Kinetic Theory of Gases

Answer: (b)

1. Mean KE $= \frac{3}{2}k_BT$, so $E \propto T$ (in kelvin).
2. $T_1 = 27 + 273 = 300$ K, so half the KE needs $T_2 = 150$ K.
3. $150 - 273 = -123^\circ$C
4. So the answer is (b) $-123^\circ$C.

Answer: (c)

1. Mean KE $\propto T$ (kelvin).
2. $T_1 = 300$ K; increasing by $300^\circ$ gives $T_2 = 600$ K.
3. $E_2 = E_1 \times \frac{600}{300} = 2E_1$
4. So the answer is (c) $2E_1$.

Answer: (c)

1. KE $\propto T$ (kelvin).
2. $T_1 = 300$ K, $T_2 = 177 + 273 = 450$ K
3. $E = E_0 \times \frac{450}{300} = \frac{3}{2}E_0$
4. So the answer is (c) $3E_0/2$.

Answer: (d)

1. Translational KE $= \frac{3}{2}k_BT$, so KE $\propto T$.
2. To double the KE, double $T$.
3. So the answer is (d) increasing T to 2T.

Answer: (c)

1. Force on the walls $F = PA$, and for a closed container (fixed $V$) $P = \frac{nRT}{V} \propto T$.
2. So $F \propto T^{1}$.
3. So the answer is (c) $T^{1}$.

Answer: (a)

1. Pressure $P = \frac{2}{3}\frac{E}{V}$ (translational KE per unit volume).
2. $\frac{E}{V} = \frac{600}{10^{-3}} = 6 \times 10^{5}$ J/m$^3$
3. $P = \frac{2}{3} \times 6 \times 10^{5} = 4 \times 10^{5}$ N/m$^2$
4. So the answer is (a) $4 \times 10^{5}$ N/m$^2$.

Answer: (b)

1. $E_1 = \frac{3}{2}k_BT = \frac{3}{2}\frac{R}{N}T$ and $E_2 = eV$.
2. $E_1 = E_2 \Rightarrow T = \frac{2eVN}{3R}$
3. So the answer is (b) $\frac{2eVN}{3R}$.

Answer: (b)

1. Average KE $\propto T$ (kelvin).
2. $T_1 = 300$ K, $T_2 = 327 + 273 = 600$ K
3. $E_2 = E \times \frac{600}{300} = 2E$
4. So the answer is (b) 2E.

Answer: (d)

1. Mean KE $\propto T$ (kelvin).
2. $T_1 = 399 + 273 = 672$ K; for $E/2$, $T_2 = 336$ K.
3. $336 - 273 = 63^\circ$C
4. So the answer is (d) $63^\circ$C.

Answer: (a)

1. Average KE $= \frac{3}{2}k_BT$, so $\frac{E_B}{E_A} = \frac{T_B}{T_A}$.
2. $\frac{420}{350} = \frac{6}{5}$
3. So the answer is (a) $6:5$.

Answer: (a)

1. $F = PA$ and, at fixed volume, $P \propto T$.
2. So $F \propto T^{1}$, giving $x = 1$.
3. So the answer is (a) 1.

Answer: (d)

1. Translational KE $= \frac{3}{2}nRT$.
2. $n = \frac{50}{44} = 1.136$ mol, $T = 17 + 273 = 290$ K
3. $KE = 1.5 \times 1.136 \times 8.3 \times 290 \approx 4102.8$ J
4. So the answer is (d) 4102.8 J.

Answer: (d)

1. Average KE $\propto T$ (kelvin).
2. $T_1 = -78 + 273 = 195$ K; for $2K$, $T_2 = 390$ K.
3. $390 - 273 = 117^\circ$C
4. So the answer is (d) $117^\circ$C.

Answer: (b)

1. Average KE of a monatomic molecule $= \frac{3}{2}k_BT$.
2. $0.414\text{ eV} = 0.414 \times 1.6 \times 10^{-19} = 6.624 \times 10^{-20}$ J
3. $T = \frac{2 \times 6.624 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = 3200$ K
4. So the answer is (b) 3200 K.

Answer: (d)

1. Force by a molecule = change of momentum per collision × collision rate: $F \propto (2mv)\left(\frac{v}{2L}\right) \propto mv^2$.
2. $mv^2 \propto T$, so $F \propto T^{1}$.
3. So the answer is (d) 1.

Answer: (a)

1. $P = \frac{2}{3}\frac{E}{V}$
2. $V = 10$ L $= 10^{-2}$ m$^3$, so $\frac{E}{V} = \frac{7.5 \times 10^{3}}{10^{-2}} = 7.5 \times 10^{5}$ J/m$^3$
3. $P = \frac{2}{3} \times 7.5 \times 10^{5} = 5 \times 10^{5}$ N m$^{-2}$
4. So the answer is (a) $5 \times 10^{5}$ N m$^{-2}$.

Answer: (b)

1. $F = PA$ and, for a closed container, $P \propto T$.
2. So $F \propto T^{1}$, $x = 1$.
3. So the answer is (b) 1.

Answer: (d)

1. Average translational KE $= \frac{3}{2}k_BT$ does not depend on molar mass, only on $T$.
2. At double the temperature the KE doubles: $2 \times 0.042 = 0.084$ eV.
3. So the answer is (d) 0.084.

Answer: (a)

1. Oxygen is diatomic: 3 translational + 2 rotational degrees of freedom, so total KE per mole $= \frac{5}{2}RT$.
2. $T = 300$ K
3. $KE = 2.5 \times 8.31 \times 300 = 6232.5$ J
4. So the answer is (a) 6232.5 J.

Answer: (d)

1. Average translational KE per molecule $= \frac{3}{2}k_BT$ depends only on temperature, not on mass or amount of gas.
2. Both gases in the mixture are at the same $T$, so $K_{Ar}/K_{H} = 1$.
3. So the answer is (d) 1.

Answer: (c)

1. KE $\propto T$ (kelvin).
2. $T_1 = 300$ K, so double KE needs $T_2 = 600$ K.
3. $600 - 273 = 327^\circ$C
4. So the answer is (c) $327^\circ$C.

Answer: (c)

1. Average KE of a molecule $= \frac{3}{2}k_BT$.
2. It depends only on absolute temperature, not on volume, pressure or type of gas.
3. So the answer is (c) proportional to absolute temperature.

Answer: (a)

1. Average translational KE per molecule $= \frac{3}{2}k_BT$, the same for every gas at the same $T$.
2. The mass ratio does not matter.
3. So the answer is (a) $1:1$.

Answer: (b)

1. $P = \frac{1}{3}\frac{N m v_{rms}^2}{V}$, so $P \propto m v^2$.
2. $m \to \frac{m}{2}$, $v \to 2v$: $mv^2 \to \frac{m}{2} \times 4v^2 = 2mv^2$
3. $P = 2p_0$
4. So the answer is (b) $2p_0$.

Answer: (d)

1. $F = PA$ and, at fixed volume, $P \propto T$.
2. So $F \propto T^{1}$, $x = 1$.
3. So the answer is (d) 1.

Answer: (d)

1. $PV = \frac{2}{3}N\bar{E}$, so $N = \frac{3PV}{2\bar{E}}$.
2. $V = 500$ cc $= 5 \times 10^{-4}$ m$^3$
3. $N = \frac{3 \times 2 \times 10^{5} \times 5 \times 10^{-4}}{2 \times 6 \times 10^{-21}} = \frac{300}{1.2 \times 10^{-20}} = 2.5 \times 10^{22}$
4. So the answer is (d) $2.5 \times 10^{22}$.

Answer: (d)

1. Average KE per molecule $= \frac{3}{2}k_BT$ depends only on $T$.
2. Both gases are at $27^\circ$C, so the ratio is $1:1$; the mass ratio $3:2$ does not matter.
3. So the answer is (d) $1:1$.

Answer: (c)

1. Molecules hit the wall and bounce back, so the momentum of each changes by $2mv$.
2. The wall supplies this change of momentum, so the molecules push back on the wall. Force per unit area is the pressure.
3. So the answer is (c) they suffer change in momentum when they hit the walls.

Answer: (b)

1. Translational KE $= \frac{3}{2}k_BT$, so KE $\propto T$.
2. Doubling the KE needs $T \to 2T$.
3. So the answer is (b) by increasing $T$ to $2T$.

Answer: (b)

1. KE per mole $E = \frac{3}{2}RT$.
2. $R = \frac{2E}{3T}$
3. So the answer is (b) $\frac{2E}{3T}$.

Answer: (d)

1. (A) $v_{rms} = \sqrt{\frac{3RT}{M}}$ → (Q)
2. (B) $P = \frac{1}{3}nm\bar{v}^2$ → (P)
3. (C) Average KE of a molecule $= \frac{3}{2}k_BT$ → (S)
4. (D) Diatomic gas has 5 degrees of freedom, $U = \frac{5}{2}RT$ per mole → (R)
5. So the answer is (d) (A)-(Q), (B)-(P), (C)-(S), (D)-(R).

Answer: (a)

1. KE of the electron $= eV$; average KE of a molecule $= \frac{3}{2}k_BT$ with $k_B = \frac{R}{N}$.
2. $\frac{3}{2}\frac{R}{N}T = eV$
3. $T = \frac{2eVN}{3R}$
4. So the answer is (a) $\frac{2eVN}{3R}$.

Answer: (c)

1. $PV = \frac{2}{3}N\bar{E}$, so $N = \frac{3PV}{2\bar{E}}$.
2. $P = 2$ atm $= 2.026 \times 10^{5}$ Pa, $V = 10^{-3}$ m$^3$, $\bar{E} = 2 \times 10^{-9}$ J
3. $N = \frac{3 \times 2.026 \times 10^{5} \times 10^{-3}}{2 \times 2 \times 10^{-9}} \approx 1.5 \times 10^{11}$
4. So the answer is (c) $1.5 \times 10^{11}$.

Answer: 500

1. $\frac{3}{2}k_BT = eV = 0.1 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-20}$ J
2. $T = \frac{2 \times 1.6 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} \approx 773$ K
3. $773 - 273 = 500^\circ$C
4. So the answer is $500^\circ$C.

Answer: (d)

1. $P = h\rho g = 2 \times 13.6 \times 980 = 26656$ dyne/cm$^2$
2. $PV = \frac{2}{3}N\bar{E}$, so $N = \frac{3PV}{2\bar{E}}$
3. $N = \frac{3 \times 26656 \times 4}{2 \times 4 \times 10^{-14}} \approx 4.0 \times 10^{18}$
4. So the answer is (d) $4.0 \times 10^{18}$.

Answer: (b)

1. Force by one molecule $\propto \frac{\Delta p}{\Delta t} = \frac{2mv}{2L/v} = \frac{mv^2}{L}$.
2. $mv^2 \propto T$, so $F \propto T^{1}$ and $q = 1$.
3. So the answer is (b) 1.

Answer: (d)

1. KE at $0^\circ$C ($T = 273$ K) $= \frac{3}{2}k_B(273)$.
2. All of this KE becomes potential energy at the top: $Mgh = \frac{3}{2}k_B \times 273$
3. $h = \frac{819k_B}{2Mg}$
4. So the answer is (d) $\frac{819k_B}{2Mg}$.

Answer: (c)

1. By kinetic theory the KE of a molecule is $\frac{3}{2}k_BT$, which is zero at $T = 0$ K.
2. Density and potential energy do not become zero at 0 K.
3. So the answer is (c) kinetic energy.

Answer: (c)

1. In kinetic theory the mean kinetic energy of a molecule is $\frac{3}{2}k_BT$.
2. At $T = 0$ K this energy becomes zero, so the molecules stop moving.
3. So the answer is (c) molecular motion stops.

Answer: (b)

1. $v_{rms} = \sqrt{\frac{3k_BT}{m}}$
2. So $v_{rms} \propto m^{-1/2}T^{1/2}$.
3. So the answer is (b) $m^{-1/2}T^{1/2}$.

Answer: (a)

1. Kinetic theory gives $P = \frac{1}{3}\rho \overline{v^2}$.
2. Kinetic energy per unit volume $E = \frac{1}{2}\rho \overline{v^2}$, so $\rho \overline{v^2} = 2E$.
3. $P = \frac{1}{3}(2E) = \frac{2}{3}E$
4. So the answer is (a) $\frac{2}{3}$ of KE per unit volume.

Answer: (d)

1. $v_{rms} = \sqrt{\frac{3RT}{M}}$, so $v^2 = \frac{3R}{M}T$.
2. For a given gas $\frac{3R}{M}$ is fixed, so $v^2 \propto T$.
3. So the answer is (d) $\frac{v^2}{T} =$ constant.

Answer: (c)

1. A: average KE $= \frac{3}{2}kT$ depends only on $T$, so equal KE means equal $T$, whatever the container size. A is true.
2. R: $v_{rms} = \sqrt{\frac{3RT}{M}}$; $M$ differs for $\text{H}_2$ (2) and $\text{O}_2$ (32), so the speeds differ. R is false.
3. So the answer is (c).

Answer: (b)

1. $v_{rms} = \sqrt{\frac{3RT}{M}}$, so $\frac{v_H}{v_O} = \sqrt{\frac{T_H}{T_O}\cdot\frac{M_O}{M_H}}$
2. $T_O = 47 + 273 = 320\text{ K}$, $\frac{M_O}{M_H} = 16$
3. $4.5^2 = \frac{16\,T_H}{320} \Rightarrow 20.25 = \frac{T_H}{20} \Rightarrow T_H = 405\text{ K}$
4. So the answer is (b) $132^\circ\text{C}$.

Answer: (d)

1. Average speed $\bar v = \frac{1+2+3+4+5+6+7}{7} = 4\text{ km/s}$
2. Mean square speed $= \frac{1+4+9+16+25+36+49}{7} = \frac{140}{7} = 20$
3. $v_{rms} = \sqrt{20} = 2\sqrt5\text{ km/s}$
4. $\frac{v_{rms}}{\bar v} = \frac{2\sqrt5}{4} = \frac{\sqrt5}{2}$
5. So the answer is (d) $\sqrt5 : 2$.

Answer: (a)

1. Same temperature, so $v_{rms} \propto \frac{1}{\sqrt M}$ (mass ratio in the flask does not matter).
2. $\frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{70}{40}} = \sqrt{\frac{7}{4}} = \frac{\sqrt7}{2}$
3. So the answer is (a) $\frac{\sqrt7}{2}$.

Answer: (b)

1. Equal $v_{rms}$ means equal $\frac{T}{M}$.
2. $T_O = 47 + 273 = 320\text{ K}$; $T_H = 320 \times \frac{2}{32} = 20\text{ K}$
3. $20 - 273 = -253^\circ\text{C}$
4. So the answer is (b) $-253$.

Answer: (a)

1. Average translational KE per molecule: $E_1 = \dfrac{3}{2}kT = \dfrac{3RT}{2N}$ (using $k = R/N$).
2. Electron accelerated through $V$: $E_2 = eV$.
3. $\dfrac{3RT}{2N} = eV \Rightarrow T = \dfrac{2NeV}{3R}$
4. So the answer is (a) $\dfrac{2VNe}{3R}$.

Answer: (b)

1. $v_{rms} \propto \sqrt{T}$
2. $\frac{v'}{x} = \sqrt{\frac{400}{100}} = 2$
3. So the answer is (b) $2x$.

Answer: (a)

1. Kinetic theory assumes molecules are perfectly elastic spheres.
2. In an elastic collision both momentum and kinetic energy are conserved.
3. So the answer is (a).

Answer: (c)

1. $v_{rms} = \sqrt{\frac{\sum v^2}{N}}$
2. $\sum v^2 = 4 + 25 + 9 + 36 + 9 + 25 = 108$
3. $v_{rms} = \sqrt{\frac{108}{6}} = \sqrt{18} \approx 4.24\text{ m/s}$
4. So the answer is (c) $4.24\text{ m/s}$.

Answer: (c)

1. Average KE per molecule $= \frac{3}{2}kT$ depends only on temperature.
2. Velocity and momentum are vectors that average to zero; their magnitudes differ from molecule to molecule.
3. So the answer is (c) kinetic energy.

Answer: (b)

1. $v_{rms} \propto \sqrt{T}$
2. $\frac{v_{800}}{v_{200}} = \sqrt{\frac{800}{200}} = 2$
3. So the answer is (b) twice the value at 200 K.

Answer: (b)

1. $v_{rms} \propto \sqrt{T}$
2. $\frac{v'}{x} = \sqrt{\frac{400}{100}} = 2$
3. So the answer is (b) $2x$.