparticlegAdy,
or dP = -mρparticlegdy.
The ideal gas law lets us express ρparticle in terms of the
temperature and the pressure.
- Details of the calculation:
(a) T is constant.
From the ideal gas law we know that P = ρparticlekT, with k being
the Boltzmann constant. We may therefore write
dP = -P(mg/kT)dy, or dP/P = -(mg/kT)dy.
This integrates to
Pf = P0e-mg(yf-y0)/(kT) = P0e-Mg(yf-y0)/(RT).
The pressure decreases exponentially with altitude.
(b) Pf = P0exp[-Mg(yf
- y0)/(RT)]
= (1.01*105)*exp[-28.8*10−3*9.8*2025/(8.314*293)]
kPa
= 7.98*104 kPa = 0.8 atm.
(c)
Consider 1 kg of water. It has a volume of 10-3 m3.
The volume of 1kg of steam is 1/0.59 m3 = 1.695 m3.
Compared to Vsteam, the volume of the liquid is negligible, so we can
set ∆V = Vsteam.
dT/dP = 373*1.695/(2264.7*103) K/Pa = 2.79*10-4 K/Pa.
∆T = -(2.79*10-4 K/Pa)*(21.2*103 Pa) = -5.92 K.
The boiling temperature is lowered by ~6
oC.
Problem 3:
In a Wilson cloud chamber at a temperature of 20 degrees C, particle tracks
are made visible by causing condensation on ions by an approximately reversible
adiabatic expansion of the volume in the ratio final volume/initial volume = 1.375.
The ratio of the specific heats of the gas at constant pressure and at
constant volume is CP/CV = 1.41. Estimate the gas
temperature after the expansion.
Solution:
- Concepts:
Specific heat,
first law if thermodynamics, the ideal gas law
- Reasoning:
We are supposed to use the definitions of the specific
heats of the gas at constant pressure and at constant volume and the given
ratio of the two to estimate the temperature.
- Details of the calculation:
Remember or derive for an adiabatic
expansion: P1V1γ = P2V2γ,
γ = Cp/CV.
[Derivation:
at constant
pressure: dQ = nCPdT
at constant volume: dQ = nCVdT
Energy conservation:
at constant pressure: dU = dQ - dW = nCPdT
- PdV = nCPdT - nRdT
at constant volume: dU = nCVdT
dU depends only on the change in temperature,
so nCPdT -
nRdT = nCVdT, CP - CV = R.
Adiabatic
expansion: dU = -PdV = nCVdT, dT = -PdV/(nCV).
Ideal gas law: PdV + VdP = nRdT = -nRPdV/(nCV) = -(CP
- CV)PdV/CV.
(Cp/CV)PdV +
VdP = 0, (Cp/CV)dV/V + dP/P = 0.
Integrate: (Cp/CV)lnV
+ lnP = 0. PVγ = constant, γ = Cp/CV.]
P1V1γ = P2V2γ.
P1 = P2(1.375) γ = 1.57 P2.
P1V1
= nRT1, 1.57 P2V1 = nRT1, P2V2
= nRT2.
Ratio: 1.57 (V1/V2) = T1/T2.
T2 = 293 K/1.14 = 257 K.
Problem 4:
An object of mass m and density ρ = (3/4)ρwater is fixed to the
bottom of an aquarium with a string and is fully submerged in water. The
aquariums sits on a truck bed and the truck is at rest.
(a) What is the tension in the string?
(b) The truck now is accelerating with acceleration of magnitude g/10
along a straight line. What is the tension in the string and what is the
angle θ the string makes with the vertical?
(c) What is the angle α the water's surface makes with the horizontal?
Solution:
-
Concepts:
Buoyancy, free-body diagrams
- Reasoning:
Let the x-axis point towards the right and the y-axis point
up.
Let the acceleration point in the positive x-direction.
When the truck is at rest the net force F on m is zero.
When the truck is at accelerating the net force F on m is mg/10
i.
- Details of the calculation:
(a) The weight of the displaced water is w = ρwaterVobjectg,
with Vobject = m/ρ = 4m/(3ρwater).
w = 4mg/3.
B - mg - T = 0. 4mg/3 - mg - T = 0, T = mg/3 is the tension in the
string.
(b)
The buoyant force is the force the surrounding water would have to exert on the
displaced water to accomplish the state of motion given in the problem in the
absence of the object.
B = 4mg/3 k + 4mg/30 i.
4mg/3 - mg - Ty = 0, Tx + 4mg/30 = mg/10.
Ty = mg/3, Tx = -mg/30, T = mg/2.985.
θ = atan(-Tx/Ty) = atan(1/10) = 5.71o.
The string makes an angle θ with the y-direction, towards the positive
x-direction.
(c) Observe a volume dV = (dx)3 in the accelerating aquarium from
an inertial frame.
Observe a volume dV = (dx)3 in the accelerating aquarium from
an inertial frame.
The pressure in the liquid depends on the depth
below the surface. The pressure on the left side is higher that the pressure on
the right side of dV by an amount ρg|dy|.
The net force on dV is F = ρg|dy|(dx)2. It accelerates the water in dV.
We have F = mwa.
ρg|dy|(dx)2 = ρ(dx)3g/10. dy/dx =
-1/10 is the
magnitude of the slope. The slope is negative.
The water slopes upward towards the back of the aquarium. The slope |dy|/dx = tan(α).
Or:
The the apparent weight of the water, -mwgj - mwai
is normal to the surface of the water. The surface makes an angle α with
the x-direction.
tan(α) = (g/10)/g = 1/10. The water slopes upward towards the back of
the aquarium.
Problem 5:
One mole of a monatomic ideal gas is driven around the cycle A B C A shown on
the PV diagram below. Step AB is isothermic, with a temperature TA
= 500 K. Step BC is isobaric, and step
CA is isochoric. The volume of the gas at point A is VA = 1
liter, and at point B is VB = 4 liter.
(a) What is the pressure PB at point B?
(b) What is the net work done by the gas in completing one cycle A B C
A?
(c) What is the entropy change SC - SB?
Provide numerical answers in SI units.
Solution:
- Concepts:
The ideal gas law PV = nRT, work done by the system W = ∫PdV,
entropy
- Reasoning:
For an isothermal process W = nRT ln(Vi/Vf).
- Details of the calculation:
(a) PBVB = nRTB = nRTA.
PB = RTA/VB = 8.31*500/(4*10-3)
Pa = 1.04*10-6 Pa.
(b) WAB = RTA ln(4). WBC = -PB3VA.
WCA = 0.
Wnet = RTA ln(4) - 3PBVA
= 8.31*500*ln(4) - 8.31*500*3/4 = 2.64*103 J.
(c) The pressure is constant.
dU = dQ - dW, dS = dQ/T. dS = dU/T + PdV/T
U = 3RT/2, dS = (3/2)RdT/T + RdV/V.
SC - SB = (3/2)R∫TBTCdT/T + R∫VBVCdV/V
= (3/2)Rln(TC/TB) + Rln(VC/VB) .
Since P is constant TC/TB = VC/VB
= VA/VB = 1/4.
SC - SB = (3/2)R ln(1/4) + R ln(1/4) = R ln(1/4)(5/2)
= -28.8 J/K.