Kinetic Theory

Given: Diameter $d = 3 \text{ \AA} = 3 \times 10^{-10} \text{ m}$. Radius $r = 1.5 \times 10^{-10} \text{ m}$.
Consider 1 mole of oxygen gas at STP.

1. Actual Molecular Volume ($V_{mol}$):
Volume of 1 molecule = $\frac{4}{3}\pi r^3$.
$V_{molecule} = \frac{4}{3} \times 3.14 \times (1.5 \times 10^{-10})^3 \approx 1.41 \times 10^{-29} \text{ m}^3$.
Total volume of molecules ($N_A$ molecules):
$V_{mol} = N_A \times V_{molecule} = 6.02 \times 10^{23} \times 1.41 \times 10^{-29} \approx 8.5 \times 10^{-6} \text{ m}^3$.

2. Actual Volume at STP ($V_{gas}$):
Molar volume at STP = $22.4 \text{ L} = 22.4 \times 10^{-3} \text{ m}^3$.

Conclusion: The molecules occupy a very tiny fraction ($\approx 0.04\%$) of the total gas volume, justifying the ideal gas assumption that molecular volume is negligible.

Conditions at STP:
Pressure $P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$.
Temperature $T = 0^\circ \text{C} = 273.15 \text{ K}$.
Gas Constant $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$.

Ideal Gas Equation: $PV = nRT$. For $n=1$:
$V = \frac{RT}{P} = \frac{8.314 \times 273.15}{1.013 \times 10^5}$
$V = \frac{2270.9}{101300} \approx 0.0224 \text{ m}^3$.

(a) Dotted plot: It represents the Ideal Gas behavior, where $PV/T = nR = \text{constant}$, independent of Pressure.

(b) Temperature Comparison: Real gases approach ideal behavior at higher temperatures. The curve $T_1$ is closer to the horizontal ideal line than $T_2$. Hence, $\mathbf{T_1 > T_2}$.

(c) Value at y-axis ($nR$):
Mass $m = 1.00 \times 10^{-3} \text{ kg} = 1 \text{ g}$. Molar mass $O_2 = 32 \text{ g/mol}$.
Moles $n = 1/32$.
Value = $nR = \frac{1}{32} \times 8.31 \approx \mathbf{0.26 \text{ J K}^{-1}}$.

(d) Hydrogen Plot:
No, the value depends on $n$. For Hydrogen, $M = 2.02 \text{ g/mol}$.
For same $PV/T$ value ($0.26$), we need same $n = 1/32$.
Mass of $H_2 = n \times M_{H2} = \frac{1}{32} \times 2.02 \approx \mathbf{0.063 \text{ g}}$.

Volume $V = 30 \text{ L} = 30 \times 10^{-3} \text{ m}^3$.
Molecular mass $M = 32 \text{ g}$.

Initial State (1):
$P_1 (\text{gauge}) = 15 \text{ atm} \Rightarrow P_1 (\text{abs}) = 16 \text{ atm} = 16 \times 1.013 \times 10^5 \text{ Pa}$.
$T_1 = 27 + 273 = 300 \text{ K}$.
Moles $n_1 = \frac{P_1 V}{R T_1} = \frac{16 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 300} \approx 19.5 \text{ mol}$.

Final State (2):
$P_2 (\text{gauge}) = 11 \text{ atm} \Rightarrow P_2 (\text{abs}) = 12 \text{ atm}$.
$T_2 = 17 + 273 = 290 \text{ K}$.
Moles $n_2 = \frac{P_2 V}{R T_2} = \frac{12 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 290} \approx 15.1 \text{ mol}$.

Mass Removed:
$\Delta n = 19.5 – 15.1 = 4.4 \text{ mol}$.
Mass $= \Delta n \times M = 4.4 \times 32 \approx \mathbf{140.8 \text{ g}} = \mathbf{0.141 \text{ kg}}$.

Initial (Bottom):
Depth $h = 40 \text{ m}$. $T_1 = 12^\circ \text{C} = 285 \text{ K}$. $V_1 = 1.0 \text{ cm}^3$.
$P_1 = P_{atm} + \rho g h = 1.01 \times 10^5 + 1000 \times 9.8 \times 40 = 1.01 \times 10^5 + 3.92 \times 10^5 = 4.93 \times 10^5 \text{ Pa}$.

Final (Surface):
$P_2 = P_{atm} = 1.01 \times 10^5 \text{ Pa}$. $T_2 = 35^\circ \text{C} = 308 \text{ K}$.

Ideal Gas Law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$
$V_2 = 1.0 \times \frac{4.93}{1.01} \times \frac{308}{285} \approx 1.0 \times 4.88 \times 1.08 \approx \mathbf{5.3 \text{ cm}^3}$.

Volume $V = 25.0 \text{ m}^3$. Temp $T = 300 \text{ K}$. $P = 1.013 \times 10^5 \text{ Pa}$.
Boltzmann Constant $k_B = 1.38 \times 10^{-23} \text{ J/K}$.

Formula: $PV = N k_B T \Rightarrow N = \frac{PV}{k_B T}$.

Average thermal energy $E = \frac{3}{2} k_B T$.

(i) Room Temp (300 K):
$E = 1.5 \times 1.38 \times 10^{-23} \times 300 \approx \mathbf{6.2 \times 10^{-21} \text{ J}}$.

(ii) Sun Surface (6000 K):
$E = 1.5 \times 1.38 \times 10^{-23} \times 6000 \approx \mathbf{1.24 \times 10^{-19} \text{ J}}$.

(iii) Star Core ($10^7$ K):
$E = 1.5 \times 1.38 \times 10^{-23} \times 10^7 \approx \mathbf{2.1 \times 10^{-16} \text{ J}}$.

1. Number of Molecules:
By Avogadro’s Hypothesis, equal volumes of gases at the same temperature and pressure contain equal number of molecules. Thus, all vessels contain the same number of molecules.

2. RMS Speed:
$v_{rms} = \sqrt{\frac{3 k_B T}{m}}$. Since $T$ is same, $v_{rms} \propto \frac{1}{\sqrt{m}}$.
Masses: Neon ($20.2$ u) < Chlorine ($71$ u) < Uranium Hexafluoride ($352$ u).
Lighter molecules move faster. Neon has the largest $v_{rms}$.

Condition: $(v_{rms})_{Ar} = (v_{rms})_{He}$.
$\sqrt{\frac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\frac{3RT_{He}}{M_{He}}} \Rightarrow \frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$.

Given: $M_{Ar} = 39.9$, $M_{He} = 4.0$.
$T_{He} = -20^\circ \text{C} = 253 \text{ K}$.

Given: $P = 2.0 \text{ atm} = 2.026 \times 10^5 \text{ Pa}$. $T = 290 \text{ K}$.
Radius $r = 1.0 \text{ \AA} \Rightarrow$ Diameter $d = 2.0 \text{ \AA} = 2 \times 10^{-10} \text{ m}$.

1. Number Density ($n$):
$n = \frac{N}{V} = \frac{P}{k_B T} = \frac{2.026 \times 10^5}{1.38 \times 10^{-23} \times 290} \approx 5.06 \times 10^{25} \text{ m}^{-3}$.

2. Mean Free Path ($\lambda$):
$\lambda = \frac{1}{\sqrt{2} \pi n d^2} = \frac{1}{1.414 \times 3.14 \times 5.06 \times 10^{25} \times (2 \times 10^{-10})^2}$
$\lambda \approx \frac{1}{8.99 \times 10^6} \approx \mathbf{1.11 \times 10^{-7} \text{ m}}$.

3. Collision Frequency ($\nu$):
$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}} \approx 508 \text{ m/s}$.
$\nu = \frac{v_{rms}}{\lambda} = \frac{508}{1.11 \times 10^{-7}} \approx \mathbf{4.58 \times 10^9 \text{ s}^{-1}}$.

4. Comparison with Free Time:
Collision time $\tau_{col} \approx d/v \approx 2 \times 10^{-10} / 500 \approx 4 \times 10^{-13}$ s.
Free time $\tau_{free} = 1/\nu \approx 2.2 \times 10^{-10}$ s.
The time between collisions is roughly 500 times larger than the duration of a collision.