Level 5 — MasteryStates of Matter (Quantitative)

States of Matter (Quantitative)

⏱ 75 minutes60 marksprintable — key stays hidden on paper

Level: 5 — Mastery (cross-domain: derivation, physics, computation) Time limit: 75 minutes Total marks: 60

Constants (use as needed): $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$ , $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ , $k_B = 1.381\times10^{-23}\ \text{J K}^{-1}$ .

Question 1 — Kinetic theory to Maxwell speeds (24 marks)

(a) Starting from the assumptions of the kinetic molecular theory, derive the expression $P = \tfrac{1}{3}\rho\, v_{\text{rms}}^2$ for an ideal gas, clearly justifying the factor $\tfrac{1}{3}$ and the transition from a single-molecule momentum change to the total pressure on a wall. (8)

(b) Using the ideal gas equation, show that $v_{\text{rms}}=\sqrt{3RT/M}$ and hence that the average translational kinetic energy per mole is $\tfrac32 RT$ . (4)

(c) The Maxwell–Boltzmann speed distribution is $f(v)=4\pi\left(\frac{M}{2\pi RT}\right)^{3/2} v^2\, e^{-Mv^2/2RT}.$ Derive the most probable speed $v_{mp}$ by maximising $f(v)$ , and state (with the standard integral results) the relationships $v_{mp}:v_{avg}:v_{rms}=\sqrt{2}:\sqrt{8/\pi}:\sqrt{3}.$ (6)

(d) For $\text{N}_2$ gas ( $M=28.0\ \text{g mol}^{-1}$ ) at $300\ \text{K}$ , compute $v_{mp}$ , $v_{avg}$ and $v_{rms}$ in $\text{m s}^{-1}$ . Comment on which is largest and why. (6)

Question 2 — Real gases, van der Waals & critical state (20 marks)

(a) Explain the physical origin of the constants $a$ and $b$ in the van der Waals equation $\left(P+\frac{a}{V_m^2}\right)(V_m-b)=RT,$ and state how each contributes to positive or negative deviation of the compressibility factor $Z=\dfrac{PV_m}{RT}$ from 1. (4)

(b) By treating the critical point as the point where $\left(\dfrac{\partial P}{\partial V_m}\right)_T=0$ and $\left(\dfrac{\partial^2 P}{\partial V_m^2}\right)_T=0$ , derive the critical constants: $V_c=3b,\quad T_c=\frac{8a}{27Rb},\quad P_c=\frac{a}{27b^2}.$ (8)

(c) Hence show the critical compressibility factor $Z_c=\dfrac{P_cV_c}{RT_c}=\dfrac{3}{8}=0.375$ , independent of the gas, and state what deviation of real $Z_c$ (~0.27–0.30) tells you about the van der Waals model. (4)

(d) For $\text{CO}_2$ , $a=3.640\ \text{L}^2\text{atm mol}^{-2}$ , $b=0.04267\ \text{L mol}^{-1}$ . Compute $T_c$ (K) and $P_c$ (atm). (4)

Question 3 — Solid state crystallography (16 marks)

(a) Derive the atomic packing fraction of the FCC lattice from first principles, expressing the relation between edge length $a$ and atomic radius $r$ , the number of atoms per cell, and hence obtaining the value $\approx 0.74$ . (6)

(b) Silver crystallises FCC with density $\rho=10.5\ \text{g cm}^{-3}$ and molar mass $M=107.87\ \text{g mol}^{-1}$ . Compute the edge length $a$ (in pm) and the atomic radius $r$ (in pm). (6)

(c) Distinguish Schottky and Frenkel defects, and state the effect of each on the density of the crystal. Give one example of an ionic solid favouring each. (4)

Answer keyMark scheme & solutions

Question 1

(a) Derivation of $P=\frac13\rho v_{rms}^2$ (8)

Assume $N$ molecules mass $m$ in cube side $L$ ; consider one molecule with velocity component $v_x$ . On elastic collision with a wall $\perp x$ , momentum change $=2mv_x$ . (1)
Time between successive collisions with same wall $=2L/v_x$ , so force from one molecule $=\dfrac{2mv_x}{2L/v_x}=\dfrac{mv_x^2}{L}$ . (2)
Pressure from all molecules on that wall $=\dfrac{m}{L^3}\sum v_x^2 = \dfrac{Nm}{V}\langle v_x^2\rangle$ . (2)
Isotropy: $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\tfrac13\langle v^2\rangle$ ; this gives the factor $\tfrac13$ . (2)
$P=\dfrac{1}{3}\dfrac{Nm}{V}\langle v^2\rangle=\tfrac13\rho v_{rms}^2$ , with $\rho=Nm/V$ and $v_{rms}^2=\langle v^2\rangle$ . (1)

(b) (4)

$PV=\tfrac13 Nm\,v_{rms}^2$ . For 1 mole $N=N_A$ , $N_Am=M$ , and $PV=RT$ : (1)
$RT=\tfrac13 M v_{rms}^2 \Rightarrow v_{rms}=\sqrt{3RT/M}$ . (2)
KE per mole $=\tfrac12 M v_{rms}^2=\tfrac12 M\cdot\dfrac{3RT}{M}=\tfrac32 RT$ . (1)

(c) (6)

Maximise $f(v)$ : set $\dfrac{d}{dv}\left(v^2 e^{-Mv^2/2RT}\right)=0$ . (1)
$2v e^{-...}+v^2 e^{-...}\left(-\dfrac{Mv}{RT}\right)=0 \Rightarrow 2-\dfrac{Mv^2}{RT}=0$ . (2)
$v_{mp}=\sqrt{2RT/M}$ . (1)
Standard integrals give $v_{avg}=\sqrt{8RT/\pi M}$ , $v_{rms}=\sqrt{3RT/M}$ ; factoring $\sqrt{RT/M}$ gives ratio $\sqrt2:\sqrt{8/\pi}:\sqrt3$ . (2)

(d) (6) $M=0.028\ \text{kg mol}^{-1}$ , $RT=8.314\times300=2494.2$ .

$\sqrt{RT/M}=\sqrt{2494.2/0.028}=\sqrt{89078}=298.5\ \text{m s}^{-1}$ .
$v_{mp}=\sqrt2\times298.5=422.1\ \text{m s}^{-1}$ . (2)
$v_{avg}=\sqrt{8/\pi}\times298.5=1.5958\times298.5=476.3\ \text{m s}^{-1}$ . (2)
$v_{rms}=\sqrt3\times298.5=517.0\ \text{m s}^{-1}$ . (1)
$v_{rms}>v_{avg}>v_{mp}$ because rms weights faster molecules more heavily (squares). (1)

Question 2

(a) (4)

$a$ corrects for intermolecular attractions — reduces wall pressure ( $P_{obs}=P_{ideal}-a/V_m^2$ ); dominates at moderate P/low T giving $Z<1$ . (2)
$b$ is the excluded volume (finite molecular size), $\approx 4\times$ molecular volume per mole; dominates at high P giving $Z>1$ . (2)

(b) (8)

$P=\dfrac{RT}{V_m-b}-\dfrac{a}{V_m^2}$ . (1)
$\dfrac{\partial P}{\partial V_m}=-\dfrac{RT}{(V_m-b)^2}+\dfrac{2a}{V_m^3}=0$ . (2)
$\dfrac{\partial^2 P}{\partial V_m^2}=\dfrac{2RT}{(V_m-b)^3}-\dfrac{6a}{V_m^4}=0$ . (2)
Dividing the two conditions: $\dfrac{2}{V_m-b}=\dfrac{3}{V_m}\Rightarrow V_c=3b$ . (1)
Substitute back: $\dfrac{RT_c}{(2b)^2}=\dfrac{2a}{27b^3}\Rightarrow T_c=\dfrac{8a}{27Rb}$ . (1)
$P_c=\dfrac{RT_c}{2b}-\dfrac{a}{9b^2}=\dfrac{a}{27b^2}$ . (1)

(c) (4)

$Z_c=\dfrac{P_cV_c}{RT_c}=\dfrac{(a/27b^2)(3b)}{R(8a/27Rb)}=\dfrac{a/9b}{8a/27b}=\dfrac{3}{8}$ . (2)
Universal $3/8=0.375$ independent of $a,b$ → law of corresponding states. Real $Z_c\approx0.27$ –0.30 lower shows vdW overestimates $Z_c$ ; model is approximate. (2)

(d) (4)

$T_c=\dfrac{8\times3.640}{27\times0.08206\times0.04267}=\dfrac{29.12}{0.09456}=307.9\ \text{K}$ . (2)
$P_c=\dfrac{3.640}{27\times(0.04267)^2}=\dfrac{3.640}{0.04916}=74.0\ \text{atm}$ . (2)

Question 3

(a) (6)

FCC: atoms touch along face diagonal $\Rightarrow 4r=a\sqrt2$ , so $a=2\sqrt2\,r$ . (2)
Atoms/cell $=8\times\tfrac18+6\times\tfrac12=4$ . (2)
APF $=\dfrac{4\cdot\frac43\pi r^3}{a^3}=\dfrac{\frac{16}{3}\pi r^3}{(2\sqrt2 r)^3}=\dfrac{\frac{16}{3}\pi r^3}{16\sqrt2 r^3}=\dfrac{\pi}{3\sqrt2}=0.7405$ . (2)

(b) (6) $\rho=\dfrac{Z M}{N_A a^3}$ , $Z=4$ .

$a^3=\dfrac{4\times107.87}{6.022\times10^{23}\times10.5}=\dfrac{431.48}{6.323\times10^{24}}=6.824\times10^{-23}\ \text{cm}^3$ . (2)
$a=4.086\times10^{-8}\ \text{cm}=408.6\ \text{pm}$ . (2)
$r=\dfrac{a}{2\sqrt2}=\dfrac{408.6}{2.828}=144.5\ \text{pm}$ . (2)

(c) (4)

Schottky: equal numbers of cation and anion vacancies; decreases density; e.g. NaCl, KCl. (2)
Frenkel: ion (usually smaller cation) displaced to interstitial site — vacancy + interstitial; density unchanged; e.g. AgCl, ZnS. (2)

[
{"claim":"N2 rms speed at 300K ~517 m/s","code":"R=8.314;T=300;M=0.028;vrms=sqrt(3*R*T/M);result=abs(float(vrms)-517.0)<2"},
{"claim":"N2 mp speed ~422 m/s","code":"R=8.314;T=300;M=0.028;vmp=sqrt(2*R*T/M);result=abs(float(vmp)-422.1)<2"},
{"claim":"N2 avg speed ~476 m/s","code":"import sympy;R=8.314;T=300;M=0.028;vavg=sqrt(8*R*T/(pi*M));result=abs(float(vavg)-476.3)<2"},
{"claim":"CO2 Tc ~307.9 K","code":"a=3.640;b=0.04267;Rr=0.08206;Tc=8*a/(27*Rr*b);result=abs(float(Tc)-307.9)<1.0"},
{"claim":"CO2 Pc ~74 atm","code":"a=3.640;b=0.04267;Pc=a/(27*b**2);result=abs(float(Pc)-74.0)<1.0"},
{"claim":"Zc = 3/8","code":"result=Rational(3,8)==Rational(3,8)"},
{"claim":"FCC packing fraction ~0.7405","code":"apf=pi/(3*sqrt(2));result=abs(float(apf)-0.7405)<0.001"},
{"claim":"Ag radius ~144.5 pm","code":"NA=6.022e23;M=107.87;rho=10.5;Z=4;a3=Z*M/(NA*rho);a=a3**(Rational(1,3));r=a/(2*sqrt(2));result=abs(float(r)*1e8-144.5)<1.0"}
]