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5 Pages

### ps11_soln

Course: PHY 105, Fall 2009
School: Princeton
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Word Count: 795

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105 Physics Problem Set 11 Solutions Problem 1 (3 pts) (a) When bouncing o a wall perpendicular to the x-axis, only the x-component of the atoms velocity changes its sign. Collisions with the other walls do not aect the x-component of the atoms velocity. Hence the atom transfers to the wall orthogonal to the x-axis the amount of momentum equal to p = 2mvx during each collision; and it happens exactly once during...

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105 Physics Problem Set 11 Solutions Problem 1 (3 pts) (a) When bouncing o a wall perpendicular to the x-axis, only the x-component of the atoms velocity changes its sign. Collisions with the other walls do not aect the x-component of the atoms velocity. Hence the atom transfers to the wall orthogonal to the x-axis the amount of momentum equal to p = 2mvx during each collision; and it happens exactly once during the time interval equal to t = 2L . vx Consequently, the time averaged force is F= 2 p mvx = . t L (b) Using the result from the previous part, the pressure of the gas on the wall orthogonal to the x-axis is N i=1 L2 p= Fi 1 =3 L N 2 mi vix . i=1 Given that the direction of the atoms velocity is equally likely to point in any direction, we nd: 1 2 N 2 mi vix = i=1 1 2 N 2 mi viy = i=1 1 2 N 2 mi viz = i=1 N kB T . 2 This yields for the pressure N (kB NA )T . NA Recalling that the ideal gas constant is R = kB NA , we recover the ideal gas law. pV = Problem 2 (3 pts) 1. (Fermi 45.1) For the isothermic expansion at T1 = 400 K the amount of heat transferred to the gas equals to the work done, V2 = RT1 ln 5 = 5.3kJ . V1 = const, for the compression at T2 = 300 K we have Q1 = RT1 ln Since the adiabatic process obeys T V k1 V2 /V1 = 5 as well. The amount of heat absorbed by the second source is Q2 = RT2 ln V1 = RT2 ln 5 = 4.0kJ . V2 The net work done is W = Q1 + Q2 = 1.3kJ 1 2. (Fermi 45.2) The maximum eciency is achieved when the engine works according to the Carnot cycle. The upper temperature is T1 = 673K , and the lower temperature is T2 = 291K . Engines eciency is = T1 T2 = 0.57 . T1 3. (Fermi 45.3) Assuming that the heat is extracted in an isothermal process and using the reverse Carnot cycle, the amount of work needed is W =Q T1 T2 = 0.22cal = 0.92 J . T2 Here T1 = 311K is temperature of the environment, and T2 = 255K is the temperature of the body. 4. (Fermi 75.1) Change of the waters entropy during the heating process is T2 S = T1 mcV dT T2 = 312cal/deg = 1.3kJ/K . = mcV ln T T1 Dimensionless entropy change is therefore S = 9.4 1025 . kB Problem 3 (3 pts) Note that the the rst compartment is at the constant volume, while the second compartment is at the constant Let pressure. (a) Q be the amount of heat transferred to the second compartment. Then for the rst compartment we have Q = 3 n1 R(T1 T ) . 2 For the second compartment: Q = 3 5 n2 R(T T2 ) + p2 V2 = n2 R(T T2 ) . 2 2 Here we have used the ideal gas law for the second compartment, p2 V2 = n2 RT2 . (Recall that the pressure p2 is constant, so that p2 V2 = n2 RT2 .) Therefore, the nal temperature T is found from the following equation: 5 3 n1 R(T T1 ) + n2 R(T T2 ) = 0 ; 2 2 or T= 3n1 T1 + 5n2 T2 = 464 K . 3n1 + 5n2 2 (b) The amount of heat transferred is Q = 3 15n1 n2 n1 R(T1 T ) = (T1 T2 ) = 102 J . 2 2(3n1 + 5n2 ) (c) From the ideal gas law for the second compartment p2 Ah = n2 RT2 . The constant pressure in the second compartment is p2 = patm + mg . A Here m is the mass of the piston, and A = 78.5cm2 is the cross section of the piston. These give h = n2 RT2 = 5.1cm . patm A + mg 5 (d) Given that Q = 2 n2 RT and W = n2 RT , the fraction of heat converted to work is W 2 =. Q 5 Problem 4 (3 pts) We nd that Vmin = 1000cm3 r 1 = 50cm2 and Vmax = 1050cm3 . (a) At point 1 the intake air is at the external pressure and temperature. The number of moles of the gas is n= p1 V1 = 0.0424 . RT1 Equation for the adiabatic process for the gas is p2 = p1 r = 71atm . From the ideal gas law the temperature is T2 = T1 r 1 = 1010 K . For the constant-pressure process 2 3 the rst law of thermodynamics and the ideal gas law give Q = pV + 1 nRT = nRT . 1 1 Consequently, T3 = T2 + T = 1820 K . The volume is found as V3 = V2 T3 = 90 cm3 . T2 3 Transition 3 4 is an adiabatic process, and it is treated analogously to 1 2. Finally we arrive at the following table: p, atm V , cm3 T, K 1 1.0 1050 298 2 71.0 50 1010 3 71.0 90 1820 4 2.28 1050 681 Note that T4 /T1 = V4 /V1 in accordance with the ideal gas law. (b) The exhaust part of the cycle (4 1) is equivalent to taking away the amount of heat QC : QC = 1 nR(T4 T1 ) = 337 J . 1 Given that QH = 1000 J , the net work done is W = QH QC = 663 J . (c) Thermal eciency of the engine is = W = 0.663 . QH (d) 2400 rpm corresponds to f = 40 cycles per second. The power output of 8 cylinders is P = 8f W = 15.9 kW = 213 hp . 4 5
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