Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 1 • 1.9: What is the volume of one mole of air, at room temperature and 1 atm pressure. From the ideal gas law, we have P V = nRT . Therefore: V V (1 mol)(8.31 J/mol/K)(300 K) nRT = P (105 N/m2 ) ≈ 0.025 m3 ≈ 25 L = • 1.10: Estimate the number of air molecules in an average-sized room. Consider that an average room is about 4 metres squared and 3 metres high. The number of air molecules (at the standard room temperature and pressure) is PV (105 N/m2 )(4 × 4 × 3 m3 ) = kB T (1.38 × 10−23 J/K)(300 K) N = 1.2 × 1027 ≈ ×1027 N = which is about 2000 moles. • 1.11: Rooms A and B are the same size, and are connected by an open door. Room A, however, is warmer (perhaps because its windows face the sun). Which room contains the greater mass of air? Explain carefully. Pressure is the same in both because they are connected by an open doorway; if the pressure were different then a wind would blow from one room to the other. The volume of both rooms is also the same. Therefore PA VA = PB VB and, given the ideal gas law, NA kTA = NB kTB as well. Thus the cooler temperature room has the higher number of molecules, and thus the greater mass. • 1.12: Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a small molecule like N2 or H2 O? The volume per molecule for an ideal gas at room temperature and atmospheric pressure is: V N V N kB T (1.38 × 10−23 J/K)(300 K) = = P 105 N/m2 = 4.1 × 10−26 m3 ≈ 41 nm3 The average distance between molecules can be estimated as V N 1/3 ≈ 3.5 nm3 The diameter of a molecule like N2 or O2 is only a few Angstroms, approximately 10× smaller than this average distance. • 1.13: A mole is approximately the number of protons in a gram of protons. The mass of a neutron is about the same as the mass of a proton, while the mass of an electron is usually negligible in comparison, so if you know the total number of protons and neutrons in in a molecule (i.e. its “atomic mass”), you know the approximate mass (in grams) of a mole Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 2 of these molecules. Referring to the periodic table, find the mass of each of the following: water, molecular nitrogen, lead, and quartz (SiO2 ) H2 O = 2(1) + 1(16) = 18 grams/mol N2 = 2(14) = 28 grams/mol Lead = Pb = 207 grams/mol SiO2 = 1(28.08) + 2(16) = 60 grams/mol • 1.14: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), O2 (21%) and Ar (1%). Use weighted average to calculate the mass of air: wi fi (wi = weights) wi 0.78mN2 + 0.21mO2 + 0.01mAr m = 0.78 + 0, 21 + 0, 01 f = i From the periodic table, mN2 = 28.014 g/mol, mO2 = 32.00 g/mol, and mAr = 39.95 g/mol Therefore, the mass of a mole of dry air: m = 0.78(28.014) + 0.21(32) + 0.01(39) m = 28.92 g/mol • 1.15: Estimate the average temperature of the air in a hot air balloon (see Fig. 1.1 in book). Assume that the total mass of the unfilled balloon and payload is 500 kg. What is the mass of the air inside the balloon? The upward buoyant force on the balloon is equal to the weight of the air displaced. Assuming that this force is approximately in balance with gravity, we can write: ρ0 V g = (M + ρV )g or ρ0 − ρ = M/V where ρ0 is the density of the surrounding air, V is the volume of the balloon, M is the mass of the unfilled balloon and payload, and ρ is the density of the air inside the balloon. According to the ideal gas law, the density of air is: ρ= mP mn = V RT where m is the mass of one mole of air (≈29 g). This formula applies either inside or outside the balloon, with the same pressure in both places but different temperatures. Therefore, teh balance of forces implies: mP mP M − = RT0 RT V where T and T0 are the temperature inside and outside the balloon, respectively. Rearranging yields: 1 1 M P =− − T T0 m RV Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 3 Assume an outside air temperature of 290 K and atmospheric pressure. The volume of the balloon can be estimated from the figure. Comparing the heights of the people standing beneath, a reasonable estimate for the diameter of the balloon in the foreground is ≈15 metres. This gives a volume, assuming a spherical balloon ( 33 πr 3 ), of about 1770 m3 . The mass of the unfilled balloon and payload is assumed to be 500 kg, so the previous expression evaluates to: 1 T 1 T 1 (500 kg) (8.31 J/K) − 5 (290 K) (0.029 kg) (10 N/m2 )(1770 m3 ) 1 1 1 = − − = (290 K) (1235 K) (379 K) = − Thus, the temperature inside the balloon must be about 379 K or just over 100◦ C. Assuming this temperature, the mass of the air inside the balloon should be roughly: mP V RT (0.029 kg)(105 N/m2 )(1770 m3 ) = = 1600 kg (8.31 J/K)(379 K) Mair = mn = Mair which s more than three times the mass of the unfilled ballon and payload. • 1.16: The Exponential atmosphere. a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air. This diagram shows the forces on the slab of air. P (z) is the atmospheric pressure as a function of height z, and M is the total mass of the slab. Because the slab of air is at rest, the net force on the slab must be zero. Thus P (z + dz)A + Mg − P (z)A = 0 ⇒ [P (z + dz) − P (z)]A = −Mg ⇒ P (z + dz) − P (z) = −Mg A Now the volume of the slab is Adz, so if the density of air is ρ(z), then M = ρ(z)Adz, and so P (z + dz) − P (z) = − ⇒ ρ(z)Agdz = −ρ(z)gdz A P (z + dz) − P (z) = −ρ(z)g dz Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 4 If dz is very small, then the left hand side is the definition of the derivative dP/dz. Thus, dP = −ρ(z)g dz b. Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. Show, then, that the pressure obeys the differential equation: mg dP =− P, dz kT called the barometric equation. We can derive this from part a by finding an expression for ρ(z). The density of the slab is equal to the mass of the slab, M divided by its volume, V . The mass, in turn, is equal to the mass per molecule, m, times the number of molecules, N. Thus: ρ= mN V The ideal gas law says that P V = NkT , so that N/V = P/kT . Thus ρ = mP/kT , and dP mP mg =− g=− P (z) dz kT kT c. Assuming that the temperature of the atmosphere is independent of height, solve the barometric equation to obtain the pressure as a function of height: P (z) = P (0)e−mgz/kT . Show also that the density obeys a similar equation. For convenience, let’s write α = mg/kT , so that dP = −αP (z) dz This is one of the most basic differential equations, but if you’ve never seen it before, we can solve it by separation of variables: put the pressure on one side, and the altitude z on the other: dP dz dP P P (z) dP P (0) P ln P (z) − ln P (0) P (z) ln P (0) P (z) P (0) P (z) = −αP (z) = −αd(z) = − z 0 αd(z) = −αz = −αz = e−αz = P (0)e−αz Substituting in the definition for α gives us P (z) = P (0)e−mgz/kT as expected. Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 5 We showed in part b that ρ = mP/kT , so ρ(z) = m P (0)e−αz , kT we could also write mP (0)/kT as ρ(0), the density at z = 0. Note that the constant α has units of m−1 ; the constant z0 = 1/α is the length scale over which the pressure and density vary. d. Estimate the pressure, in atmospheres, at Mt. Everest. (Assume that the pressure at sea level is 1 atm.) We can write: P (z) = P (0)e−z/z0 where z0 = kT /mg and it is convenient to calculate this quantity first. Assume that T = 280K. Air is 80% N2 , which has an atomic mass of 28, and 20% O2 which has an atomic mass of 32. An atomic mass unit is 1.67 × 10−27 kg (roughly the mass of a proton), so the average mass per molecule is: m = 80%(28)(1.67 × 10−27 kg) + 20%(32)(1.67 × 10−27 kg) = 4.8 × 10−26 kg and so z0 = (1.38 × 10−23 J/K)(280K) = 8200 m = 8.2 km (4.8 × 10−26 kg)(9.8 m/s2 ) (Because this is so large, pressure and density does not vary appreciably inside a room). Now P (0) is the pressure at sea level, thus 1 atm, and so P (z) = (1atm)ez/8200m , and thus Mt. Everest = 8850m, pressure = 0.34 atm • 1.18: Calculate the rms speed of a nitrogen molecule at room temperature. We know: vrms = 3kT m −27 kg One nitrogen atom has mass 14.00 u. 1.66×10 = 2.3 × 10−26 kg, so a nitrogen molecule 1u has (approximately) twice that mass, or m = 4.6 × 10−26 kg. Thus vrms = 3(1.38 × 10−23 J/K)(300K) = 5200 m/s 4.6 × 10−26 kg • 1.19: Suppose you have a gas containing hydrogen and oxygen molecules, in thermal equilibrium. Which molecules are moving faster, on average? By what factor? The rms speed of the molecules in an ideal gas is vrms = hydrogen and oxygen at the same temperature is: vH = vO 3kT . m The ratio of the speeds of mO mH Now √ oxygen molecules are sixteen times more massive than hydrogen molecules, so vH /vO = 16 = 4. So hydrogen molecules will be moving four times faster than oxygen molecules. Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 6 • 1.21: During a hailstorm, hailstones with an average mass of 2 g and a speed of 15 m/s strike a windowpane at a 45◦ angle. The area of the window is 0.5 m2 and the hailstones hit it at a rate of 30 /s. What average pressure do they exert on the window? How does this compare to the pressure of the atmosphere? The hailstones strike the window at intervals of 1/30 s on average. During this time period, the average force exerted by the window on the hailstone must be: → F x= m ∆vx ∆t where ∆vx is the change in the component of the hailstone’s velocity perpendicular to the window. Assuming elastic collisions and a velocity of 15m/s at 45◦ , this change in velocity is 2vcos(45◦) = 21m/s. The average pressure is then → m∆vx (0.002 kg)(21 m/s) F = = = 2.5 N/m2 = 2.5 Pa P= A A∆t (0.5 m2 )(0.033 s) → This is smaller than atmospheric pressure by a factor of 40,000. (However, the instantaneous pressure during a collision is much higher, as the force is localized in time and space; this is what might cause the window to break.) • 1.22: If you poke a hole in a container full of gas, the gas will start leaking out. In this problem you will make a rough estimate of the rate at which gas escapes through a hole. a. Consider a small portion (area A) of the inside wall of a container full of gas. Show that → the number of molecules colliding with this surface in a time interval ∆t is P A∆t/(2m v x ), → where P is the pressure, m the average molecular mass, and v x is the average x velocity of those molecules that collide with the wall. As in Eq. 1.9, Newton’s Laws imply that each molecule colliding with the surface exerts an average pressure of: → P= − m∆vx A∆t For an elastic collision, ∆vx = −2vx . If there are N molecules, the total pressure is the sum of N terms of this form, one for each molecule; the sum over vx values can be written as N → times the average, or N v x . Therefore → m(2N v x ) P A∆t P = ⇒N = → A∆t 2m v x → b. It’s not easy to calculate v x , but a good enough approximation if (vx2 )0.5 , where the bar now represents an average over all molecules in the gas. Show that (vx2 )0.5 = kT /m. This result follows directly from Eq. 1.15: kT = mvx2 ⇒ vx2 = kT /m c. If we now take away this small part of the wall, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number N of molecules inside the container as a function of time is governed by the differential equation: dN A =− dt 2V kT N m Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 7 Solve this equation (assuming constant T ) to obtain a formula of the form N(t) = N(0)e−t/τ where τ is the “characteristic time” for N (and P ) to drop by a factor of e. What we called N above is now −∆N, the change in the number of molecules in the container. Substituting this into part (a), and bringing in part (b), gives us: −∆N = P A∆t 2m m kT Now use the ideal gas law to eliminate P , and divide through by ∆t, taking the limit ∆t → 0 to get the derivative: −δN = (NkT /V )A∆t 2m kT A m ∆N = −N kT δt 2mV A m dN =− kT dt 2V 1 kT N =− N m τ where τ = (2V /A) m/kT . Thus N(t) is a function whose derivative is equal to −1/τ times itself, which makes it an exponential: N(t) = N0 e−t/τ d. Calculate the characteristic time for air at room temperature to escape from a 1-litre container punctured by a 1mm2 hole. To calculate τ we start with the quantity kT /m = RT /M, where M is the molar mass of the gas (that is, the mass of a mole of air). Assuming the gas is near room temperature: RT = M (8.3 J/mol/K)(300 K) = 293 m/s ≈ 300 m/s 0.029 kg/mol A one liter bottle has volume 10−3m3 , and 1 mm2 = 10−6 m2 , so τ= 2V A RT /M = 2(0.001 m3 ) =7s (10−6 m2 )(300 m/s) e. The tire “going flat” I assume to mean that et/τ ≈ 0 in the shortest time. For example, e−3 ≈ .05, which is close to 0. That means: t 2V − − 3 → t = 3τ t = 3τ = 3 τ A m 6V = (3.42 × 10−3) kB T A Again, I assume V = 1 L and 1 hour = 3600 s. 3600s = 6(10−3 m3 ) (3.42 × 10−3 s/m A Calculate the area A; A = 5.7 × 10−9 m2 = 5.7 × 10−3 mm2 f. Assuming that in time ∆t, all particles within the half sphere of radius vx ∆t surrounding the hole escape, since vx is on the order of several hundreds of meters per second on average it seems unlikely that the travelers could toss the corpse out the window in a short enough time for an insignificant number of particles to escape. Of course, this hinges on the the definition of significant- if the space ship was very large, the percentage of air compared to the total volume of the ship would be very small and perhaps termed “insignificant.” Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 8 • 1.26: A battery is connected in series to a resistor, which is immersed in water (to prepare a nice hot cup of tea). Would you classify the flow of energy from the battery to the resistor as “heat” or “work”? What about the flow of energy from the resistor to the wire? Energy flowing from the battery to the resistor should be classified as work, since it is a process that would not happen spontaneously. The flow of energy from the resistor to the water is heat, since it is a spontaneous process. • 1.27: Give an example of a process in which no heat is added to a system but its temperature increases. Then give an example of the opposite: a process in which heat is added to a system but its temperature does not change. An example of a process in which no heat is added to a system but the temperature increases is pumping a tire full of air. A process in which heat is added to a system but its temperature remains constant is a phase change, melting a piece of ice for example. • 1.28: Estimate how long it would take to bring a cup of water to boiling temperature in a typical 600-watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water). Explain why no heat is involved in this process. There are a few ways to approach this problem. ∆U = Q + W where, for this problem, Q = 0 so ∆U = W . Looking at the change in thermal energy (eqn 1.23) for a system of N molecules with f degrees of freedom: 1 ∆U = Nf k∆T 2 Assuming that the water is at 20◦ C at the start and it boils at 100◦ C, then ∆T = (100 20) =80◦ C. H2 O = 2H + O ⇒ 2 g/mol + 16 g/mol ⇒ 18 g/mol. A typical cup of water is about 200g, and so: (200 g) (6.02 × 1023 molecules) N= (18 g/mol) (1 mol) N = 6.7 × 1024 molecules The number of degrees of freedom, f , is not so clearly defined in the problem. There are certainly 3 translational degrees of freedom and there are 3 rotational. We do not know if the vibrational degrees of freedom are frozen out, but if we assume that they are then f = 6. ∆U ≈ 6.7 × 1024 6 1.38 × 10−23 73 J∆U ≈ 2.02 × 104 J 2 For a 600 W oven, the “heating” rate is 600 J/s, so: ∆t ≈ ∆U ≈ 34 s 600 J/s Alternatively, we can look at the definition of a calorie. 1 calorie = 4.186 J is the amount of energy required to raise the temperature of 1 g of water by 1◦ C. So, for 200 g of water we need: U = (4.186 J/g/◦ C)(200 g)(80 ◦ C) U = 6.11 × 104 J Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 9 So, ∆t ≈ 6.11 × 104 ≈ 102 s 600 Clearly, f = 6 is not correct for liquid water and there must be a larger number of degrees of freedom. Remember that flow of heat is a spontaneous process from higher T to lower T . The microwave overn is an example of electro-magnetic work. • 1.31: Imagine some helium in a cylinder with an initial volume of 1 litre and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 litres, in such a way that the pressure rises in direct proportion to its volume. (a) Sketch a graph of pressure vs volume for this process. (b) Calculate the work done on the gas during this process, assuming that there are no “other” types of work being done. The work done is just minus the area under the graph shown in (a). The easiest way to compute the area is to note that the average pressure during this process is 2 atm and: W = −P ∆V = −(2 atm)(2 l) W = −(2 × 105 Pa)(2 × 10−3 m3 ) W = −400 J The minus sign indicates that 400 J of work is done by the gas on its surroundings. (c) Calculate the change in the helium’s energy content during this process Each helium atom has three degrees of freedom, so at any point the thermal energy of the helium is U = 32 NkT = VP . The change in energy during this process is: 3 [Pf Vf − Pi Vi ] 2 3 ∆U = [(3 atm)(3 l) − (1 atm)(1 l)] 2 ∆U = 12 atm l = 1200J ∆U = Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 10 (d) Calculate the amount of heat added to or removed from the helium during this process. By the first law, Q = ∆U − W Q = 1200 − (−400) = 1600 J This amount of heat enters the gas. (e) Describe what you might do to cause the pressure to rise as the helium expands? To cause such an increase in pressure (and temperature) as the gas expands, you must provide heat, for example, by holding a flame under the cylinder and letting the piston out slowly enough to allow the pressure to rise as desired. • 1.33: An ideal gas is made to undergo the cyclic process shown in the figure. For each of the steps A, B, and C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas, (b) the change in the energy content of the gas, and (c) the heat added to the gas. Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish (in real-world terms)? The work done on the gas depends entirely on the change in volume; decreasing volume means that work is positive, while increasing volume means that work is negative. The thermal energy of the gas is directly proportional to the temperature, which is in turn proportional to P V . The heat must then be accounted for via the equation ∆U = W + Q. For branch A, the volume increases, so W < 0. P V increases as well, so the thermal energy is increasing. Since the energy increases but work flows out, there must be a flow of heat inward to compensate, and Q > 0. For branch B, there is no change in volume so W = 0. P V increases, so ∆U > 0. Because the energy is increasing, there must be a flow of heat inward. For branch C, volume and pressure are both decreasing, so W > 0 and ∆U < 0. Because work flows in and the energy still decreases, there must be a net flow of heat outward, so Q < 0. To summarize: Branch A B C energy increases increases decreases W negative zero positive Q positive positive negative During one complete cycle, P V comes back to its original value, so the thermal energy does not change. The work done on this system is positive (because branch C has a greater area Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 11 under it than branch A, and thus has more work flowing). Because the system’s energy doesn’t change, Q must be negative: heat flows out of the system. In short, this cycle transforms work into heat; it could, for instance, be some sort of space heater. If we note, however, that this system both absorbs heat (in steps A and B) and emits heat, we can think of a more sophisticated use: in particular, we can absorb heat from one room and emit it into the other. Similar cycles can act as refrigerators, but this particular cycle would not make a practical refrigerator because the heat absorption stages are not at a lower temperature than the emission stages. • 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the figure below. Assume that the temperature is always such that the rotational degrees of freedom are active but that the vibrational modes are “frozen out.” Also assume that the only type of work done on the gas is quasidiabatic compressionexpansion work. (a) For each of the four steps A through D, compute the work done om the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of P1 , P2 , V1 and V2 )Hint: compute ∆U before Q using the ideal gas law and the equipartition theorem.) For an ideal diatomic gas at room temperature, U = 25 NkB T = 52 P V (since P V = NkB T ) Step A: 5 5 V ∆P = V1 (P2 − P1 ) 2 2 = 0 5 = ∆UA − WA = V1 (P2 − P1 ) 2 ∆UA = WA QA Step B: ∆UB = 5 5 P ∆V = P2 (V2 − V1 ) 2 2 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar V2 WB = − V1 12 P dV = −P2 (V2 − V1 ) = P2 (V1 − V2 ) 7 5 QB = ∆UB − WB = P2 (V2 − V1 ) + P2 (V1 − V2 ) = P2 (V2 − V1 ) 2 2 Step C: 5 5 5 V ∆P = V2 (P1 − P2 ) = − V2 (P2 − P1 ) 2 2 2 = 0 5 = ∆UC − WC = − V2 (P2 − P1 ) 2 ∆UC = WC QC Step D: ∆UD = 5 5 P ∆V = − P1 (V2 − V1 ) 2 2 WD = − V1 V2 P dV = −P1 (V1 − V2 ) = P1 (V2 − V1 ) 5 7 QD = ∆UD − WD = − P1 (V2 − V1 ) − P2 (V2 − V1 ) = − P1 (V2 − V1 ) 2 2 (b) Describe in words what is physically being done during each of the four steps; for example, during Step A, heat is added to the gas (from an external flame or something) while the piston is held fixed. In step A, the volume is held constant at V1 , and heat is added to the gas in the cylinder, thus raising its temperature and pressure. In step B, the piston is moved out to increase the volume from V1 to V2 , and, at the same time, heat is added so as to maintain constant pressure at P2 . In step C, the volume is held constant at V2 , and heat is removed from the cylinder to lower the pressure from P2 to P1 . Finally, in step D, the gas is compressed from V2 to V1 , while continuing to remove heat, so that the pressure is maintained at P1 . (c) Compute the net work done on the gas, the net heat added to the gas, and the net change in the energy of the gas during the entire cycle. Are the results as you expected? Explain briefly. For the cyclical process, we have: ∆Ucycle = ∆UA + ∆UB + ∆UC + ∆UD 7 5 7 5 ∆UD = V1 (P2 − P1 ) + P2 (V2 − V1 ) + − V2 (P2 − P1 ) + − P1 (V2 − V1 ) 2 2 2 2 ∆Ucycle = 0 Similarly, Wcycle = −P2 (V2 − V1 ) + P1 (V2 − V1 ) Wcycle = −(P2 − P1 )(V2 − V1 ) and 5 7 5 7 V1 (P2 − P1 ) + P2 (V2 − V1 ) − V2 (P2 − P1 ) − P1 (V2 − V1 ) 2 2 2 2 5 7 = (P2 − P1 )(V1 − V2 ) + (P2 − P1 )(V2 − V1 ) 2 2 = (P2 − P1 )(V2 − V1 ) Qcycle = Qcycle Qcycle Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 13 Yes, the results are as expected. For a complete cycle, the net change in the internal energy U should be zero, since U is a function of state. Furthermore, from the first law, we have ∆U = Q + W ⇒ Q = −W if ∆U = 0. • 1.38: Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly so that no heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully. Both bubbles are identical to begin with, so both have the same number of air molecules N inside. The pressure inside each bubble must be more-or-less equal to the pressure of the water (we ignore the effects of surface tension here), so both bubbles end up at the same pressure as well. Therefore the volume of each bubble, V = NkT /P , depends entirely on temperature: the warmer bubble will be bigger. That also means that the larger bubble will have the larger thermal energy. Now both bubbles do work on the surrounding water as they expand (W < 0) Because bubble A does not exchange heat with its environment, it loses energy and so cools down. Bubble B, on the other hand, remains at a constant temperature. Therefore, by our reasoning above, bubble B, the one that absorbs heat, will be larger. • 1.40: In problem 1.16 you calculated the pressure of the earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10-15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT /dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling. (a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation: dT 2 T = dP f +2P From the book: P V γ = const. V T f /2 = const. Using these and solving for V gives: V = 1 (f +2)/2 T = const. P This can be written in terms of natural logarithms as follows: ln T (f +2)/2 − ln P = const. f +2 ln T − ln P = const. 2 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 14 Now if we take a derivative: f +2 2 dT dP − =0 T P Simplifying this gives the desired differential equation dT 2 T = dP f +2P (b)Assume that dT /dz is just at the critical value fro convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of problem 1.16(b) to find a formula for dT /dz in this case. This result should be a constant, independent of temperature and pressure, which evaluates to approximately -10◦/km. This fundamental meteorological constant is known as the dry adiabatic lapse rate. From Problem 1.16 (b): mg dP =− dz kB T From the answer in part (a), dT /dP can be decomposed into the product dT dT dz = dP dz dP and therefore dT dz 2 T = dz dP f +2P Substituting in the expression for dP/dz: dT dz − kB T mg 1 2 T = P f +2P The T s and P s cancel, and we are left with (after solving for dT /dz) dT mg = dz kB 2 f +2 Using the mass of air, mair = 4.84 × 10−26 kg/molecule. Substituting in all the numbers: dT dz dT dz (4.84 × 10−26 kg/molecule)(9.8 m/s2 ) 2 1.38 × 10−23 J/K 5+2 K K = .0098 ≈ .01 m m = To change to K/km we just have to mutliply by 1000 m/km, and we get dT K = 10 dz km Which was to be shown. Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 15 • 2.1: Suppose you flip four fair coins. (a) Make a list of all the possible outcomes, as in Table 2.1. The number of outcomes is TTTT THTT HTTT TTTH THTH HTTH TTHT THHT HTHT TTHH THHH HTHH 24 = 16. HHTT HHTH HHHT HHHH (b) Make a list of all the different “macrostates” and their probabilities. 0 1 2 3 4 Heads Heads Heads Heads Heads Ω Ω Ω Ω Ω = = = = = 1 4 6 4 1 P P P P P = = = = = 1/16 4/16 5/16 4/16 1/16 (c) Compute the multiplicity of each macrostate using the combinatorial formula (2.6), and check that these results agree with what you got by brute force counting. Ω(0, 4) = Ω(1, 4) = Ω(2, 4) = Ω(3, 4) = Ω(4, 4) = 4! 0! 4! 4! 1! 3! 4! 2! 2! 4! 3! 1! 4! 4! 0! Note : 0! = 1 So, Ω = 1, P=1/16. So, Ω = 24 , P=4/16. 6 So, Ω = 24 , P=6/16. 4 24 So, Ω = 6 , P=4/16. So, Ω = 1, P=1/16. • 2.2: Suppose you flip 20 coins. (a) How many possible outcomes “macrostates” are there? For only 2 possible states ⇒ 220 microstates. 220 = 1,048,576. Note that this is Ω(all) or Ω(total) (b) What is the probability of getting the sequence HTHHTTTHTHHHTHHHHTHT (in exactly that order)? Simply, this is only one macrostate out of 220 . Since each state is equally probable, the possibility of getting this particular one is 1/220 ⇒ less than one in a million. (c) What is the probability of getting 12 heads and 8 tails (in any order)= This is another macrostate: P = Ω(20, 12) 20! 125970 = = = 0.12 (20−12)! Ω(all) 1048576 12! 1048576 So, the probability is about 12% • 2.3: Suppose you flip 50 coins. (a) How many possible outcomes (microstates) are there? There are two outcomes for each coin, so total number of macrostates: Ω(all) = 250 = 1.13 × 1015 (b) How many ways are there of getting exactly 25 heads and 25 tails? Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 16 To get exactly 25 heads: Ω(q, N) = Ω(25, 50) = N! (N − q)! q! 50! = 1.26 × 1014 (50 − 25)! 25! (c) What is the probability of getting exactly 25 heads and 25 tails? P (25) = P (25) = Ω(25, 50) Ω(all) 1.26 × 1014 = 0.112 250 So, not so large. (d) What is the probability of getting exactly 30 heads and 20 tails? P (30) = P (30) = Ω(30, 50) Ω(all) 1 50! = 0.042 50 2 30! 20! (e) What is the probability of getting exactly 40 heads and 10 tails? P (40) = P (40) = Ω(40, 50) Ω(all) 1 50! = 0.0000091 = 9.1 × 10−6 50 2 40! 10! (f) What is the probability of getting exactly 50 heads and 0 tails? P (50) = 1 = 8.9 × 10−16 Ω(all) (g) Plot a graph of the probability of getting n heads as a function of n • 2.4: Calculate the number of possible five-card poker hands, dealt from a deck of 52 cards. (The order of cards in a hand does not matter.) A royal flush consists of the five highestranking cards (ace, king, queen, jack, 10) of any one of the four suits. What is the probability of being dealt a royal flush (on the first deal)? The number of hands: N= 52! = 2598960 5! 47! The probability of getting a royal flush: P = 4! = 2598960 5! 47! Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 17 • 2.5: For an Einstein solid with each of the following values of N and q, list all of the possible microstates, count them, and verify formula 2.9 −1)! (a) N = 3, q = 4 : Ω(N, q) = (q+N = 15 q! (N −1)! Oscillator Oscillator 1 2 3 1 2 3 4 0 0 0 1 3 0 4 0 2 2 0 0 0 4 2 0 2 15 states. 3 1 0 2 1 1 3 0 1 0 2 2 1 3 0 1 2 1 0 3 1 1 1 2 1 0 3 −1)! (b) N = 3, q = 5 : Ω(N, q) = (q+N = 21 q! (N −1)! Oscillator Oscillator 1 2 3 1 2 3 5 0 0 3 1 1 0 5 0 2 3 0 0 0 5 0 3 2 4 1 0 1 3 1 4 0 1 2 0 3 21 states. 1 4 0 0 2 3 0 4 1 1 1 3 1 0 4 2 2 1 0 1 4 2 1 2 3 2 0 1 2 2 3 0 2 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar −1)! (c) N = 3, q = 6 : Ω(N, q) = (q+N = 28 q! (N −1)! Oscillator Oscillator 1 2 3 1 2 3 6 0 0 1 4 1 0 6 0 2 0 4 0 0 6 0 2 4 5 1 0 1 1 4 5 0 1 3 3 0 1 5 0 3 0 3 28 states. 0 5 1 3 2 1 1 0 5 3 1 2 0 3 5 3 3 0 4 2 0 0 3 3 4 0 2 2 3 1 4 1 1 1 3 2 2 4 0 2 1 3 0 4 2 1 2 3 (d) N = 4, q = 2 : Ω(N, q) = Oscillator 1 2 3 4 2 0 0 0 0 2 0 0 0 0 2 0 0 0 0 2 10 states. 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 (q+N −1)! q! (N −1)! = 10 −1)! (e) N = 4, q = 3 : Ω(N, q) = (q+N = 20 q! (N −1)! Oscillator Oscillator 1 2 3 4 1 2 3 4 3 0 0 0 1 0 2 0 0 3 0 0 0 1 2 0 0 0 3 0 0 0 2 1 0 0 0 3 1 0 0 2 20 states. 2 1 0 0 0 1 0 2 2 0 1 0 0 0 1 2 2 0 0 1 1 1 1 0 1 2 0 0 1 1 0 1 0 2 1 0 1 0 1 1 0 2 0 1 0 1 1 1 (f) N = 1, q = anything Ω(N, q) = (q+N −1)! q! (N −1)! = q Oscillator 1: q 18 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 19 (g) N = anything, q = 1 Ω(N, q) = (q+N −1)! q! (N −1)! =N • 2.6: Calculate the multiplicity of an Einstein solid with 30 oscillators and 30 units of energy. (Do not attempt to list all the microstates) Ω(30, 30) = (q+N −1)! q! (N −1)! = 59! 30! 29! = 5.91 × 1016 • 2.7: For an Einstein solid with four oscillators and two units of energy, represent each possible microstate as a series of dots and lines, as used in the text to prove equation 2.9 N = 4, q = 2, Ω(N, q) = Oscillator 1 2 3 4 2 0 0 0 0 2 0 0 0 0 2 0 0 0 0 2 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 (q+N −1)! q! (N −1)! = 10 Dots and lines 1 2 3 4 ·· ·· ·· ·· 10 states. · · · · · · · · · · · · • 2.8: Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed. (a) How many different macrostates are available to this system? Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 different macrostates are available to the system of weakly coupled A and B solids sharing 20 units of energy. (b) How many different microstates are available to this system? The combined system has 20 oscillators and 20 units of energy. Therefore, Ω(N, q) = (q + N − 1)! 39! = = 6.89 × 1010 q!(N − 1)! 20! 19! (c) Assuming that this system is in thermal equilibrium, what is the probability of finding all of the energy in solid A? For the macrostate with all energy in solid A: Ω = ΩA ΩB = ΩA (10, 20)ΩB (10, 0) = 29! = 1 × 107 20! 9! The probability of this macrostate is: P = 1 × 107 = 1.45 × 10−4 10 6.89 × 10 (d) What is the probability of finding exactly half the energy in solid A? Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 20 For the macrostate with exactly half the energy in solid A: Ω = ΩA ΩB = ΩA (10, 10)ΩB (10, 10) = P = 8.53 × 109 = 0.124 6.89 × 1010 19! = 8.53 × 109 10! 9! (e) Under what circumstances would this system exhibit irreversible behaviour? If the energy was initially all in A, the system would evolve toward half energy in A and half energy in B, and this distribution of energy would be irreversible. In other words, once the energy is evenly distributed, it is highly unlikely that the system would find itself with all the energy back in A. • 2.16: Suppose you flip 1000 coins. (a) What is the probability of getting exactly 500 heads and 500 tails? (Hint: First write down a formula for the total number of possible outcomes. Then, to determine the “multiplicity” of the 500-500 “macrostate”, use Sterling’s approximation. If you have a fancy calculator that makes Stirling’s approximation unnecessary, multiply all the numbers in this problem by a sufficient factor that you have to manually make this approximation. First, note that the number of possible outcomes (microstates) is 21000 . The probability of getting 500 heads and 500 tails is (from Sterling’s approximation): 1000! (500!)2 √ 10001000 e−1000 2π 1000 21000 √ Ω(500) ≈ = √ 2 500π 500500 e−500 2π 500 1000 500 Ω(500) = = The probability is Ω(500)/Ω(all) : Ω(500) = 1 21000 √ √ = = 0.025 21000 500π 500π So the chance of getting exactly 500 heads is about 2.5%, or 1 in 40. (b)What is the probability of getting exactly 600 heads and 400 tails? From Sterling’s approximation: Ω(600) = 1000 600 = 1000! 600! 400! √ 10001000 e−1000 2π 1000 10001000 √ √ √ Ω(600) ≈ = 600600 e−600 2π 600 400400 e−400 2π 400 600600 400400 480π The probability is Ω(600)/Ω(all) : 10001000 5001000 500600 500400 √ √ √ = = 21000 600600 400400 480π 600600 400400 480π 600600 400400 480π 5 600 5 400 1 √ Ω(600) = 6 4 480π Ω(600) = So the chance of getting exactly 600 heads is 4.6 × 10−11 , much smaller than P (500) Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 21 • 2.21: Use a computer to plot formula 2.22 directly, as follows. Define z = qa /q such that (1 − z) = qb /q. Then, aside from a few constants that can be ignored, the multiplicity function is [4z(1 − z]N , where z ranges from 0 to 1 and the factor 4 ensures that the height of the peak is equal to 1 for any N. Plot this function for N=1, 10, 100, 1000, and 10,000. Observe how the width of the peak decreases as N increases. • 2.23: Consider a two-state paramagnet with 1023 elementary dipoles, with half the total energy fixed at zero so that exactly half the dipole point up and half point down. (a) How many microstates are “accessible” to this system? With N dipoles of which exactly N/2 point up, the multiplicity of the paramagnet is: Ω= N N 2 ⇒ N! N 2 ! N 2 ! So, taking the Sterling approximation: Ω= √ N N e−N 2πN √ N/2 N −N/2 πN e 2 2 = 2N 2 πN 23 For N = 1023 , then, the multiplicity is roughly: 210 /4 × 1011 Since the denominator is 23 merely “large,” we could just as well neglect it and say that Ω = 210 , which is the number of microstates if we allow any number of dipoles to be pointing up. (b)Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)? A year is about 3 × 107 s, so 10 billion years is about 3 × 101 7 s, or 3 × 102 6 ns. If the microstate of the system changes once every nanosecond, this is how many microstates the system will explore in the age of the universe. But this is a tiny fraction of the total number of microstates. In fact, the fraction is so small that the ratio of states not explored to states 23 explored is 210 . 22 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar (c) Is it correct to say that, if you wait long enough, a system will eventually be found in ev ery “accessible” microstate? Explain your answer, and discuss the meaning of the word “accessible.” Even if we wait for the age of the universe, the fraction of all “accessible” microstates that are actually explored by this system is so tiny that it might be more accurate to say that the system explores none of its “accessible” microstates. When we call a microstate “accessible,” therefore, we should not think that the system will ever actually be in that microstate. So what do we mean? One of the best interpretations is in terms of our ignorance of which microstates the system will actually explore in the future. F orallweknow, the system might soon be found in any of its “accessible” microstates, even though the probability of its being found in any of them is vanishingly small. • 2.24: For a single large two-state paramagnet, the multiplicty function is very sharply peaked about N↑ = N/2 (a) Use Sterling’s approximation to estimate the height of the peak in the multiplicity function. If N = N↑ , then N↑ = N↓ = N 2 and the multiplicity is: Ωmax = N! = N↑ ! N↓ ! N! N ! 2 2 √ N N e−N 2πN Ωmax ≈ N/2 N 2 e−N/2 2π N2 2 = 2N 2 πN Using Sterling’s approximation: √ NN N! N N e−N 2πN ≈ N↑ = Ω = N N N N↑ ! N↓ ! N↑ ↑ N↓ ↓ N↑ e−N↑ 2πN↑ N↓ ↓ e−N↓ 2πN↓ N 2πN↑ N↓ (b) Use the methods discussed in the lectures to derive a formula for the multiplicity function in the vicinity of the peak, in terms of x ≡ N↑ − (N/2). Check that your formula agrees with your answer to part (a) when x = 0 Using x ≡ N↑ − (N/2), we can rearrange for N↑ : N↑ = N +x 2 and solve for N↓ : N↓ = N − N↑ ⇒ N↓ = N − N N − x ⇒ N↓ = −x 2 2 Using the equation obtained from Sterlings approximation: Ω= NN N N↑ ↑ N N↓ ↓ N 2πN↑ N↓ setting N↑ = (N/2) + x and N↓ = (N/2) − x: Ω= NN N 2 +x N/2+x N N 2 −x N/2−x 2π N 2 +x N 2 −x 23 Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar Rearranging . . . N Ω= N 2 2 N/2 − x2 N 2 N +x x N 2 −x −x 2π 2 N 2 − x2 Apply the logarithmic approximation to this equation: N ln Ω ≈ N ln N − ln 2 N 2 N 1 − ln 2π 2 N 2 + ln 2 − x2 − x ln N N + x + x ln −x 2 2 2 − x2 Nothing yet has been assumed about the size of x relative to N but if x ≪ N, the logarithms containing two terms can be expanded: ln N 2 ln 2 − x2 N ±x 2 N 2 N = ln 2 2 = ln + ln 1 − + ln 1 ± 2x N 2 N 2x − 2 N 2x ± N 2 ≈ 2 ln 2x N ≈ ln N 2 Using these expressions, we get: 2x2 N 2x2 N 2x2 N + − x ln − + x ln − 2 N 2 N 2 N 2 N N 2x − ln + 2 2π 2 N ln Ω = N ln N − N ln + ln ln Ω = N ln 2 − 2x2 + ln N 2 2x2 − 2 πN N The last term is clearly negligible and can be neglected. Taking the exponential: √ 2 2 ln Ω N ln 2− 2x +ln N πN e ≈e 2 −2x2 /N Ω = 2N e for x ≪ N πN This is a Gaussian function, peaked at x = 0. Clearly, when x = 0: Ω = 2N 2 πN which is the same result obtained in part (a). (c) How wide is the peak in the multiplicity function? The Gaussian function falls off to 1/e of its peak value when: √ 2x2 N =1 ⇒ x= N 2 √ But the width of the peak will be 2x so: width = 2N Richard Thomas - FK4005 “Thermodynamik” - Solutions to R¨akn¨ovningar 24 (d) If you flip one million coins. Would you be surprised to obtain 501,000 heads and 499,999 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain. Here, N = 106 x = N2 + 103 . The half-width of the peak in the multiplicity function would √ be 500000 ≈ 700. So an excess of 1000 heads is only a little beyond the point where the Gaussian has fallen off to 1/e of its maximum value. It would nit be too surprising to obtain approximately this many heads, though it would be surprising to obtain an excess of exactly 1000. Conversely, an excess of 10,000 heads lies fay outside the peak in the multiplicity function. At this point the Gaussian has fallen off to e200 ≈ 10−87 of its maximum value. If a result close to this was achieved, it’s likely that there is a problem with the coins!
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