From the equation (18-6) of textbook, we have the Maxwell distribution of speeds as: v 2 e 1 mv 2. 2 kt. 2 4kT m. v p = and f(2vp) as: f(v p ) = 1

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1 Solution Set 9 1 Maxwell-Boltzmann distribution From the equation 18-6 of textbook, we have the Maxwell distribution of speeds as: fv 4πN In example 18-5, the most probable speed v p is given as m v e 1 mv kt. 1.1 πkt v p kt m 1. using this, we can compute the ratios fvp/ fv p and fvp fv p as: fv p / fv p e 1 mv p 4kT 1 4e 1 mvp 4 exp 1 mvp 4kT + 1 mvp 1. kt kt 1 4 exp 1 mvp 1 1 4kT 4 exp m kt 1.4 4kT m 1 4 exp fv p fv p 4e 1 4mvp kt e 1 mvp kt 4 exp 1 4 exp 1 4mvp kt + 1 mvp 4 exp 1 kt mv p kt m kt kt m exp giancoli

2 a glass tumber with water and ice cubes The final status will be where tumbler, water, and ice cubes all are in thermal equilibrium with one another at a same temperature. It is important to determine whether this final temperature is below or above the melting point of the ice, because there is latent heat involved with phase transition between solid ice cube and liquid water phases. see example 19-5 in textbook for a similar approach First we compute an energy release required to bring tumbler and water from 4 C to 0 C. see tables 19-1 and mL of water weighs 00grams [m T c T + m W c W ] 4 0 [0.05kg 840J/kg C kg 4186J/kg C] 4 C kJ 1kJ.1 Now compute amount of energy needed to bring an ice cube from - C into 0 C, m ice c ice kg 100J/kg C C 0.189kJ 0.19kJ. and amount of energy needed to bring an ice cube at 0 C into water at 0 C Since m ice L F 0.00kg kj/kg 9.99kJ 10kJ kJ > 0.189kJ kJ.4 the final phase of water will be in liquid state, with final temperature T > 0 Q mc T Q [mw c W + m T c T ] T 4 + m ice c ice 0 + m ice L F + m ice c W T Solving for T : [m W c W + m T c T ] T 4 + m ice c ice 0 + m ice L F + m ice c W T 0.7 [m W c W + m T c T + m ice c W ] T kJ kJ kJ.8 [0.05kg 840J/kg C + 0.0kg 4186J/kg C] T kJ.9 [ kJ/ C] T kJ.10 T C 11 C.11 4 Nitrogen gas Ideal gas law textbook equation 17- with universal gas constant R The first law of thermodynamics where W is work done by gas. P V nrt 4.1 R 8.14J/mol K 4. E int Q W 4.

3 denotes initial state and the volume is computed as: V nrt P mol R 9K 1.0 mol 8.14J/mol K 9K 5.0atm N/m m m 4.6 Through adiabatic expansion we go to B state. See section 19-9 for adiabatic expansion gas. During a quasistatic adiabatic process of a and Q B P V γ constant 4.8 with γ 1.40 for N gas. Therefore, we have following relations between volumes before and after the adiabatic expansion V B V PB P 1 γ 1 1 γ 1 5 γ Both temperature and volume change, and change in volume corresponds to work. dw P dv. Work can be computed in a similar manner as example 19-1 in this problem we are computing work done by the gas dw P dv, whereas in that example, they compute work done on gas, with opposite sign W B B P V γ The temperature at B would be B P dv P V γ V γ dv P V γ γ γ 1 γ P V 5 1 γ γ 1 γ V 1 γ B γ + 1 V 1 γ 4.10 V 1 γ N/m m kJ 4.14 T B P BV B nr P /5.1569V.1569 P V nr 5 nr.1569 T K K this number looks rather small, however, the nitrogen is still in gas phase because the phase transition to liquid will be at 77 K at 1atm [wikipedia]

4 Next, this goes through Heating/compressing at a constant pressure: or isobaric process, see p.508 from B to C. Since T C K T and P C P B P /5, Boyle s law gives us Since the pressure does not change the work is given simply as V C 5V 4.16 W BC P V N/m 5V.1569V N/m m kJ 4.19 as in equation 19-9a. From textbook p.51, diatomic molecules have internal energy E int 5 nrt 4.0 E int 5 nr T mol 8.14J/mol K K J J kJ 4.5 heat is now Q BC E int + W BC kj kJ 4.8 and then we go through heating at a constant volume isovolumetric from C to D T D 5T C 1465K 4.9 W CD 0 since volume does not change. Q CD E int J 4.60kJ 4.1 table 19-4

5 Finally, we go through isobaric compression compression at constant pressure from D to Since the pressure does not change the work is given simply as W D P V N/m V 5V N/m m kJ 4.4 as in equation 19-9a. From textbook p.51, diatomic mo lecules have internal energy E int 5 nrt 4.5 E int 5 nr T Q CD 4.60kJ heat is now Q D E int + W D kj kJ 4.8 a PV diagram b work W B + W BC + W CD + W D kj kJ -6.6 k J

6 c heat absorbed Q B + Q BC + Q CD + Q D kj kJ -6.6 k J 4.4 lternatively, from the first law of thermodynamics E int Q W, we can argue that the answers to b and c have to equal to each other since E int only depends on temperature, and E int 0 when you come back to the same temperature. 5 oxygen and helium gases a See section 19-9 for adiabatic expansion of a gas. During a quasistatic adiabatic process and with γ 1.67 for Helium gas and γ 1.40 for O gas. Q B P V γ constant 5. Therefore, we have following relations between volumes before and after the adiabatic expansion P f Vf V i γ P i 5. P f,helium P f,oxygen atm atm atm atm 5.5 b temperature T P V nr 5.6 T f Vf V i 1 γ T i 5.7 T f,helium T f,oxygen K K K K 5.9 6

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