Level 2 — RecallThermodynamics

Thermodynamics

40 marksprintable — key stays hidden on paper

Level 2 (Recall & Standard Problems)

Time: 30 minutes Total marks: 40

Use R=8.314 Jmol1K1R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}, σ=5.67×108 Wm2K4\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}, kB=1.38×1023 J/Kk_B = 1.38\times10^{-23}\ \mathrm{J/K} where needed.


Q1. State the Zeroth law of thermodynamics and explain briefly why it justifies the use of a thermometer. (3 marks)

Q2. State the first law of thermodynamics, defining each symbol and its sign convention using the form dU=dQdWdU = dQ - dW. (3 marks)

Q3. A copper block of mass 0.50 kg0.50\ \mathrm{kg} is heated from 20C20^\circ\mathrm{C} to 70C70^\circ\mathrm{C}. Given specific heat capacity c=385 Jkg1K1c = 385\ \mathrm{J\,kg^{-1}K^{-1}}, calculate the heat supplied. (3 marks)

Q4. How much heat is required to convert 200 g200\ \mathrm{g} of ice at 0C0^\circ\mathrm{C} into water at 0C0^\circ\mathrm{C}? (Latent heat of fusion Lf=3.34×105 Jkg1L_f = 3.34\times10^5\ \mathrm{J\,kg^{-1}}.) Then how much additional heat raises this water to 30C30^\circ\mathrm{C} (take cwater=4186 Jkg1K1c_{water}=4186\ \mathrm{J\,kg^{-1}K^{-1}})? (4 marks)

Q5. A metal rod of length 1.00 m1.00\ \mathrm{m} has linear expansion coefficient α=1.2×105 K1\alpha = 1.2\times10^{-5}\ \mathrm{K^{-1}}. Find its increase in length when heated by 80 K80\ \mathrm{K}. State the volumetric expansion coefficient in terms of α\alpha. (4 marks)

Q6. (a) Write the ideal gas law PV=nRTPV = nRT. (b) Calculate the pressure of 2.0 mol2.0\ \mathrm{mol} of an ideal gas in a 0.050 m30.050\ \mathrm{m^3} container at 300 K300\ \mathrm{K}. (4 marks)

Q7. For one mole of an ideal gas undergoing an isothermal expansion from volume V1V_1 to V2V_2 at temperature TT, derive the expression for the work done by the gas. (4 marks)

Q8. (a) Write the relation for the RMS speed of gas molecules in terms of temperature and molar mass. (b) Compute vrmsv_{rms} for oxygen (M=0.032 kg/molM = 0.032\ \mathrm{kg/mol}) at 300 K300\ \mathrm{K}. (4 marks)

Q9. A Carnot engine operates between a hot reservoir at 500 K500\ \mathrm{K} and a cold reservoir at 300 K300\ \mathrm{K}. (a) Find its efficiency. (b) State how efficiency depends on the reservoir temperatures. (4 marks)

Q10. State the Clausius definition of entropy change and compute the entropy change when 500 J500\ \mathrm{J} of heat is transferred reversibly to a reservoir at 250 K250\ \mathrm{K}. (3 marks)


End of paper

Answer keyMark scheme & solutions

Q1. (3 marks)

  • Statement: If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. (2)
  • Justification: this transitivity means a thermometer (the "third system") gives a consistent temperature reading; two bodies at the same thermometer reading are in equilibrium. (1)

Q2. (3 marks)

  • dU=dQdWdU = dQ - dW. (1)
  • dUdU = change in internal energy; dQdQ = heat added to the system (positive when absorbed); dWdW = work done by the system (positive when gas expands). (2)

Q3. (3 marks)

  • Q=mcΔTQ = mc\Delta T (1)
  • ΔT=50 K\Delta T = 50\ \mathrm{K}; Q=0.50×385×50Q = 0.50 \times 385 \times 50 (1)
  • Q=9625 J9.6 kJQ = 9625\ \mathrm{J} \approx 9.6\ \mathrm{kJ} (1)

Q4. (4 marks)

  • Melting: Q1=mLf=0.200×3.34×105=6.68×104 JQ_1 = mL_f = 0.200 \times 3.34\times10^5 = 6.68\times10^4\ \mathrm{J} (2)
  • Heating: Q2=mcΔT=0.200×4186×30=25116 J2.51×104 JQ_2 = mc\Delta T = 0.200 \times 4186 \times 30 = 25116\ \mathrm{J} \approx 2.51\times10^4\ \mathrm{J} (2)

Q5. (4 marks)

  • ΔL=αL0ΔT=1.2×105×1.00×80\Delta L = \alpha L_0 \Delta T = 1.2\times10^{-5}\times1.00\times80 (1)
  • ΔL=9.6×104 m=0.96 mm\Delta L = 9.6\times10^{-4}\ \mathrm{m} = 0.96\ \mathrm{mm} (2)
  • Volumetric coefficient β=3α=3.6×105 K1\beta = 3\alpha = 3.6\times10^{-5}\ \mathrm{K^{-1}} (1)

Q6. (4 marks)

  • (a) PV=nRTPV = nRT (1)
  • (b) P=nRTV=2.0×8.314×3000.050P = \dfrac{nRT}{V} = \dfrac{2.0\times8.314\times300}{0.050} (2)
  • P=99768 Pa9.98×104 PaP = 99768\ \mathrm{Pa} \approx 9.98\times10^4\ \mathrm{Pa} (1)

Q7. (4 marks)

  • W=V1V2PdVW = \int_{V_1}^{V_2} P\,dV (1)
  • Isothermal: P=nRTVP = \dfrac{nRT}{V} (from PV=nRTPV=nRT, TT const) (1)
  • W=nRTV1V2dVVW = nRT\int_{V_1}^{V_2}\dfrac{dV}{V} (1)
  • W=nRTln ⁣(V2V1)W = nRT\ln\!\left(\dfrac{V_2}{V_1}\right) (with n=1n=1: W=RTln(V2/V1)W = RT\ln(V_2/V_1)) (1)

Q8. (4 marks)

  • (a) vrms=3RTMv_{rms} = \sqrt{\dfrac{3RT}{M}} (1)
  • (b) vrms=3×8.314×3000.032v_{rms} = \sqrt{\dfrac{3\times8.314\times300}{0.032}} (2)
  • =233831483 m/s= \sqrt{233831} \approx 483\ \mathrm{m/s} (1)

Q9. (4 marks)

  • (a) η=1TCTH=1300500=0.40\eta = 1 - \dfrac{T_C}{T_H} = 1 - \dfrac{300}{500} = 0.40 (40%) (2+1)
  • (b) Efficiency increases as THT_H rises or TCT_C falls; maximum when TC0T_C\to0. (1)

Q10. (3 marks)

  • dS=dQrevTdS = \dfrac{dQ_{rev}}{T} (1)
  • ΔS=500250\Delta S = \dfrac{500}{250} (1)
  • ΔS=2.0 J/K\Delta S = 2.0\ \mathrm{J/K} (1)
[
{"claim":"Q3 heat for copper block = 9625 J","code":"result = (0.50*385*50 == 9625)"},
{"claim":"Q4 total heat = 66800 + 25116 J","code":"Q1=0.200*3.34e5; Q2=0.200*4186*30; result = (abs(Q1-66800)<1e-6) and (abs(Q2-25116)<1e-6)"},
{"claim":"Q6 pressure approx 99768 Pa","code":"P=2.0*8.314*300/0.050; result = abs(P-99768)<1"},
{"claim":"Q8 v_rms oxygen approx 483 m/s","code":"v=sqrt(3*8.314*300/0.032); result = abs(float(v)-483)<1"},
{"claim":"Q9 Carnot efficiency = 0.40","code":"result = (1 - Rational(300,500) == Rational(2,5))"},
{"claim":"Q10 entropy change = 2.0 J/K","code":"result = (Rational(500,250) == 2)"}
]