Noesis
Level 5 — MasteryThermodynamics

Thermodynamics

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Answer keyMark scheme & solutions

Question 1

(a) [6]

  • Assumptions: large N, random motion, negligible molecular volume, elastic collisions, no intermolecular forces. [1]
  • A molecule hitting wall (⊥ x) reverses vxv_x: momentum change 2mvx2mv_x. Time between wall hits 2L/vx2L/v_x; force per molecule =mvx2/L=m v_x^2/L. [2]
  • Sum over molecules, pressure P=1V∑mvx,i2=NVm⟨vx2⟩P=\frac{1}{V}\sum m v_{x,i}^2 = \frac{N}{V}m\langle v_x^2\rangle. Isotropy ⟨vx2⟩=13⟨v2⟩\langle v_x^2\rangle=\tfrac13\langle v^2\rangle gives P=13NVm⟨v2⟩P=\tfrac13\frac{N}{V}m\langle v^2\rangle. [2]
  • Compare PV=NkBTPV=Nk_BT: 13Nm⟨v2⟩=NkBT⇒12m⟨v2⟩=32kBT\tfrac13 Nm\langle v^2\rangle = Nk_BT \Rightarrow \tfrac12 m\langle v^2\rangle=\tfrac32 k_BT. [1]

(b) [9]

  • vpv_p: set dfdv=0\frac{df}{dv}=0. From v2e−mv2/2kBTv^2 e^{-mv^2/2k_BT}: 2v−v2⋅mvkBT=0⇒vp=2kBT/m2v - v^2\cdot\frac{mv}{k_BT}=0\Rightarrow v_p=\sqrt{2k_BT/m}. [3]
  • ⟨v⟩=∫0∞vf(v)dv=8kBT/πm\langle v\rangle=\int_0^\infty v f(v)dv=\sqrt{8k_BT/\pi m} (using ∫0∞x3e−ax2dx=1/2a2\int_0^\infty x^3 e^{-ax^2}dx=1/2a^2). [3]
  • vrms=⟨v2⟩=3kBT/mv_{rms}=\sqrt{\langle v^2\rangle}=\sqrt{3k_BT/m} (using ∫x4e−ax2\int x^4e^{-ax^2}). [1]
  • Ratios: vp:⟨v⟩:vrms=2:8/π:3=1.414:1.596:1.732v_p:\langle v\rangle:v_{rms}=\sqrt2:\sqrt{8/\pi}:\sqrt3=1.414:1.596:1.732. Ordering confirmed. [2]

(c) [7]

  • vrms=3RT/M=3⋅8.314⋅3000/0.028=2.673×106=1635 m/sv_{rms}=\sqrt{3RT/M}=\sqrt{3\cdot8.314\cdot3000/0.028}=\sqrt{2.673\times10^{6}}=1635\ \mathrm{m/s}. [3]
  • Fraction =∫v0∞f(v) dv=\int_{v_0}^\infty f(v)\,dv with m/2kBT=M/2RTm/2k_BT=M/2RT. Let a=M/2RT=0.028/(2⋅8.314⋅3000)=5.614×10−7a=M/2RT=0.028/(2\cdot8.314\cdot3000)=5.614\times10^{-7}. [2]
  • Numeric estimate: dimensionless u0=v0/vpu_0=v_0/v_p, vp=2RT/M=1335 m/sv_p=\sqrt{2RT/M}=1335\ \mathrm{m/s}, u0=2.247u_0=2.247. Fraction =4π∫u0∞u2e−u2du≈0.077=\frac{4}{\sqrt\pi}\int_{u_0}^\infty u^2e^{-u^2}du\approx 0.077 (~7.7%). [2]

Question 2

(a) [6]

  • First law dU=dQ−dWdU=dQ-dW; adiabatic dQ=0⇒dU=−P dVdQ=0\Rightarrow dU=-P\,dV. [1]
  • dU=nCVdTdU=nC_V dT, so nCVdT=−P dVnC_V dT=-P\,dV. [1]
  • PV=nRT⇒nR dT=P dV+V dPPV=nRT\Rightarrow nR\,dT=P\,dV+V\,dP; substitute ndT=(PdV+VdP)/RndT=(PdV+VdP)/R. [1]
  • CV(PdV+VdP)/R=−PdV⇒CVVdP=−(CV+R)PdV=−CPPdVC_V(PdV+VdP)/R=-PdV\Rightarrow C_V VdP=-(C_V+R)PdV=-C_P PdV. [1]
  • dPP=−γdVV\frac{dP}{P}=-\gamma\frac{dV}{V}, γ=CP/CV\gamma=C_P/C_V; integrate ⇒PVγ=\Rightarrow PV^\gamma=const. [2]

(b) [8]

  • Four steps: isothermal expansion at THT_H (QHQ_H absorbed), adiabatic expansion, isothermal compression at TCT_C (QCQ_C rejected), adiabatic compression. [2]
  • Entropy over isotherms: QH/TH=QC/TCQ_H/T_H=Q_C/T_C (net ΔScycle=0\Delta S_{cycle}=0), so QC/QH=TC/THQ_C/Q_H=T_C/T_H. [2]
  • η=1−TC/TH=1−300/600=0.5\eta=1-T_C/T_H=1-300/600=0.5 (50%). [2]
  • W=ηQH=0.5⋅1200=600 JW=\eta Q_H=0.5\cdot1200=600\ \mathrm{J}; QC=QH−W=600 JQ_C=Q_H-W=600\ \mathrm{J}. [2]

(c) [6]

  • Free expansion: W=0W=0 (no external pressure), Q=0Q=0 (isolated) ⇒ ΔU=0\Delta U=0 ⇒ ΔT=0\Delta T=0. [2]
  • Entropy is a state function; use reversible isothermal path: ΔS=nRln⁡(V2/V1)=1⋅8.314⋅ln⁡2=5.76 J/K\Delta S=nR\ln(V_2/V_1)=1\cdot8.314\cdot\ln2=5.76\ \mathrm{J/K}. [2]
  • ΔS>0\Delta S>0 though Q=0Q=0 because process is irreversible: Clausius inequality ΔS≥∫dQ/T\Delta S\ge\int dQ/T; here ∫dQ/T=0<ΔS\int dQ/T=0<\Delta S. [2]

(d) [4]

  • Microstates W∝VNW\propto V^N (each molecule independently accesses volume). Doubling: W2/W1=2NW_2/W_1=2^N. [2]
  • ΔS=kBln⁡(2N)=NkBln⁡2\Delta S=k_B\ln(2^N)=Nk_B\ln2. For N=NAN=N_A: ΔS=NAkBln⁡2=Rln⁡2=5.76 J/K\Delta S=N_A k_B\ln2=R\ln2=5.76\ \mathrm{J/K}, matching (c). [2]

Question 3

(a) [7]

  • Fourier: Q=kA dT/LQ=kA\,dT/L. Series resistances Ri=Li/(kiA)R_i=L_i/(k_iA) add. [2]
  • R1=0.10/(0.5⋅1)=0.20R_1=0.10/(0.5\cdot1)=0.20; R2=0.20/(2.0⋅1)=0.10R_2=0.20/(2.0\cdot1)=0.10; Rtot=0.30 K/WR_{tot}=0.30\ \mathrm{K/W}. [2]
  • Q=ΔT/Rtot=100/0.30=333.3 WQ=\Delta T/R_{tot}=100/0.30=333.3\ \mathrm{W}. [2]
  • Interface: Tint=400−QR1=400−333.3⋅0.20=333.3 KT_{int}=400-Q R_1=400-333.3\cdot0.20=333.3\ \mathrm{K}. [1]

(b) [7]

  • Scheme: Tin+1=Tin+αΔtΔx2(Ti+1n−2Tin+Ti−1n)T_i^{n+1}=T_i^n+\frac{\alpha\Delta t}{\Delta x^2}(T_{i+1}^n-2T_i^n+T_{i-1}^n); loop over time, apply BCs. [3]
  • Stability: αΔtΔx2≤12\frac{\alpha\Delta t}{\Delta x^2}\le \tfrac12. [1]
  • Expansion: ΔL=αlinL0ΔT=1.2×10−5⋅1⋅50=6×10−4 m=0.6 mm\Delta L=\alpha_{lin}L_0\Delta T=1.2\times10^{-5}\cdot1\cdot50=6\times10^{-4}\ \mathrm{m}=0.6\ \mathrm{mm}. [2]
  • Fractional volume change ≈3αlinΔT=3⋅6×10−4=1.8×10−3\approx 3\alpha_{lin}\Delta T=3\cdot6\times10^{-4}=1.8\times10^{-3} (0.18%). [1]
[
  {"claim":"Carnot efficiency 600/300K = 0.5, W=600J, QC=600J","code":"TH=600;TC=300;QH=1200;eta=1-TC/TH;W=eta*QH;QC=QH-W;result=(eta==Rational(1,2)) and (W==600) and (QC==600)"},
  {"claim":"Free expansion entropy nR ln2 = 5.76 J/K","code":"import sympy as sp;R=sp.Float(8.314);dS=R*sp.log(2);result=abs(float(dS)-5.763)<0.01"},
  {"claim":"MB speed ratios vp:vmean:vrms","code":"import sympy as sp;vp=sp.sqrt(2);vm=sp.sqrt(sp.Rational(8,1)/sp.pi);vr=sp.sqrt(3);result=(float(vp)<float(vm)<float(vr)) and abs(float(vm)-1.5958)<0.001"},
  {"claim":"Composite wall Q=333.3W, interface=333.3K","code":"R1=sp.Rational(1,5);R2=sp.Rational(1,10);Rtot=R1+R2;Q=100/Rtot;Tint=400-Q*R1;result=abs(float(Q)-333.333)<0.01 and abs(float(Tint)-333.333)<0.01"},
  {"claim":"vrms N2 3000K approx 1635 m/s","code":"import sympy as sp;vr=sp.sqrt(3*sp.Float(8.314)*3000/sp.Float(0.028));result=abs(float(vr)-1635)<5"}
]