Homework Chapter 1 Solutons 1.7 1.13 1.17 1.19 1.6 1.33 1.45 1.51 1.71 page 1
Problem 1.7 A mole sample of oxygen gas s confned to a 5 lter vessel at a pressure of 8 atm. Fnd the average translatonal knetc energy of the oxygen molecules under these condtons. Soluton The temperature of the gas s PV = nrt (8 101,35 Pa)(5 10 3 m 3 ) = ( mol)(8.314 T = 43.75 K At ths temperature, the average knetc energy of a molecules s K = 5 k B T = 8.4130 10 1 )T At ths temperature, the total knetc energy above s dstrbuted to all of the degrees of freedom evenly and there are 5. The translatonal knetc energy s only 3/5 of ths amount. K translatonal = 3 5 8.4130 10 1 = 5.0478 10 1 page
Problem 1.13 A sample of a datomc deal gas has pressure P and volume V. When the gas s warmed, ts pressure trples and ts volume doubles. Ths warmng process has two steps. The frst s at constant pressure, and the second s at constant volume. Determne the amount of heat transferred to the gas. Soluton The P-V dagram looks lke ths. 3P, V P, V P, V Through the frst process, the change n the nternal energy s We know ths. So The work s just The heat s E = 5 nr T nr T = PV PV = PV E = 5 PV =.5PV W = P V = PV Q = E W =.5PV + PV = 3.5PV Through the second process, the work s zero so The total heat s 13.5PV. Q = E = 5 nr T = 5 (6PV PV ) = 10PV page 3
Problem 1.17 A vertcal cylnder wth a heavy pston contaned ar at 300 K. The ntal pressure s 10 5 Pa and the ntal volume s 0.35 m³. Take the molar mass of ar as 8.9 g/mole and assume CV s (5/)R. Fnd the specfc heat of ar at constant volume. Calculate the mass of ar n the cylnder. Supposed the pston s held fxed, how much energy nput s requred to rase the temperature of the ar to 700 K. What would t be f the pressure were held constant? Soluton The the specfc heat of ar at constant volume s The number of moles s Ths has a mass of PV = nrt n = PV RT = ( 105 Pa)(0.35 m 3 ) = 8.065 mol (8.314 )(300 K) 8.9 g 8.065 mol = 811.08 g = 0.81108 kg 1 mol To ncrease the temperature to 700 K, the change n the temperature s 400 K. The volume s held constant, so If the process s made sobarc, C V = 5 R = 5 (8.314 ) = 0.785 1 mol 8.9 10 3 kg = 719.0 Q = nc V T = (0.81108 mol)(719.0 )(400 K) = 33.33 k kg C P = 7 R = 7 (8.314 ) = 9.099 1 mol 8.9 10 3 kg = 1,006.9 kg Q = nc P T = (0.81108 mol)(1,006.9 )(400 K) = 36.67 k kg kg K page 4
Problem 1.19 Durng the compresson stroke of a certan gasolne engne, the pressure ncreases from 1 atm to 0 atm. If the process s adabatc n the ar-fuel mxture behaves as a datomc deal gas, by what factors do the volume and temperature change? If there were 0.016 mole of gas at 7 C, what are the heat, the change n the nternal energy, and the work that characterze the process? Soluton The pressure and volume are related through = γ V f γ V = 1 V f V γ = V f = P V 1 γ = 5 1 atm 7 = 0.11768 0 atm The temperature and volume are related through T V γ 1 γ 1 = T f T f γ 1 T V γ 1 = 1 T f T 1 5 = 1 =.3535 0.11768 The heat s obvously zero. The change n the nternal energy and the work are the same and they are E nt = 5 nr T = V γ 1 γ 1 V = V f The ntal temperature s 300 K and the fnal temperature s γ 1 7 T f T =.3535 T f =.3535T =.3535(300 K) = 706.06 K E nt = 5 (0.016 mol)(8.314 )(706.06 K 300 K) = 135.04 page 5
Problem 1.6 An deal gas wth specfc heat rato gamma confned to a cylnder s put through a closed cycle. Intally, the gas s a P, V, and T. Frst, ts pressure s trpled under constant volume. It then expands adabatcally to ts orgnal pressure and fnally s compressed sobarcally to ts orgnal volume. Draw the P-V dagram dagram. Determne the volume at the end of the adabatc expanson. Fnd the temperature of the gas at the start of the adabatc expanson. Fnd the temperature at the end of the cycle. What was the net work done on the gas for ths cycle? Soluton Here s the P-V dagram. Ths ncludes the obvous relatonshps between the states. 3P, V, TB B sochor adabat P, VC, TC Use deal gas equatons for the states and the frst law for the processes. V = nrt 3 V = nrt B V C = nrt C The frst and second gve 3(nRT ) = nrt B T B = 3T The second and thrd are related by And 3 = V C V C = (3V γ ) 1 γ = 3 1 γ V V C = nrt C T C = V C nr = 31 γ V nr = 31 γ T The work done by the cycle s Here, W total = W adabat +W sobar P, V, T A sobar W adabat = f nr T = f nr(t C T B ) = f nr(31 γ T 3T ) = 1 γ 1 nrt (31 γ 3) C page 6
W adabat = (31 γ 3) γ 1 V And So W sobar = V = (V V C ) = (V 3 1 γ V ) = (3 1 γ 1) V W total = (31 γ 3) γ 1 V + (31 γ 1) V = (31 γ 3) + (3 1 γ 1) γ 1 V page 7
Problem 1.33 One cubc meter of atomc hydrogen at 0 C at atmospherc pressure contans approxmately.70x10 5 atoms. The frst excted state of the hydrogen atom has an energy of 10. ev above that of the lowest energy state (ground state). Use the Boltzmann factor to fnd the number of atoms n the frst excted state at 0 C and at 10 4 C. Soluton Accordng to the Boltzmann factor, n(e) = n o e E k B T The number at the energy of 10. ev when the temperature s 73 K s n(10. ev ) = (.7 10 5 )e (10. ev ) (1.38 10 3 )(73 K ) K We have to convert electron volts to joules frst. 10. ev 1.60 10 19 1 ev = 1.63 10 18 n(10. ev ) = (.7 10 5 )e (1.63 10 18 ) (1.38 10 3 )(73 K ) K = 10 163 We can safely call ths zero. The number at 1073 K s n(10. ev ) = (.7 10 5 )e (1.63 10 18 ) (1.38 10 3 )(1073 K ) K =.7 10 0 page 8
Problem 1.45 Soluton () The gas s heated at constant pressure to 400 K. (a)at the ntal state, the pressure s 1.00x10 5 Pa. At the fnal state, the pressure s the same, 1.00x10 5 Pa. (b)the fnal volume s = nrt f (1.00 10 5 Pa) = (.00 mol)(8.314 )(400 K) V f = 0.06651 m 3 (c)the fnal temperature s 400 K. (d)because the constant volume specfc heat s 7R/, there are 7 degrees of freedom. change n the nternal energy s ΔE nt = 7 nrδt = 7 (.00 mol)(8.314 (e)the heat appled to the gas s, for an sobarc process, Q = 9 nrδt = 9 (.00 mol)(8.314 (f) The work s done on the gas s W = nrδt = (.00 mol)(8.314 Or you can use the frst law drectly. ()The gas s heated at constant volume to 400 K. (a)at the ntal state, the volume s At the fnal state, the volume s the same. (b)the fnal volume s 0.049884 m 3. (c)the fnal temperature s 400 K. )(100 K) = 5819.8 )(100 K) = 748.6 )(100 K) = 166.8 V = nrt (1.00 10 5 Pa)V = (.00 mol)(8.314 )(300 K) V = 0.049884 m 3 = nrt f (0.06651 m 3 ) = (.00 mol)(8.314 )(400 K) P f = 133,330 Pa The page 9
(d)the change n the nternal energy s stll ΔE nt = 5819.8 (e)the heat appled to the gas s, for an sochorc process, Q = nc V ΔT = 7 nrδt = 7 (.00 mol)(8.314 (f) The work s done on the gas s zero. ()The gas s compressed at constant temperature to 1.0x10 5 Pa. (a)the fnal pressure s 1.0x10 5 Pa. (b)at the fnal state, the volume s, snce the fnal temperature s unchanged, (c)the fnal temperature s 300 K. (d)the change n the nternal energy s zero. )(100 K) = 5819.8 = nrt f (1.0 10 5 Pa) = (.00 mol)(8.314 )(300 K) V f = 0.041570 m 3 (e)the heat appled to the gas s, for an sothermal process, Q = nrt ln V f = (.00 mol)(8.314 V 0.049984 m3 )ln 0.041570 m 3 = 3.0649 (f) The work s done on the gas s 3.0649. (v)the gas s compressed adabatcally to 1.0x10 5 Pa. (a)the fnal pressure s 1.0x10 5 Pa. (b)at the fnal state, the followng s true. = (1.00 10 5 Pa)(0.049884 m 3 ) 9 7 = (1.0 10 5 9 7 Pa) = 0.04389 m 3 (c)the fnal temperature s = nrt f (1.0 10 5 Pa)(0.04389 m 3 ) = (.00 mol)(8.314 (d)the change n the nternal energy s ΔE nt = 7 nrδt = 7 (.00 mol)(8.314 (e)the heat appled to the gas s zero. (f) The work s done on the gas s 14.733. )(31.40 K 300 K) = 14.733 )T f T f = 31.40 K page 10
Problem 1.51 Soluton Here s what the stuaton looks lke. 50 cm 0.03 cm P 10 m/s 1 cm 3 We can treat the ar as an deal gas. V = nrt = nrt f The expanson s adabatc. The states of the gas are connected adabatcally. = T V γ 1 γ 1 = T f The total work done on the gas s negatve of the work done by the gas expandng aganst the atmosphere and the knetc energy of the bullet. The queston says to gnore the ar so the pressure from the atmosphere s gnored. W = 1 m v = 1 (1.10 10 3 kg)(10 m/s) = 7.9 The work done on the gas under an adabatc expanson process s also equal to the change n the nternal energy s W = ΔE nt = 5 nr T = 7.9 Here, gamma s 1.40 = 7/5, whch means f = 5. Ths can be wrtten n terms of the pressures and the volumes. nrδt = V page 11
So W = 5 nr T = 5 ( V ) = 7.9 The two pressures are unknown. But we do know ths also about the two pressures. = Solve for the fnal pressure to get the ntal pressure. The fnal volume s 1 cm 3 + (50 cm)(0.03 cm ) = 13.5 cm 3. W = 7.9 = 5 ( V ) = W + P 5 V = ( 7.9 ) + P 5 (1 10 6 m 3 ) =.3467 10 5 Pa + 0.88888P 13.5 10 6 m 3 The ntal pressure s = = V γ = (.3467 10 5 Pa + 0.88888 ) 13.5 10 6 m 3 1 10 6 m 3 1.4 =.7674 10 5 Pa + 1.048 = 5.7379 10 6 Pa = 56.69 atm page 1
Problem 1.71 Soluton There are two states here and each state have two gases. Let the ntal volumes be V. P 1 V 1 = nrt 1 P 1 V = nr(550) P V = nrt P V = nr(50) Let the fnal pressures be P. P 1f V 1f = nrt 1f PV 1f = nrt 1f V f = nrt f PV f = nrt f No heat flows between these gases and the outsde so the ntal state and the fnal state for both sdes are on ther own adabats. Ths means that the total nternal energy for both sdes s conserved. Ths also means that the total work done on the two gases add up to zero. ΔE 1 + ΔE = 0 W 1 +W = 0 Here s the algebra. The rato of the ntal state equatons s ths. P 1 V P V = nr(550) nr(50) P 1 P = 550 50 = 11 5 The states on each sde are connected adabatcally. The adabatc ndex s 1.4. P 1 1 = P 1f V 1f P 1 = PV 1f P = V f P = PV f Ther rato s ths. P 1 = V 1f P V f γ = 11 5 V 1f = 11 V f 5 1/γ = 1.7563 page 13
The rato of the fnal state equatons s ths. PV 1f PV f = nrt 1f nrt f V 1f V f = T 1f T f = 1.7563 T 1f = 1.7563T f The energy equaton says ths ΔE 1 + ΔE = 0 ΔT 1 + ΔT = 0 T 1f T 1 +T f T = 0 T 1f +T f = 800 K 1.7563T f +T f =.7563T f = 800 K T f = 90.4 K T 1f = 1.7563T f = 1.7563(90.4 K) = 509.76 K Here s what the states of the processes look lke. P left (1) rght () Above, I sad ths. W 1 +W = 0 V Ths s clearly not the area under the adabats. The only way that ths can be consstent s f the processes were not adabatc. The adabats connect only the end states of the two gases, t s not the path the gas actually took gong from the ntal to the fnal states. Here s one possble example of how to connect the end states that would make the work equal n magntude. P left (1) rght () The queston states that the pston s allowed to move slowly. Ths means a process that s quas-statc. It mples reversblty. In ths case, we can, at most, say that the process s quasstatc and that s lkely untrue. It s defntely not reversble. V page 14