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PhysicsKinetic Theory of Gases
A container has two chambers of volumes $V_1 = 2$ L and $V_2 = 3$ L separated by a thermal insulator. Chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm. When partition is removed, the equilibrium pressure is:
Options
1
1.8 atm
2
1.3 atm
3
1.6 atm
4
1.4 atm
Correct Answer
1.6 atm
Solution
1

Conservation of gas molecules (isothermal process):

$$n_1 + n_2 = n_{total}$$$$\frac{p_1 V_1}{RT} + \frac{p_2 V_2}{RT} = \frac{P(V_1+V_2)}{RT}$$
2
$$p_1 V_1 + p_2 V_2 = P(V_1 + V_2)$$$$(1)(2) + (2)(3) = P(2+3)$$$$2 + 6 = 5P \implies P = \frac{8}{5} = \boxed{1.6 \text{ atm}}$$
$$P = \frac{p_1 V_1 + p_2 V_2}{V_1 + V_2} = \frac{2+6}{5} = 1.6 \text{ atm}$$
Theory: Kinetic Theory of Gases
1. Ideal Gas Law

$$PV = nRT$$ where $P$ = pressure, $V$ = volume, $n$ = moles, $R = 8.314$ J mol$^{-1}$K$^{-1}$, $T$ = temperature in Kelvin. For fixed $T$: $PV = nRT = $ constant. When two chambers are combined isothermally: $n_{total} = n_1 + n_2 = rac{p_1V_1 + p_2V_2}{RT}$. Final equilibrium pressure: $P = rac{(p_1V_1 + p_2V_2)}{V_1+V_2}$.

2. Dalton Law of Partial Pressures

For a mixture of ideal gases, total pressure equals sum of partial pressures: $$P_{total} = P_1 + P_2 + P_3 + ...$$ Partial pressure of gas $i$: $P_i = x_i P_{total}$ where $x_i = n_i/n_{total}$ is the mole fraction. Each gas behaves as if it alone occupies the container. No intermolecular forces between different gas molecules (ideal assumption).

3. Kinetic Theory Assumptions

Ideal gas assumptions: (1) Gas molecules are point masses (negligible volume). (2) No intermolecular forces except during collisions. (3) Collisions are perfectly elastic. (4) Molecules move randomly in all directions. (5) Average kinetic energy $\propto T$: $\langle KE angle = rac{3}{2}k_BT$ per molecule. Pressure arises from molecular collisions with walls: $P = rac{1}{3} ho ar{v^2} = rac{1}{3} rac{mN}{V}v_{rms}^2$.

4. Molecular Speeds

Three important speeds from Maxwell distribution: RMS speed: $v_{rms} = \sqrt{ rac{3RT}{M}}$. Mean speed: $ar{v} = \sqrt{ rac{8RT}{\pi M}}$. Most probable speed: $v_p = \sqrt{ rac{2RT}{M}}$. Ratio: $v_p : ar{v} : v_{rms} = 1 : 1.128 : 1.225$. All increase with $\sqrt{T}$ and decrease with $\sqrt{M}$. Lighter gases (H$_2$, He) move faster than heavier ones (O$_2$, CO$_2$).

5. Mean Free Path

Average distance between collisions: $\lambda = rac{1}{\sqrt{2}\pi d^2 n}$ where $d$ = molecular diameter, $n$ = number density. Increases with decreasing pressure. At atmospheric pressure and room temperature: $\lambda pprox 70$ nm for air. Mean free path determines: thermal conductivity, viscosity, diffusion. At high altitudes (low pressure): $\lambda$ increases, atmosphere thins, convection changes.

6. Degrees of Freedom and Equipartition

Each degree of freedom contributes $ rac{1}{2}k_BT$ to average energy (equipartition theorem). Monatomic gas (He, Ar): 3 translational DOF, $U = rac{3}{2}nRT$, $C_v = rac{3}{2}R$. Diatomic gas (N$_2$, O$_2$): 3 translational + 2 rotational DOF, $U = rac{5}{2}nRT$, $C_v = rac{5}{2}R$. At very high $T$: vibrational modes activated, $C_v$ increases. $\gamma = C_p/C_v$: monatomic $= 5/3$, diatomic $= 7/5$.

7. Real Gases — Van der Waals Equation

$$\left(P + rac{an^2}{V^2} ight)(V - nb) = nRT$$ $a$ = intermolecular attraction correction (reduces pressure). $b$ = finite molecular volume correction. At high $P$ or low $T$: real gas deviates significantly from ideal. Compression factor $Z = PV/nRT$. Ideal gas: $Z = 1$ always. Real gas: $Z < 1$ at moderate $P$ (attraction dominates), $Z > 1$ at high $P$ (repulsion dominates). Boyle temperature: $T_B = a/Rb$, where $Z pprox 1$ for a range of pressures.

8. Specific Heat at Constant Volume and Pressure

$C_v$ = molar specific heat at constant volume: $Q = nC_v\Delta T$, all heat goes to internal energy. $C_p$ = molar specific heat at constant pressure: $Q = nC_p\Delta T$, heat goes to internal energy plus work. Mayer relation: $C_p - C_v = R$. The extra $R$ accounts for work done by gas expanding at constant pressure. $\gamma = C_p/C_v$. For monatomic: $\gamma = 5/3 pprox 1.67$. For diatomic: $\gamma = 7/5 = 1.4$. Higher $\gamma$ = more adiabatic compression heating (diesel engines use this).

Frequently Asked Questions
1. Why is the formula $P = \frac{p_1V_1 + p_2V_2}{V_1+V_2}$? ⌄
Starting from ideal gas law $PV = nRT$: $n_1 = p_1V_1/RT_1$ and $n_2 = p_2V_2/RT_2$. If both chambers are at same temperature $T$ (given: thermal insulator means no heat exchange, so both remain at temperature $T$), then $n_{total} = (p_1V_1 + p_2V_2)/RT$. After partition removed: $P \cdot (V_1+V_2) = n_{total}RT = p_1V_1 + p_2V_2$. Therefore $P = (p_1V_1 + p_2V_2)/(V_1+V_2)$. This is simply conservation of the total number of gas molecules.
2. What would happen if the partition were conducting (not insulating)? ⌄
With a conducting partition: both chambers equilibrate to the same temperature eventually. The formula $P = (p_1V_1 + p_2V_2)/(V_1+V_2)$ still holds if both gases started at the same temperature. If they started at different temperatures $T_1$ and $T_2$, the final temperature would be $T_f = (n_1T_1 + n_2T_2)/(n_1+n_2)$ (energy conservation). Then $P = n_{total}RT_f/(V_1+V_2)$. The answer 1.6 atm assumes both chambers were at the same initial temperature.
3. What is Dalton law of partial pressures in this context? ⌄
After mixing, gas from chamber 1 (originally at 1 atm) has partial pressure $P_1 = p_1V_1/(V_1+V_2) = (1)(2)/5 = 0.4$ atm. Gas from chamber 2 (originally 2 atm) has partial pressure $P_2 = p_2V_2/(V_1+V_2) = (2)(3)/5 = 1.2$ atm. Total pressure = $P_1 + P_2 = 0.4 + 1.2 = 1.6$ atm. This is an alternative way to get the same answer using Dalton law.
4. What is the compressibility of an ideal gas mixture? ⌄
For ideal gas mixture: $PV = n_{total}RT = (n_1+n_2)RT$. Compressibility factor $Z = PV/(n_{total}RT) = 1$. Real gas mixtures deviate from this. Mixing of ideal gases is spontaneous (entropy increases) but involves no enthalpy change ($\Delta H_{mix} = 0$ for ideal gases). Mixing of real gases may involve small heat effects (due to different intermolecular forces).
5. How does temperature change when partition is removed? ⌄
For ideal gas mixing with thermal insulator partition: internal energy of each gas depends only on temperature. When mixed isothermally, no heat is exchanged and no work is done (rigid container). So total internal energy is conserved: $U_1 + U_2 = U_{final}$. $n_1C_vT + n_2C_vT = n_{total}C_vT_f$. If both gases are at same initial temperature $T$: $T_f = T$. Temperature does not change! This confirms the isothermal assumption used in the pressure calculation.
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