UNIVERSITY OF SOUTHAMPTON VERY IMPORTANT NOTE. Section A answers MUST BE in a separate blue answer book. If any blue

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1 UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2011/12 ENERGY AND MATTER SOLUTIONS Duration: 120 MINS VERY IMPORTANT NOTE Section A answers MUST BE in a separate blue answer book. If any blue answer booklets contain work for both Section A and B questions - the latter set of answers WILL NOT BE MARKED. Answer all questions in Section A and two and only two questions in Section B. Section A carries 1/3 of the total marks for the exam paper and you should aim to spend about 40 mins on it. Section B carries 2/3 of the total marks for the exam paper and you should aim to spend about 80 mins on it. A Sheet of Physical Constants will be provided with this examination paper. An outline marking scheme is shown in brackets to the right of each question. Only university approved calculators may be used. Number of Copyright 2012 c University of Southampton Pages??

2 Section A 2 PHYS1013W1 A1. (a) For Cu block: S = T 1 T 2 cdt T = c ln(t 1/T 2 ), heat given up at T. For lake: S = c(t 2 T 1 ) T 1, heat delivered at T 1. [2 marks] ( ) T S Universe = c 2 T 1 T 1 + ln(t 1 /T 2 ) = 150 ( ln(283/273)) = 6.28 J K 1. (b) For Cu block: S = 0. potential energy mgh delivered at T 1, S Universe = mgh T 1 = 1.4 J K 1. problem based on previously seen problem worked through in lecture A2. Flux at solar surface = σt 4 S Flux at dust grain = 10 4 σt 4 S, where 10 4 = 1 (2 50) 2 Energy intercepted = πr 2 }{{} C/S area So, T d = T S = 424 K σts 4 = }{{} 4πr2 σtd 4 sur f ace area in equilibrium. At Earth, substitute 10 4 for , gives T=290 K

3 3 PHYS1013W1 In this case complicated by geothermal activity, i.e. not all heat comes from Sun, or similar. problem based on worked example in lecture notes A3. A reversible process is: Will give for any of the following: performed very slowly quasi-static where no energy is lost to e.g. turbulence, friction An adiabatic change is thermally isolated or where there is no heat flow, i.e. Q = 0. An isothermal change is one where there is no change in temperature or for an ideal gas, no change in internal energy, i.e. U = 0. problem based on question set in mid-semester exam A4. van der Waals provides an equation of state for a real gas. b represents reduction in volume due to finite size of molecules: V (V b) TURN OVER

4 4 PHYS1013W1 a represents increase in pressure due to attractive forces: p (p + a V 2 ). new question A5. Clausius: No process possible whose sole effect is to transfer heat from cold to hot. Kelvin: No process possible whose sole effect is to convert heat into work. [2 marks] Will accept either: Suppose Clausius false, and set up Carnot engine which delivers Q L : Net effect is to convert heat completely into work, which violates Kelvin. or: Suppose Kelvin false: use W from Kelvin-violator to drive heat pump:

5 5 PHYS1013W1 Net effect is Clausius-violator. Therefore if Kelvin is false, Clausius is false. Conclusion: Kelvin Clausius. [3 marks] Equivalence of the two statements has been covered at length in lectures. question based on derivation worked through in lectures TURN OVER

6 Section B 6 PHYS1013W1 B1. a.) Will give for any/each of the following to a maximum of 4 marks: A gas is composed of a large number of molecules. The molecules are small compared to their separation. The molecules are uniformly distributed and move randomly. The molecules obey Newtons laws of motion. The molecules feel no force except during collisions with other molecules or the walls of the container; they are hard spheres. Molecules collide elastically, and the walls are smooth. b.) For an ideal gas: p = 1 3nm < v2 > and internal energy for ideal gas is just K.E., therefore u = 1 2 nm < v2 >. therefore, u = 1 2 3p = 3 2 p

7 7 PHYS1013W1 c.) Piston moving with velocity, v, where we assume v << c (c is speed of molecules, not speed of light). Change in volume is therefore: dv dt = Av number of particles hitting piston per second in (c, c + dc), (θ, θ + dθ) J = n Ac sin θ 2π cos θdθ 4π p(c)dc Note: θ is defined in question as angle between incident particle and piston, rather than between incident particle and the normal to the piston. [3 marks] (d) recoils with velocity, [(c sin θ + 2v) 2 + (c cos θ) 2 ] 1/2 [c 2 + 4cv sin θ] 1/2 [2 marks] So, KE = 1 2m 4cv sin θ = 2mcv sin θ. (e) sin θ cos θdθ rate of change of internal energy: nac 2 p(c)dc 2mcv sin θ. π/2 0 sin 2 θ cos θdθ = 1 3, c 2 p(c)dc =< c 2 > TURN OVER

8 8 PHYS1013W1 du dt = 1 3 na < c2 > mv = 1 3 nav 3k BT = T Avnk B (where 1 2 m < c2 >= 3 2 k BT ) U = 3 du 2RT dt = 3 2 RdT dt = T Avnk B = T( dv dt ) R V (where n = N A V ). du dt = dv dt RT V [2 marks] (f) 3 2 RdT dt = dv dt RT V, cancel dt on both sides and re-arrange: dt T + 2 dv 3 V = 0 TV2/3 = const. problem based on previously seen problem in revision lecture

9 9 PHYS1013W1 B2. a.) [3 marks] for fully labelled p V diagram of Carnot cycle, Fig.?? Figure 1: Carnot Cycle. Will give for any of the following to a maximum of 1 mark: Carnot s theorem: Q 2 /Q 1 is the same for any reversible engine operating between T 1 and T 2. Can be used to define empirical temperature scale: Q 2 /Q 1 T 2 /T 1 Q 2 /Q 1 = T 2 /T 1 therefore thermodynamic temperature the same as ideal gas temperature (up to some factor). b.) Stirling Cycle, Fig.??. [4 marks] A B: Q = c v (T 2 T 1 ) ( W = 0) B C: W = pdv = RT V dv = RT 2 ln(v 2 /V 1 ) out TURN OVER

10 10 PHYS1013W1 Figure 2: Stirling Cycle. = Q in C D: Q = c v (T 2 T 1 ) out D A: W = pdv = RT V dv = RT 1 ln(v 2 /V 1 ) in = Q out [4 marks] ɛ = W out W in Q in = R(T 2 T 1 ) ln(v 2 /V 1 ) c v (T 2 T 1 )+RT 2 ln(v 2 /V 1 ) 1 ɛ = T 2 T 2 T 1 + c v/r ln(v 2 /V 1 ) = T 2 T 2 T 1 + 3/2 ln(v 2 /V 1 ) c.) Efficiency of a Carnot Cycle: ɛ = T 2 T 1 T 2. Therefore, for a Carnot Cycle, 1 ɛ = T 2 T 2 T 1. V 2 > V 1, therefore 1 ɛ stirling > 1 ɛ carnot.

11 11 PHYS1013W1 Therefore ɛ carnot > ɛ stirling, a Carnot Cycle is more efficient than a Stirling Cycle. d.) work extracted = W B C W D A = RT 2 ln(v 2 /V 1 ) RT 1 ln(v 2 /V 1 ) = R(T 2 T 1 ) ln(v 2 /V 1 ) = 8.31 (200 20) ln(3) = 1.64 kj/mol [3 marks] problem based on previously seen problem from problem sheets TURN OVER

12 12 PHYS1013W1 B3. a.) dp dt = Lp RT 2 Re-arrange: dp p = p RT 2 dt Integrate: ln p = L RT + const. approximation: if L f (T). Therefore, p v = p 0 exp( L/RT). b.) ρ liquid = 71 kg m 3, therefore molar volume: = m 3 ρ solid = 81 kg m 3, therefore molar volume: = m 3 Using T = p/3.3: liquid-solid: dp dt = 3.3 MPa K 1 = S V = L/T with T = 14 K. Gives L = 161 J/mol S = L T = 11.5 J/K/mol

13 13 PHYS1013W1 c.) S liquid S solid = 11.5 J/K/mol S vapour S liquid = L T = = 72.1 J/K/mol Therefore, S vapour S solid = 83.6 J/K/mol and L = T S = 1170 J/mol. d.) V vapour V solid V vapour = RT p v = m 3. Using, p v = 90[MPa] exp( 1010/RT) = kpa; T = 14 K. dp dt = S V = = 11.1 kpa K 1. e.) dp dt = Lp RT 2 (p here is p v ) ln p = L RT + const. Treat as linear function (i.e. y = mx + c, with y = ln p and x = 1/T ), gradient is: (ln p)/ (1/T) = L R. TURN OVER

14 14 PHYS1013W1 Re-write table as below. T ln p v 1/T (ln p v )/ (1/T) [2 marks] Boiling point: ln p v = 0, 1 T = 77 C. = ( ) 10 3, therefore T = 350 K [2 marks] f.) L R = 3647 ( ) 7 20 = 3637, therefore L = 30.2 kj/mol. [2 marks] new question

15 15 PHYS1013W1 B4. a.) Flux of photon particles hitting small area of wall da per second at an angle of incidence, θ, is nc cos θda, where n is the number density. The probability of hitting the area da is p(da)= dω 4πr 2, which can be expressed as p(θ)dθ = 1 2 sin θdθ, for a particle with velocity in direction θ (given in question). Momentum of a photon is related to energy as ɛ = Pc. Change in momentum when hitting wall (complete reflection) is: P = 2 ɛ c cos θ Pressure = rate of change of momentum. Integrate: 2 ɛ c p = 1 3 nɛ. 1 cos θ nc cos θ 2 sin θdθ to get [2 marks] nɛ = u, therefore p = 1 3 u will also accept pressure on absorption being p = 1 6u and then being doubled for pressure exerted by cavity wall to maintain equilibrium. derivation using reflection and derivation using absorption both worked through in lectures b.) du = TdS pdv, adiabatic therefore ds = 0; U = uv = 3pV TURN OVER

16 16 PHYS1013W1 3pdV + 3Vdp = 0 pdv 4dV V Integrate to give: V 4 p 3 =const. + 3dp p = 0 pv 4/3 =const. problem based on worked problem in lecture c.) TdS = 4pdV + 3Vdp S = sv; T(sdV + Vds) = 4pdV + 3Vdp ds = ( ) 4p st V dv + 3 T dp But, S is extensive so s should be independent of V, therefore 4p st = 0. So, s = 4p T. problem based on question in problem sheets d.) From (c.) ( s p ) V = 3 T = 4 T 4p ( T T 2 p ) V [2 marks] dp p = 4dT T p T 4

17 17 PHYS1013W1 new question e.) G = U TS + pv Therefore, G = 3pV T 4p T V + pv = 0 new question f.) c V = T ( ) s T T ( ) S T = TV ( ) s V T V s = 4p T = 41 AT 4 3 T = 4 3 AT 3 Therefore, T ( ) S T = TV 4 V 3 A3T 2 = 3s problem based on question sheet problem

18 END OF PAPER 18 PHYS1013W1