Kinetic Theory

Given:
- Diameter of oxygen molecule, $d = 3$ Å $= 3 \times 10^{-10}$ m
- At STP, 1 mole of any ideal gas occupies $V_{\text{actual}} = 22.4$ L $= 22.4 \times 10^{-3}$ m³
- Number of molecules in 1 mole, $N_A = 6.023 \times 10^{23}$

Step 1: Volume of one oxygen molecule (treated as a sphere)
$$V_{\text{molecule}} = \frac{4}{3}\pi\left(\frac{d}{2}\right)^3 = \frac{4}{3}\pi\left(\frac{3\times10^{-10}}{2}\right)^3$$
$$= \frac{4}{3}\pi (1.5\times10^{-10})^3 = \frac{4}{3}\times 3.14159 \times 3.375\times10^{-30}$$
$$= 1.414 \times 10^{-29} \text{ m}^3$$

Step 2: Total molecular volume for 1 mole
$$V_{\text{mol}} = N_A \times V_{\text{molecule}} = 6.023\times10^{23} \times 1.414\times10^{-29}$$
$$= 8.52 \times 10^{-6} \text{ m}^3$$

Step 3: Fraction of molecular volume to actual volume
$$\text{Fraction} = \frac{V_{\text{mol}}}{V_{\text{actual}}} = \frac{8.52\times10^{-6}}{22.4\times10^{-3}}$$
$$\boxed{\approx 3.8 \times 10^{-4}}$$

Conclusion: The molecular volume is only about $3.8 \times 10^{-4}$ times the actual volume occupied by the gas, confirming that molecules in a gas are very far apart compared to their own size.

Given:
- Pressure at STP: $P = 1$ atm $= 1.013 \times 10^5$ Pa
- Temperature at STP: $T = 0°C = 273$ K
- Number of moles: $\mu = 1$
- Universal gas constant: $R = 8.314$ J mol⁻¹ K⁻¹

Using the ideal gas equation:
$$PV = \mu RT$$

Solving for V:
$$V = \frac{\mu RT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 10^5}$$

$$V = \frac{2269.7}{1.013 \times 10^5}$$

$$V = 2.240 \times 10^{-2} \text{ m}^3$$

$$\boxed{V = 22.4 \text{ litres}}$$

(Since $1 \text{ m}^3 = 10^3$ litres, $2.240 \times 10^{-2}$ m³ $= 22.4$ litres.)

This confirms that 1 mole of any ideal gas at STP occupies 22.4 litres.

(a) Significance of the dotted plot:

For an ideal gas, $PV = \mu RT$, so $\frac{PV}{T} = \mu R =$ constant, independent of pressure $P$. The dotted plot (a horizontal straight line) represents the behaviour of an ideal gas. It signifies that $PV/T$ does not vary with pressure — this is the ideal gas limit.

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(b) Which temperature is higher, T₁ or T₂?

Real gases deviate from ideal behaviour, and the deviation is less at higher temperatures (gas behaves more ideally at higher $T$). The curve that is closer to the dotted (ideal) line corresponds to the higher temperature. From the typical shape of such plots, the curve at $T_1$ is closer to the ideal (dotted) line than the curve at $T_2$.

\boxed{T_1 > T_2}

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(c) Value of PV/T at the y-axis (P → 0):

At very low pressures, real gases behave ideally. So:
$$\frac{PV}{T} = \mu R$$

Mass of oxygen $= 1.00 \times 10^{-3}$ kg $= 1$ g

Molecular mass of O₂ $= 32.0$ g/mol

Number of moles:
$$\mu = \frac{1}{32} = 3.125 \times 10^{-2} \text{ mol}$$

Therefore:
$$\frac{PV}{T} = \mu R = 3.125 \times 10^{-2} \times 8.31$$
$$\boxed{\frac{PV}{T} = 0.26 \text{ J K}^{-1}}$$

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(d) For 1.00 × 10⁻³ kg of hydrogen:

Molecular mass of H₂ $= 2.02$ g/mol

Number of moles of H₂:
$$\mu_{H_2} = \frac{1}{2.02} = 0.495 \text{ mol}$$

$$\frac{PV}{T} = \mu_{H_2} R = 0.495 \times 8.31 = 4.11 \text{ J K}^{-1}$$

This is not the same as 0.26 J K⁻¹ obtained for oxygen.

Mass of hydrogen that gives the same PV/T = 0.26 J K⁻¹:

Required $\mu = \frac{0.26}{8.31} = 3.125 \times 10^{-2}$ mol

Mass of H₂:
$$m = \mu \times M_{H_2} = 3.125 \times 10^{-2} \times 2.02$$
$$\boxed{m \approx 6.3 \times 10^{-2} \text{ g} = 6.3 \times 10^{-5} \text{ kg}}$$

So, $6.3 \times 10^{-5}$ kg of hydrogen would yield the same value of $PV/T$ at the y-axis intercept.

Given:
- Volume of cylinder: $V = 30$ L $= 30 \times 10^{-3}$ m³ (constant)
- Initial gauge pressure: $P_1 = 15$ atm; absolute pressure $= 15 + 1 = 16$ atm $= 16 \times 1.013 \times 10^5 = 1.621 \times 10^6$ Pa
- Initial temperature: $T_1 = 27°C = 300$ K
- Final gauge pressure: $P_2 = 11$ atm; absolute pressure $= 11 + 1 = 12$ atm $= 12 \times 1.013 \times 10^5 = 1.216 \times 10^6$ Pa
- Final temperature: $T_2 = 17°C = 290$ K
- $R = 8.31$ J mol⁻¹ K⁻¹, $M = 32$ g/mol

Step 1: Find initial number of moles $\mu_1$

Using $P_1 V = \mu_1 R T_1$:
$$\mu_1 = \frac{P_1 V}{R T_1} = \frac{1.621\times10^6 \times 30\times10^{-3}}{8.31 \times 300}$$
$$= \frac{48630}{2493} = 19.51 \text{ mol}$$

Step 2: Find final number of moles $\mu_2$

Using $P_2 V = \mu_2 R T_2$:
$$\mu_2 = \frac{P_2 V}{R T_2} = \frac{1.216\times10^6 \times 30\times10^{-3}}{8.31 \times 290}$$
$$= \frac{36480}{2409.9} = 15.14 \text{ mol}$$

Step 3: Moles of oxygen withdrawn
$$\Delta\mu = \mu_1 - \mu_2 = 19.51 - 15.14 = 4.37 \text{ mol}$$

Step 4: Mass of oxygen withdrawn
$$m = \Delta\mu \times M = 4.37 \times 32 = 139.8 \text{ g}$$

$$\boxed{m \approx 141 \text{ g} \approx 0.141 \text{ kg}}$$

Given:
- Initial volume: $V_1 = 1.0$ cm³ $= 1.0 \times 10^{-6}$ m³
- Depth of lake: $h = 40$ m
- Temperature at bottom: $T_1 = 12°C = 285$ K
- Temperature at surface: $T_2 = 35°C = 308$ K
- Atmospheric pressure: $P_0 = 1.013 \times 10^5$ Pa
- Density of water: $\rho = 1000$ kg/m³, $g = 9.8$ m/s²

Step 1: Pressure at the bottom of the lake
$$P_1 = P_0 + \rho g h = 1.013\times10^5 + 1000 \times 9.8 \times 40$$
$$= 1.013\times10^5 + 3.92\times10^5 = 4.933\times10^5 \text{ Pa}$$

Step 2: Pressure at the surface
$$P_2 = P_0 = 1.013\times10^5 \text{ Pa}$$

Step 3: Apply the combined gas law
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

$$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$$

$$V_2 = 1.0\times10^{-6} \times \frac{4.933\times10^5}{1.013\times10^5} \times \frac{308}{285}$$

$$V_2 = 1.0\times10^{-6} \times 4.870 \times 1.0807$$

$$V_2 = 1.0\times10^{-6} \times 5.263$$

$$\boxed{V_2 \approx 5.3 \times 10^{-6} \text{ m}^3 = 5.3 \text{ cm}^3}$$

The bubble grows to approximately 5.3 cm³ when it reaches the surface.

Given:
- Volume of room: $V = 25.0$ m³
- Temperature: $T = 27°C = 300$ K
- Pressure: $P = 1$ atm $= 1.013 \times 10^5$ Pa
- $k_B = 1.38 \times 10^{-23}$ J K⁻¹

Using the ideal gas equation in terms of number of molecules:
$$PV = N k_B T$$

Solving for N:
$$N = \frac{PV}{k_B T} = \frac{1.013\times10^5 \times 25.0}{1.38\times10^{-23} \times 300}$$

$$N = \frac{2.5325\times10^6}{4.14\times10^{-21}}$$

$$\boxed{N \approx 6.1 \times 10^{26} \text{ molecules}}$$

The room contains approximately $6.1 \times 10^{26}$ air molecules.

Helium is a monatomic gas with 3 translational degrees of freedom. By the law of equipartition of energy, the average thermal (kinetic) energy per atom is:
$$E = \frac{3}{2} k_B T$$

where $k_B = 1.38 \times 10^{-23}$ J K⁻¹.

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(i) Room temperature: $T = 27°C = 300$ K
$$E = \frac{3}{2} \times 1.38\times10^{-23} \times 300$$
$$= \frac{3}{2} \times 4.14\times10^{-21}$$
$$\boxed{E \approx 6.21 \times 10^{-21} \text{ J}}$$

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(ii) Surface of the Sun: $T = 6000$ K
$$E = \frac{3}{2} \times 1.38\times10^{-23} \times 6000$$
$$= \frac{3}{2} \times 8.28\times10^{-20}$$
$$\boxed{E \approx 1.24 \times 10^{-19} \text{ J}}$$

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(iii) Core of a star: $T = 10^7$ K
$$E = \frac{3}{2} \times 1.38\times10^{-23} \times 10^7$$
$$= \frac{3}{2} \times 1.38\times10^{-16}$$
$$\boxed{E \approx 2.07 \times 10^{-16} \text{ J}}$$

Part 1: Do the vessels contain equal number of molecules?

All three vessels have the same volume $V$, same temperature $T$, and same pressure $P$.

From the ideal gas equation:
$$PV = Nk_BT \implies N = \frac{PV}{k_BT}$$

Since $P$, $V$, and $T$ are identical for all three vessels, $N$ is the same for all three.

Yes, all three vessels contain equal number of molecules (Avogadro's law).

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Part 2: Is $v_{\text{rms}}$ the same in all three cases?

The rms speed is given by:
$$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$$

where $m$ is the mass of one molecule. Since the three gases have different molecular masses, their $v_{\text{rms}}$ values will be different.

- Neon (Ne): $M = 20.2$ u (monatomic)
- Chlorine (Cl₂): $M = 71$ u (diatomic)
- Uranium hexafluoride (UF₆): $M = 238 + 6\times19 = 352$ u (polyatomic)

Since $v_{\text{rms}} \propto \frac{1}{\sqrt{m}}$, the lighter the molecule, the larger the rms speed.

Neon has the smallest molecular mass, so:
$$\boxed{v_{\text{rms}} \text{ is largest for Neon (Ne)}}$$

Order: v_{\text{rms}}(\text{Ne}) > v_{\text{rms}}(\text{Cl}_2) > v_{\text{rms}}(\text{UF}_6)

Given:
- Temperature of helium: $T_{He} = 20°C = 293$ K
- Atomic mass of Ar: $M_{Ar} = 39.9$ u
- Atomic mass of He: $M_{He} = 4.0$ u
- Let the required temperature of argon be $T_{Ar}$.

Formula for rms speed:
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$

Setting $v_{\text{rms}}$ of Ar equal to $v_{\text{rms}}$ of He:
$$\sqrt{\frac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$$

Squaring both sides:
$$\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$$

$$T_{Ar} = T_{He} \times \frac{M_{Ar}}{M_{He}}$$

$$T_{Ar} = 293 \times \frac{39.9}{4.0}$$

$$T_{Ar} = 293 \times 9.975$$

$$\boxed{T_{Ar} \approx 2923 \text{ K}}$$

The argon gas must be at approximately 2923 K for its rms speed to equal that of helium at 293 K.

Given:
- Pressure: $P = 2.0$ atm $= 2.0 \times 1.013 \times 10^5 = 2.026 \times 10^5$ Pa
- Temperature: $T = 17°C = 290$ K
- Radius of N₂ molecule: $r = 1.0$ Å, so diameter $d = 2r = 2.0$ Å $= 2.0 \times 10^{-10}$ m
- Molecular mass: $M = 28.0$ g/mol $= 28.0 \times 10^{-3}$ kg/mol
- $k_B = 1.38 \times 10^{-23}$ J K⁻¹, $R = 8.31$ J mol⁻¹ K⁻¹

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Step 1: Number density $n$
$$n = \frac{P}{k_B T} = \frac{2.026\times10^5}{1.38\times10^{-23}\times290}$$
$$= \frac{2.026\times10^5}{4.002\times10^{-21}} = 5.06 \times 10^{25} \text{ m}^{-3}$$

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Step 2: Mean free path $l$
$$l = \frac{1}{\sqrt{2}\, n\pi d^2}$$
$$= \frac{1}{\sqrt{2} \times 5.06\times10^{25} \times \pi \times (2.0\times10^{-10})^2}$$
$$= \frac{1}{1.414 \times 5.06\times10^{25} \times 3.14159 \times 4.0\times10^{-20}}$$
$$= \frac{1}{1.414 \times 5.06\times10^{25} \times 1.2566\times10^{-19}}$$
$$= \frac{1}{1.414 \times 6.358\times10^{6}}$$
$$= \frac{1}{8.99\times10^{6}}$$
$$\boxed{l \approx 1.11 \times 10^{-7} \text{ m} \approx 1.1 \times 10^{-7} \text{ m}}$$

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Step 3: rms speed of N₂
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.31 \times 290}{28.0\times10^{-3}}}$$
$$= \sqrt{\frac{7230.3}{0.028}} = \sqrt{258225} \approx 508 \text{ m/s}$$

(We use $v_{\text{rms}}$ as an estimate for the average speed; more precisely $\bar{v} = \sqrt{8k_BT/\pi m} \approx 0.92\, v_{\text{rms}} \approx 467$ m/s. We use $\approx 508$ m/s for the estimate.)

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Step 4: Collision frequency $f$
$$f = \frac{v_{\text{rms}}}{l} = \frac{508}{1.11\times10^{-7}}$$
$$\boxed{f \approx 4.6 \times 10^{9} \text{ s}^{-1}}$$

(About $4.6 \times 10^9$ collisions per second.)

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Step 5: Compare collision time with free travel time

The diameter of a nitrogen molecule is $d = 2.0 \times 10^{-10}$ m.

Time for a collision (time to traverse one molecular diameter):
$$t_{\text{collision}} = \frac{d}{v_{\text{rms}}} = \frac{2.0\times10^{-10}}{508} \approx 3.9 \times 10^{-13} \text{ s}$$

Time between collisions (free travel time):
$$t_{\text{free}} = \frac{l}{v_{\text{rms}}} = \frac{1.11\times10^{-7}}{508} \approx 2.2 \times 10^{-10} \text{ s}$$

Ratio:
$$\frac{t_{\text{free}}}{t_{\text{collision}}} = \frac{2.2\times10^{-10}}{3.9\times10^{-13}} \approx 564 \approx 500$$

$$\boxed{\frac{t_{\text{free}}}{t_{\text{collision}}} \approx 500}$$

Conclusion: The time a molecule spends in free travel between collisions is about 500 times the time spent during a collision. This confirms that molecules spend most of their time moving freely, and collisions are very brief events.