Thermodynamics

Question 1 (12)

(a) Consider molecule with velocity component $v_x$ hitting wall perpendicular to x. Momentum change per collision $= 2mv_x$ (elastic bounce reverses $v_x$). (1) Time between collisions with same wall $= 2L/v_x$. (1) Force from one molecule $= \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$. (1) Total force $= \frac{m}{L}\sum v_{x,i}^2 = \frac{Nm\overline{v_x^2}}{L}$. (1) Pressure $P = F/L^2 = \frac{Nm\overline{v_x^2}}{L^3} = \frac{Nm\overline{v_x^2}}{V}$. (1) Isotropy: $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\tfrac13\overline{v^2}$. (1) Hence $P = \frac{1}{3}\frac{Nm\overline{v^2}}{V}$. (1)

(b) $PV = \frac{1}{3}Nm\overline{v^2}$. Compare $PV = Nk_BT$: $\frac{1}{3}m\overline{v^2}=k_BT \Rightarrow \frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT$. (2) Temperature $\propto$ mean translational KE. (1)

(c) Assumptions: point molecules (negligible volume); no intermolecular forces except elastic collisions; walls perfectly elastic / large N so distribution steady. (2)

Question 2 (10)

(a) Adiabatic: $dQ=0$, first law $dU = -dW = -P\,dV$. (1) $nC_V\,dT = -P\,dV$. (1) Ideal gas $PV=nRT \Rightarrow P\,dV + V\,dP = nR\,dT$. (1) So $dT = \frac{P\,dV+V\,dP}{nR}$. Substitute: $\frac{C_V}{R}(P\,dV+V\,dP) = -P\,dV$. (1) $C_V V\,dP + (C_V+R)P\,dV = 0$; use $C_P=C_V+R$: $C_V V\,dP + C_P P\,dV=0 \Rightarrow \frac{dP}{P}+\gamma\frac{dV}{V}=0$. (1) Integrate: $\ln P + \gamma\ln V = \text{const} \Rightarrow PV^\gamma=\text{const}$. (1)

(b) Sub $P=nRT/V$: $\frac{nRT}{V}V^\gamma = TV^{\gamma-1}\cdot nR=\text{const}\Rightarrow TV^{\gamma-1}=\text{const}$. (2)

(c) $T_f/T_i = (V_i/V_f)^{\gamma-1} = (1/2)^{2/3} = 0.630$. (2)

Question 3 (11)

(a) Two isotherms (top at $T_H$, bottom at $T_C$) joined by two adiabats forming a closed loop; label 1→2 isothermal expansion at $T_H$, 2→3 adiabatic expansion, 3→4 isothermal compression at $T_C$, 4→1 adiabatic compression. (2)

(b) Isothermal $T_H$: $Q_H = nRT_H\ln(V_2/V_1)$. (1) Isothermal $T_C$: $Q_C = nRT_C\ln(V_3/V_4)$. (1) Adiabats: $T_H V_2^{\gamma-1}=T_C V_3^{\gamma-1}$, $T_H V_1^{\gamma-1}=T_C V_4^{\gamma-1}$. (1) Divide: $V_2/V_1 = V_3/V_4$. (1) Therefore $\frac{Q_C}{Q_H}=\frac{T_C}{T_H}$. (1) $\eta = 1 - Q_C/Q_H$. (1) $\Rightarrow \eta = 1 - T_C/T_H$. (1)

(c) $\eta = 1-300/600 = 0.5$. $W = 0.5\times900 = 450\text{ J}$; $Q_C = 900-450 = 450\text{ J}$. (2)

Question 4 (9)

(a) Kelvin–Planck: no process can convert heat entirely into work with no other effect (no single-reservoir engine). Clausius: heat cannot flow spontaneously from cold to hot without external work. (2)

(b) $\Delta S = Q/T = mL_f/T = (1)(3.34\times10^5)/273.15 = 1223\ \text{J/K}$. (3)

(c) $\Delta S = \int_{373.15}^{273.15}\frac{mc\,dT}{T} = mc\ln(T_f/T_i)$ $= (1)(4186)\ln(273.15/373.15) = 4186\times(-0.3119) = -1305\ \text{J/K}$. (4) (Negative because water loses entropy; environment gains more, total > 0.)

Question 5 (10)

(a) Fourier: $\dot Q = -kA\,dT/dx$. Stefan–Boltzmann: $P = \sigma A e T^4$ (emitted). (2)

(b) Area $=4\pi r^2 = 4\pi(0.01)=0.12566\ \text{m}^2$. (1) Net $P = \sigma A (T^4 - T_s^4) = 5.67\times10^{-8}\times0.12566\times(500^4-300^4)$. (1) $500^4 = 6.25\times10^{10}$, $300^4 = 8.1\times10^{9}$; diff $=5.44\times10^{10}$. (1) $P = 5.67\times10^{-8}\times0.12566\times5.44\times10^{10} \approx 387.5\ \text{W}$. (1)

(c) $\dot Q = kA\Delta T/L = 200\times2\times10^{-4}\times100/0.5 = 8\ \text{W}$. (2)

(d) Radiation scales as $T^4$ while conduction/convection scale roughly linearly with $\Delta T$; so at high $T$ the $T^4$ term grows far faster and dominates. (2)

Question 6 (8)

(a) The distribution $f(v)\propto v^2 e^{-mv^2/2k_BT}$ has a $v^2$ factor (rising) times a Gaussian (falling), producing an asymmetric, right-skewed curve with a long high-speed tail. Ordering: $v_{mp}<\overline{v}<v_{rms}$, specifically $v_{mp}=\sqrt{2k_BT/m}$, $\overline v=\sqrt{8k_BT/\pi m}$, $v_{rms}=\sqrt{3k_BT/m}$. The skew arises because speed is non-negative and the tail extends to infinity. (4)

(b) (1 mark each: correct formula, R & M usage, sorting, output)

import numpy as np
R = 8.314
def speeds(T, M):   # M in kg/mol
    v_rms = np.sqrt(3*R*T/M)
    v_mp  = np.sqrt(2*R*T/M)
    v_avg = np.sqrt(8*R*T/(np.pi*M))
    return sorted([v_mp, v_avg, v_rms])
print(speeds(300, 0.028))  # N2

(4)

[
  {"claim":"Q2c adiabatic T ratio (1/2)^(2/3)=0.630","code":"r=Rational(1,2)**Rational(2,3); result=abs(float(r)-0.630)<0.001"},
  {"claim":"Q3c Carnot work=450J, Qc=450J","code":"eta=1-Rational(300,600); W=eta*900; Qc=900-W; result=(W==450 and Qc==450)"},
  {"claim":"Q4b entropy of melting ice =1223 J/K","code":"dS=1*3.34e5/273.15; result=abs(dS-1223)<2"},
  {"claim":"Q4c water cooling entropy approx -1305 J/K","code":"import sympy as sp; dS=4186*sp.log(273.15/373.15); result=abs(float(dS)+1305)<3"},
  {"claim":"Q5c conduction current=8W","code":"Q=200*2e-4*100/0.5; result=abs(Q-8)<1e-9"},
  {"claim":"Q5b net radiation approx 387.5W","code":"P=5.67e-8*4*pi*0.01*(500**4-300**4); result=abs(float(P)-387.5)<3"}
]