Problem 1:
Two balloons have been filled up with air under atmospheric pressure to volumes V1 and V2, respectively. They are now submerged under water. A thin string of length L, which is run through a pulley at a fixed depth H, connects the balloons. (The radii of the pulley and the balloons are much smaller than the length of the string.) By setting the initial positions of the balloons, one can achieve a state of equilibrium. Neither balloon is rising or going down. Determine the difference in the depth of the balloons (in terms of H and L) under those conditions. The mass of the balloon skins, of the string, and of the air is negligible. The temperature of the water is constant and equal to the temperature of the air.
Solution:
 | Concepts, principles, relations
that apply to the problem: The buoyant force, the ideal gas law, pressure at depth h, Ph = Ptop + rgh The ideal gas law: PV = NkT |
| Why do they apply? The buoyant force is equal to the weight of the displaced water. To experience the same buoyant force, the two balloons must have the same volume under water. |
| How do they apply? At atmospheric pressure and at the same temperature we have V1/V2 = N1/N2 from the ideal gas law. To have the same volume under water at the same temperature we need P1/P2 = N1/N2 according to the ideal gas law. We therefore need P1/P2 = V1/V2. (Ptop + rgh1)/( Ptop + rgh2) = V1/V2. Ptop + rgh1 = (V1/V2)( Ptop + rgh2). (1 – V1/V2) Ptop + rgh1= (V1/V2)rgh2. h2 = ((V2/V1) - 1)(Ptop/rg) + (V2/V1)h1 h2 – h1 = ((V2/V1) - 1)[(Ptop/rg) + h1)] |
| Details h1
+ h2 = 2H - L |
Problem 2:
The mass of a hot air balloon and its cargo (not including the air inside) is 200 kg. The air outside is at 10 oC and 101 kPa. The volume of the balloon is 400 m3. To what temperature must the air in the balloon be heated before the balloon will lift off. (Air density at 10 oC is 1.25 kg/m3.)
Solution:
| Concepts, principles, relations
that apply to the problem: The buoyant force |
| Why do they apply? For the balloon to lift off, the buoyant force B must be greater than its weight. |
| How do they apply? The buoyant force is equal to the weight of the displaced air at 10 oC = 283 K. B = (1.25 kg/m3)(400 m3)(9.8 m/s2) = 4900 N. The weight of the balloon is 200 kg(9.8 m/s2) + weight of hot air. The hot air therefore must weigh less than 4900 N - 1960 N = 2940 N. Its mass must be less than 2940 N/(9.8 m/s2) = 300 kg. Its density must be less than r = 300 kg/(400 m3) = 0.75 kg/m3. At constant pressure, the volume of a gas is proportional to the absolute temperature. (Law of Gay-Lussac) The pressures on the inside and outside of the inflated balloon are nearly equal. The pressure on the outside is the constant atmospheric pressure. The Law of Gay-Lussac therefore applies. Since the volume of a gas at constant pressure is proportional to its temperature, its density r = m/V is proportional to 1/T. We have r1/r2 = T2/T1. r1T1/r2 = T2. (1.25 kg/m3)(283 K)/(0.75 kg/m3) = 472 K = T2. The air in the balloon must be heated to more than 472 K = 199 oC. |
| Details of the calculation: None |
Problem 3:
A Styrofoam slab has a thickness h and a density robject. What is the area of the slab, if it floats with its top surface just awash in fresh water when a swimmer of mass m is on top?
Solution:
| Concepts, principles, relations
that apply to the problem: The buoyant force |
| Why do they apply? When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight. |
| How do they apply? The magnitude of the buoyant force is equal to the magnitude of the weight wwater of the displaced water. wwater = rwaterAhg, where A is the area of the slab. The the magnitude of the weight of the object is wobject = robjectAhg + mg. We need wwater = wobject. rwaterAhg = robjectAhg + mg. rwaterAh - robjectAh = m. A = m/(rwaterh - robjecth) |
| Details of the calculation: None |
Problem 4:
An experimentalist makes independent measurements of the length and height of a rectangular feature. The values and their standard deviations are 10.62 ± 0.46 microns and 12.46 ± 0.52 microns. Calculate the perimeter P and the area A of the feature including the standard deviation of each.
Solution:
| Concepts, principles, relations
that apply to the problem: Propagation of errors |
| Why do they apply? We are asked to propagate experimental errors. |
| How do they apply? P = 2 ´ (10.62 + 12.46) ± 2 ´ (0.462 +0.522) 1/2 = 46.16 ± 1.39 microns (In addition and subtraction, the sum or the difference cannot be stated to more places after the decimal than the term with the least number of places after the decimal.) A = 10.62 ´ 12.46 ± ((0.46/10.62)2 + (0.52/12.46)2) 1/2 ´ 10.62 ´ 12.46 = 132.33 ± 7.96 = 132.3 ± 8.0 square microns (When numbers are multiplied or divided, the number of significant figures in the product or quotient cannot exceed that of the least precise number used in the calculation.) |
| Details of the calculation: More rules: When using logarithms, the significant figures are those in the mantissa (the figures after the decimal); the characteristic (figure before the decimal) indicates the placement of the decimal in the original number: e.g. log(2731) = 3.4363. |
Problem 5:
A cylinder contains a mixture of helium and
argon gas in equilibrium at 150 oC.
(a) What is the average kinetic energy of each gas molecule?
(b) What is the root-mean-square speed of each type of molecule?
mHe = 4 u, mAr = 39.9 u.
1 u = 1 atomic mass unit = 1.66´10-27
kg.
Solution:
| Concepts, principles, relations
that apply to the problem: Kinetic theory Why do they apply? The temperature is a direct measure of the average translational molecular kinetic energy, (1/2)m<v2> = (3/2)kBT |
| How do they apply? (a) The average kinetic energy of each molecule is (3/2)kBT = (3/2)1.38´10-23 J/K(423 K) = 8.76´10-21 J. The average kinetic energy is the same for both types of atoms. |
| Details of the calculation: (b) vrms2 = 2´8.76´10-21 J/m. vrms(He) = 1.62´103 m/s. vrms(Ar) = 514 m/s. The more massive molecules have a lower average speed. |
Problem 6:
How much does the entropy of a mole of a monatomic ideal gas change as it expands isothermally from V1 to V2?
Solution:
| Concepts, principles, relations
that apply to the problem: Entropy, the ideal gas law |
| Why do they apply? Change in entropy:  ,
where the subscript r denotes a reversible path.Isothermal expansion can be a reversible process. For isothermal expansion DS = DQr/T. We find DQ from the ideal gas law. |
| How do they apply? An isothermal process occurs at constant temperature. Since the internal energy of a gas is only a function of its temperature, DU = 0 for an isothermal process. For the isothermal expansion of an ideal gas we have
W is positive if V2 > V1. Since
DU = 0, the heat transferred to the gas is DQ
= W. |
| Details of the calculation: None |
Problem 7:
Give two statements of the second law of thermodynamics that are not obviously identical in conceptual content but are in fact equivalent.
Solution:
| Concepts, principles, relations
that apply to the problem: The second law of thermodynamics |
| Why do they apply? We are supposed to express the second law of thermodynamics in two not obviously identical ways. |
| How do they apply? (1) Heat cannot be taken in at a certain temperature with no other change in the system and converted into work. (2) The total entropy of a closed system is always increasing. |
| Details of the calculation: None |
Problem 8:
Consider ideal gas particles of mass m at temperature T. What is the average speed <v> (not the rms <v2>1/2 ) of the particle in terms of m, T and the Boltzmann constant kB?
Solution:
| Concepts, principles, relations
that apply to the problem: Boltzmann statistics |
| Why do they apply? The Boltzmann distribution gives the the probability of finding particles with energy E at a given temperature T. |
| How do they apply? Boltzmann statistics: Ni/N = giexp(-Ei/kT)/Si(giexp(-Ei/kT)) Here gi is the degeneracy. For this problem: n(v)dv/N = P(v)dv = g(v)exp(-mv2/(2kT))dv/ò0¥g(v’)exp(-mv’2/(2kT))dv’ = v2exp(-mv2/(2kT))dv/ò0¥v'2exp(-mv’2/(2kT))dv’, since g(v) is proportional to v2. 1/ò0¥v'2exp(-mv’2/(2kT))dv’ = (m/(2kT))3/2(4/p1/2). So <v> = ò0¥ vP(v)dv = (m/(2kT))3/2(4/p1/2) ò0¥v3exp(-mv2/(2kT))dv = (m/(2kT))3/2(2/p1/2) ò0¥v2exp(-mv2/(2kT))dv2 = (8kT/(mp))1/2. |
| Details of the calculation: Some additional remarks: The Boltzmann distribution describes the distribution of energy among classical (distinguishable) particles. P(E) = A exp(-E/(kT)) It can be used to evaluate the average energy per particle in the circumstance where there is no energy-dependent density of states to skew the distribution. To represent the probability for a given energy, it must be normalized to a probability of 1. ò0¥ Aexp(-E/(kT))dE = 1 ==> A = 1/(kT). This normalized distribution function can then be used to evaluate average energy. <E> = (kT)-1ò0¥ Eexp(-E/(kT))dE = kT. This is the average energy if the energy is randomly distributed among the available energy states and here is no energy-dependent density of states. Note that this average energy for randomly distributed energy is not the same as the average kinetic energy The average kinetic energy is found by evaluating <v2> = ò0¥ v2P(v)dv = (m/(2kT))3/2(4/p1/2) ò0¥v4exp(-mv2/(2kT))dv = 3kT/m. (1/2)m<v2> = (3/2)kT. We have an energy dependent density of states. We can use E instead of v as the variable we integrate over. g(v)dv = g(E)dE = g(E)mvdv, g(E) = g(v)/(mv). Since g(v) is proportional to v2, g(E) is proportional to E1/2. We can then rewrite all the integrals in terms of E instead of v. |
Problem 9:
(a) Consider molecules of mass m in the atmosphere above a
spherical planet of mass M and radius R. Find the escape velocity of the
molecules in terms of M, R, and the gravitational constant G.
(b) The escape velocity from part (a) is independent of the molecular mass m.
Why is it that molecules with smaller masses are lost from the planetary
atmosphere first?
Solution:
| Concepts, principles, relations
that apply to the problem: Energy conservation |
| Why do they apply? The gravitational force is a conservative force. |
| How do they apply? (a) E = T + U = 1/2mv2 - GMm/R = 0. v = (2GM/R)1/2. (b) The average speed of molecules with smaller masses is higher. <v2> = 3kT/m. If the average speed is higher, the probability the a molecule will move with a radial velocity component greater than the escape velocity away from the center of the planet is higher. |
| Details of the calculation: None |
Problem 10:
Due to the presence everywhere of the cosmic background radiation the minimum possible temperature of a gas in interstellar space is 2.7 K. This implies that a significant fraction of molecules in space may be in low-level excited states. Consider a hypothetical molecule with one possible excited state. What would the excitation energy have to be for 20% of the molecules to be in the excited state?
Solution:
| Concepts, principles, relations
that apply to the problem: Boltzmann statistics |
| Why do they apply? Boltzmann statistics tells us that Nexcited/Nground = e-DE/kT, if the degeneracy of both states is 1. |
| How do they apply? 0.2/0.8 = exp(-DE/(kT)) ln(0.25) = -DE/(kT) DE = -ln(0.25)*(kT) = -ln(0.25)*8.617*10-5*2.7 eV = 3.2*10-4 eV = 0.32 meV |
| Details of the calculation: None |
Problem 11:
An aluminum calorimeter of mass 100 g
contains 250 g of water. The calorimeter and water are in thermal
equilibrium at 10 oC. Two metallic blocks are placed in the
water. One is a 50 g piece of copper at 80oC. The other has a
mass of 70 g and is originally at a temperature of 100 oC. The
entire system stabilizes at a final temperature of 20 oC.
(a) Determine the specific heat of the unknown sample.
(b) Guess the material of the unknown sample.
Solution:
| Concepts, principles, relations
that apply to the problem: Specific heat capacity: c = DQ/(mDT) The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius. |
| Why do they apply? Energy is conserved. Knowing the specific heat capacity of the substances we can calculate how it is shared between the substances. |
| How do they apply? The temperature of the aluminum calorimeter and the temperature of the water are raised by 10 oC. The amount of energy gained by an object whose temperature is raised by DT is DQ = mcDT. The water gains DQ = 0.25 kg´1 kcal/(kgoC)´10 oC = 2.5 kcal. The aluminum gains DQ = 0.1kg´0.215 kcal/(kgoC)´10 oC = 0.215 kcal. The temperature of the copper drops 60oC. The copper looses DQ = 0.05 kg´0.092 kcal/(kgoC)´60 oC = 0.276 kcal. The unknown object therefore looses DQ=(2.5+0.215-0.276) kcal = 2.439 kcal. Its specific heat is c = DQ/(mDT) = 2.439/(0.07´80) kcal/(kgoC) = 0.436 kcal/(kgoC). The unknown material is probably beryllium. |
| Details of the calculation: None |
Problem 12:
(a) By considering the forces acting on a thin concentric shell of atmosphere at a distance r from the center of the earth, show that the variation of pressure P with height in terms of the local density of the atmosphere r and the local gravitational acceleration g is given by
dP/dr = -rg.
(b) If the temperature is 27o C at the earth’s surface and the atmosphere is assumed to be an ideal gas composed approximately of 80% diatomic 14N and 20% diatomic 16O, estimate the vertical distance over which the pressure falls to 1/e of its value at the surface (this is called the scale height of the atmosphere). To simplify your estimate you may assume the temperature to remain constant with height above the surface (which typically introduces an error of ~ 20% compared to a more realistic variation of temperature).
Solution:
| Concepts, principles, relations
that apply to the problem: Pressure, the ideal gas law |
| Why do they apply? Consider a volume of air of height h in equilibrium. It does not rise or fall. The net force on it must be zero. Therefore Pbottom - Ptop = rhg. The ideal gas law relates the density to the pressure. |
| How do they apply? (a) Since REarth >> h, the height of the atmosphere, we may consider the concentric shells locally flat. Then Fnet = PbottomA - PtopA - rADrg = 0. Ptop - Pbottom = -rDrg. dP/dr = -rg. (b)
dP/dr = -(0.2mO + 0.8mN)rparticleg
= -mrparticleg. |
| Details of the calculation: None |
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