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9B-C Physics [18.32] a) 100 Assignment #9 Cole s = ! si pi = 10 s = 54.6 b) 100 i =10 11 12 24 15 19 10 12 20 17 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 150 150 150 150 150 150 150 150 150 150 s 2 = ! si 2 pi = i =10 11 12 24 15 19 10 12 20 17 10 + 20 2 + 30 2 + 40 2 + 50 2 + 60 2 + 70 2 + 80 2 + 90 2 + 100 2 150 150 150 150 150 150 150 150 150 150 2 2 s = 3730 srms = s = 61.1 Note that the rms value is higher than the mean because the larger values squared (like 100) carry more weight in the rms calculation. 10 2 [18.48] m = a) vmp = M 44 g/mol = = 7.3 10-26 kg 23 N A 6.02 10 molecules/mol 2(1.38 10 "23 J/K )(300 K ) 7.3 10 -26 kg = 3.4 102 m/s 2 kT = m 8(1.38 10 "23 J/K )(300 K ) 8kT = = 3.8 102 m/s b) vav = -26 #m 7.3 10 kg# 3(1.38 10 "23 J/K )(300 K ) 3kT = = 4.1 102 m/s c) vrms = m 7.3 10 -26 kg Note that the most probable occurs at the peak in the M-B distribution, but because of the Boltzmann tail, both the average and the rms speeds are larger. [18.83] a) We interpret probability as the frequency of occurance. For a discrete system, pi = ni N where ni is the number of particles in the ith state, so ni = Npi . For the continuous system, the probability of finding the particle with a speedbetween v and v + dv is f(v) dv. Hence, the number of particles in this range is dni = N f(v) dv. Summing $N = N % 3/ 2 mv 2 v + $v v f (v)dv. mp 2 kT 4 & m ( v 2 e 2 kT = 4# & m ( & 2 kT ( e "1 = b) vmp = f (vmp ) = 4# . mp ' 2#kT ) ' 2#kT ) ' m ) e # vmp m For oxygen gas at 300 K, vmp = 3.95 102 m/s, and N = N f(v) v = 0.0421N. c) Increasing v by a factor of 7 changes f by a factor of 72e-48 , and N f(v) v = (2.94 10-21)N. d) Multiplying the temperature by a factor of 2 increases the most probable speed by a factor of 2 , and the answers are decreased by ; 0.0297 and 2.08 10-21. e) When the temperature is one-half what it was in parts (b) and (c ), the fractions increase by 2 to 0.0595 and 4.15 10-21. f) At lower temperatures, the distribution is more sharply peaked about the maximum (the most probable speed), as is shown in Fig. (16-17). 3/ 2 [19Q.9] When the pressure is constant dQ =CpdT. This is not the change in internal energy because some of this heat goes into doing work. From the first law, CpdT = dU + PdV. For an ideal gas, it is always true that dU = CVdT. Assigment #9, 12/3/04 Page 1 of 3 UC Davis [19Q.15] As the air rises, the pressure decreases so the air rises doing work as it expands. This work comes by lowering the internal energy; hence, the temperature drops. Essentially, this effect is a natural air conditioner ( an air conditioner uses an adiabatic expansion of the refrigerant to generate a lower temperautre). [19P.14] a) Work is the area under the curve in the P-V diagram, so along path (1) the system does the greatest amount of work. Note because the system expands, it does positive work. Along path (3) the system does the least amount of work because the average pressure is least. b) Use the first law: Q = U + W. Both U and W are positive. Since U is a state variable, U is independent of the path. The heat must be absorbed along all three paths to raise U and do W; hence, path (1) has the greatest heat absorbed. [19.18] & a) b): Going around a loop clockwise has the system doing positive work because the expansion occurs at higher pressure than the compression part of the loop. The work done is the area inside the loop. Loop I does positive work; loop II does negative work (counter-clockwise). Loop I has greater area so a net positive work is done. c) For each closed loop cycle U = 0 because U is a state variable. a) Q = W and since in the cycle the system does a net positive work equal to the area inside the curve, it must absorb a net Q to do the work. d) Loop I is net positive work so heat flows into the system to do the work. Loop II is net negative work so heat must be exhausted from the system. [19.29] = 1.127 9 C 9 9 9R R or p = 2 Or and would indicate a system with 7 degrees 7 7 8 8R CV 2 R 7 of freedom.or 16 degrees of freedom. CV = (8.3 J/mol * K) = 29. J/mol K or 2 16 CV = (8.3 J/mol K) = 66 J/mol K. 2 Or one could just use the formulas CP = +CV = 1.127Cv and CP = CV + R 1.127CV = CV + R R CV = = 65.4 J/mol K 0.127 [19.45] a) W = % P dV : Wab = 0 because V is constant, Wbc = Pc(Vc - Va ), Wcd = 0, Wda = Pa(Va - Vc). b) Using the first law, Qab = Uab + Wab = Ub - Ua . Qbc = Uc - Ub + Pc(Vc - Va). Qcd = Ud - Uc. Qda = Ua - Ud + Pa(Va - Vc). c) Wabc = Pc(Vc - Va) while Wadc = Pa (Vc - Va). So the work done along abc is larger because of the larger pressure. Qabc = Uc - Ua + Wabc = Uc - Ua + Pc(Vc - Va) and Qadc = Uc - Ua + Wadc = Uc - Ua + Pa(Vc - Va ). The change in internal energy is the same for both paths because the internal energy is a state variable and only depends on the end points; however, path abc requires more work so the system must absorb more heat to get the energy to do this work. d) As discussed above, the change in internal energy is the same, but the work depends on path. abc has more area under the curve and hence requires more work to be done. This extra work must be offset by absorbing more heat. The heat flow Q is thus dependent on the path. [19.54] a) This is just thermal expansion from Ch 15: 3 $V = ,Vo$T = (5.1 10 "5 C -1 )(0.02 m) (70 C) = 2.86 10-8m3 Assigment #9, 12/3/04 Page 2 of 3 UC Davis b) W = P V = (105 Pa)(2.86 10-8m3 ) = 2.88 10-3J c) Q = mc T = V c T = (8.9 103 kg/m3 )(8 10-6m3)(390 J/kg K)(70 C) = 1944 J. d) U = Q - W but W is insignificant compared to Q. U = 1944 J . e) The work to expand the copper is negligible so there is little difference between Cp and CV . [19.58] b) W = W12 + W23 = a) p 1 W = P (2V - V ) + ( Po (2Vo ) " P3 (4Vo )) + "1 Where for the adiabat, P (2V ) = P3(4V ) P3 = P 2- p 3 0 1 W = PVo /1 + 2 " 2 2 " + )2 ( o . + "1 1 To T3 c) Use the ideal gas law = T3 = 4(2 "+ )To = (2 2 " + )To PVo P3 ( 4Vo ) o T T 2 T V 2V 3 4V d) For the adiabat no heat flows. For the isobar, QP = CP (T2 - T ) where QP = CP T . Find 3 = T2 T = o 2Vo Vo PV PVo o Qp = o o Cp This is positive so the heat flows into the gas. R RTo [19.61] Use Q = U + W. a) Isothermal: U = 0 Q = W = 300 J b) Adiabatic: Q = 0 U = -W = -300 J c) Isobaric: Q = U + W where Qp = Cp T, 5 3 5 U = Cv T, 3 R$T = 3 R$T + 300 J R T = 300 J. Qp = 300 J = 750 J 2 2 2 3 and U = 300 J = 450 J 2 Assigment #9, 12/3/04 Page 3 of 3 UC Davis

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