States of Matter (Quantitative)

Q1 (12)

(a) A molecule with $x$-velocity $v_x$ hits a wall, momentum change $=2mv_x$ (1). Time between successive hits on same wall $=2L/v_x$ (1), so collision frequency $=v_x/2L$ (1). Force from one molecule $=2mv_x\cdot\frac{v_x}{2L}=\frac{mv_x^2}{L}$ (1).

(b) Total force $=\frac{m}{L}\sum v_{x,i}^2=\frac{Nm\langle v_x^2\rangle}{L}$ (2). Pressure $P=\frac{F}{L^2}=\frac{Nm\langle v_x^2\rangle}{L^3}=\frac{Nm\langle v_x^2\rangle}{V}$ (2). Using isotropy $\langle v_x^2\rangle=\frac13\langle v^2\rangle$: $PV=\frac13 Nm\langle v^2\rangle$ (1).

(c) $\rho=Nm/V$, so $P=\frac13\rho\langle v^2\rangle=\frac13\rho v_{rms}^2$ (1). For 1 mol, $PV=\frac13 M v_{rms}^2=RT\Rightarrow v_{rms}=\sqrt{3RT/M}$ (1). The $\frac13$ arises because motion is shared equally among 3 dimensions; only the $x$-component drives pressure on the $x$-wall (1).

Q2 (10)

(a) $v_p=\sqrt{2RT/M}$, $\bar v=\sqrt{8RT/\pi M}$, $v_{rms}=\sqrt{3RT/M}$ (3). Ratio $v_p:\bar v:v_{rms}=\sqrt2:\sqrt{8/\pi}:\sqrt3=1:1.128:1.225$ (1).

(b) $v_{rms}=\sqrt{3(8.314)(300)/0.032}=\sqrt{233\,base}$; compute: $3\times8.314\times300=7482.6$; $/0.032=233{,}831$; $\sqrt{}=483.6\text{ m s}^{-1}$ (3).

(c) Raising $T$ shifts the peak to higher speed and lowers/broadens it (flattens) (2); total area = total number of molecules (probability =1), which is conserved, so a wider curve must be shorter (1).

Q3 (10)

(a) $a$ = intermolecular attraction (corrects pressure downward); $b$ = excluded volume of molecules (finite molecular size) (2).

(b) $P=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$. First derivative $=-\frac{RT}{(V_m-b)^2}+\frac{2a}{V_m^3}=0$ (1). Second $=\frac{2RT}{(V_m-b)^3}-\frac{6a}{V_m^4}=0$ (1). Dividing the two conditions: $\frac{2}{V_m-b}=\frac{3}{V_m}\Rightarrow V_c=3b$ (2). Sub into first: $T_c=\frac{8a}{27Rb}$ (1); then $P_c=\frac{a}{27b^2}$ (1).

(c) $Z_c=\frac{P_cV_c}{RT_c}=\frac{(a/27b^2)(3b)}{R(8a/27Rb)}=\frac{3/27b}{8/27b}=\frac38=0.375$ (2).

Q4 (10)

(a) $p_{N_2}=\frac{0.20(0.0821)(300)}{10}=0.4926$ atm (1.5); $p_{He}=\frac{0.30(0.0821)(300)}{10}=0.7389$ atm (1.5); $P_{tot}=1.2315$ atm (1).

(b) $\frac{r_X}{r_Y}=\sqrt{M_Y/M_X}=1.5\Rightarrow M_X=M_Y/1.5^2=64/2.25=28.4$ g mol$^{-1}$ (3).

(c) At moderate P, attractive forces dominate → molar volume smaller than ideal → $Z<1$ (1.5); at high P, repulsion/finite volume ($b$) dominates → $Z>1$ (1.5).

Q5 (10)

(a) BCC body diagonal $=\sqrt3 a=4r\Rightarrow r=\frac{\sqrt3}{4}a$ (2). Atoms per cell $=2$ (1). PF $=\frac{2\cdot\frac43\pi r^3}{a^3}=\frac{2\cdot\frac43\pi(\sqrt3 a/4)^3}{a^3}=\frac{\sqrt3\pi}{8}=0.680$ (2).

(b) FCC: $Z=4$. $\rho=\frac{ZM}{a^3N_A}$. $a=0.405\text{ nm}=4.05\times10^{-8}$ cm; $a^3=6.643\times10^{-23}$ cm³ (2). $\rho=\frac{4\times27}{6.643\times10^{-23}\times6.022\times10^{23}}=\frac{108}{40.00}=2.70$ g cm$^{-3}$ (3).

Q6 (8)

(a) (3)

import math
def vrms(M, T, R=8.314):
    v = math.sqrt(3*R*T/M)
    print(v)
    return v
vrms(0.032, 300)  # -> ~483.6

(b) Schottky: equal number of cation+anion vacancies → density decreases; Frenkel: ion displaced to interstitial → density unchanged (3). (c) Group-15 atom (e.g. P) has 5 valence electrons; 4 bond to Si, the 5th is free → extra mobile electrons → n-type conduction (2).

[
{"claim":"O2 vrms at 300K ≈ 483.6 m/s","code":"import math; v=math.sqrt(3*8.314*300/0.032); result = abs(v-483.6)<1.0"},
{"claim":"Zc = 3/8 = 0.375","code":"result = simplify(Rational(3,8)) == Rational(375,1000)"},
{"claim":"MX from Graham = 28.44","code":"MX=64/1.5**2; result = abs(MX-28.444)<0.05"},
{"claim":"BCC packing fraction = sqrt(3)*pi/8 ≈ 0.680","code":"import math; pf=math.sqrt(3)*math.pi/8; result = abs(pf-0.680)<0.002"},
{"claim":"FCC density Al ≈ 2.70 g/cm3","code":"a=4.05e-8; rho=4*27/(a**3*6.022e23); result = abs(rho-2.70)<0.05"},
{"claim":"Ptotal of gas mixture = 1.2315 atm","code":"P=(0.20+0.30)*0.0821*300/10; result = abs(P-1.2315)<0.001"}
]