Gas Laws and KMT

Kinetic Molecular Theory -- Postulates

The kinetic theory explains gas behavior through a microscopic model based on Elastic Collision and random molecular motion.

Large Number: A finite volume of gas consists of a very large number of molecules.

Negligible Size: Molecular dimensions are much smaller than the average separation between molecules.

Random Motion: Gas molecules move randomly and change direction after every collision.

Elastic Collisions: Collisions between molecules and with container walls are perfectly elastic -- kinetic energy is conserved.

No Intermolecular Forces: Molecules exert no force on each other except during a collision.

Pressure of a Gas from KMT

Pressure is the Momentum transferred to container walls per second per unit area due to continuous molecular collisions.

$$P = \frac{1}{3}\rho \langle v^2 \rangle$$

Gas pressure equals one-third the product of density and mean square velocity.

$P$=Pressure of the gas(Pa)

$\rho$=Gas density ($mN/V$)(kg/m³)

$\langle v^2 \rangle$=Mean square velocity of molecules(m²/s²)

In terms of $N$ and $V$

→

$P = \frac{mN}{3V}\langle v^2 \rangle = \frac{2}{3}\frac{N}{V}\langle \frac{1}{2}mv^2 \rangle$

Momentum Change: Each elastic collision with a wall reverses the velocity component, giving $\Delta p = 2mv_x$ per collision.

Collision Rate: A molecule bouncing between opposite faces separated by $l$ hits a given face $v_x / 2l$ times per second.

Force per Molecule: Rate of momentum transfer gives $F = mv_x^2 / l$ for one molecule on one face.

The 1/3 Factor: Since motion is equally likely in x, y, z directions, each component contributes $\frac{1}{3}$ of $\langle v^2 \rangle$.

Key Result: $P \propto \langle KE \rangle$ -- pressure is directly proportional to the average translational kinetic energy.

Temperature as Average Kinetic Energy

Absolute Temperature is directly proportional to the average translational Kinetic Energy of gas molecules -- temperature is the macroscopic face of microscopic motion.

$$\langle KE \rangle = \frac{3}{2}kT$$

Average translational kinetic energy per molecule equals (3/2)kT.

$\langle KE \rangle$=Average translational kinetic energy per molecule(J)

$k$=Boltzmann constant(J/K)

$T$=Absolute temperature(K)

$T = 0$ K

→

All translational kinetic energy is zero -- molecules cease translational motion.

Doubling $T$

→

$\langle KE \rangle$ doubles; $v_{rms}$ increases by $\sqrt{2}$.

Derivation: Equating $PV = NkT$ with $PV = \frac{2}{3}N\langle \frac{1}{2}mv^2 \rangle$ gives $T = \frac{2}{3k}\langle \frac{1}{2}mv^2 \rangle$.

Mass Independence: At the same $T$, all gas molecules (light or heavy) have the same average KE -- heavier molecules simply move slower.

RMS Speed: $v_{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$ where $M$ is molar mass.

Proportionality Shortcut: $v_{rms} \propto \sqrt{T/M}$ -- useful for comparing speeds of different gases or temperatures.

The Ideal Gas Law

The Ideal Gas Law unifies Pressure, Volume, Absolute Temperature, and Moles into a single equation of state.

$$PV = nRT$$

For n moles of an ideal gas, the product PV is proportional to absolute temperature.

$P$=Pressure(Pa or atm)

$V$=Volume(m³ or L)

$n$=Number of moles(mol)

$R$=Universal gas constant(8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))

$T$=Absolute temperature(K)

Per-molecule form

→

$PV = NkT$ where $N$ is total number of molecules and $k = R/N_A$

STP

→

$P = 1$ atm, $T = 273$ K → 1 mol occupies 22.4 L

Density Form: $\rho = PM/(RT)$ -- derived by substituting $n = m/M$ and $\rho = m/V$.

Dimensional Check: $[Pa][m^3] = [mol][J/(mol \cdot K)][K] = [J]$ ✓

Derivation of Gas Laws from KMT

Boyle's Law: At constant temperature, Pressure and Volume are Inversely Proportional.

$$P_1V_1 = P_2V_2$$

From KMT: $PV = \frac{2}{3}N\langle \frac{1}{2}mv^2 \rangle$. At constant T, the right side is constant, so $PV = \text{const}$.

$P_1, V_1$=Initial pressure and volume(various)

$P_2, V_2$=Final pressure and volume(same as initial)

Volume halved

→

Pressure doubles -- molecules hit walls twice as often in half the space.

Microscopic Reason: Smaller volume means shorter distance between walls, increasing collision frequency and hence pressure.

Process Name: An Isothermal process -- constant temperature.

Graph: $P$ vs $V$ is a rectangular hyperbola; $P$ vs $1/V$ is a straight line through the origin.

Charles's Law: At constant pressure, Volume is Directly Proportional to Absolute Temperature.

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

From KMT: $V = \frac{2N}{3P}\langle \frac{1}{2}mv^2 \rangle$. At constant P, $V \propto \langle KE \rangle \propto T$.

$V_1, T_1$=Initial volume and temperature(L, K)

$V_2, T_2$=Final volume and temperature(L, K)

Temperature doubled

→

Volume doubles -- faster molecules push the piston out to maintain constant pressure.

Microscopic Reason: Higher $T$ means higher $\langle KE \rangle$; molecules hit walls harder, so volume must expand to keep $P$ constant.

Process Name: An Isobaric process -- constant pressure.

Internal Energy

Internal Energy ($U$) is the sum of all molecular kinetic and potential energies. For an ideal gas, it is entirely translational kinetic energy.

$$U = \frac{3}{2}nRT$$

For a monatomic ideal gas, internal energy equals (3/2)nRT since there is no potential energy between molecules.

$U$=Internal energy of the gas(J)

$n$=Number of moles(mol)

$R$=Universal gas constant (8.314 J/(mol·K))(J/(mol·K))

$T$=Absolute temperature(K)

Ideal gas

→

$U$ depends ONLY on temperature, not on pressure or volume.

State Function: Internal energy depends only on the state (initial and final $T$), not on the path taken between states.

Energy Addition: Internal energy can increase via heat transfer OR mechanical work (e.g., friction, compression).

Diatomic Molecules: Have additional rotational and vibrational energy -- $U = \frac{5}{2}nRT$ at moderate temperatures.

Work Done by a Gas

When a gas expands against a piston, it does Work on the surroundings. Work done by the system is positive; work done on the system is negative.

$$W = P\Delta V$$

At constant pressure, work equals pressure times the change in volume.

$W$=Work done by the gas(J)

$P$=Constant pressure(Pa)

$\Delta V$=Change in volume ($V_f - V_i$)(m³)

$\Delta V = 0$ (constant volume)

→

$W = 0$ -- no work done. All heat goes to internal energy.

Expansion ($\Delta V > 0$)

→

$W > 0$ -- gas does positive work on surroundings.

Compression ($\Delta V < 0$)

→

$W < 0$ -- surroundings do work on the gas.

PV Diagram: Work equals the area under the curve on a P-V graph.

Sign Convention: Heat IN is positive ($+Q$), work OUT by the system is positive ($+W$).

First Law: $Q = \Delta U + W$ -- heat added equals change in internal energy plus work done by the gas.

Molar Specific Heats of a Gas

Gases have two Molar Specific Heat values: $C_v$ (at constant Volume) and $C_p$ (at constant Pressure), because heating at constant pressure also requires work for expansion.

$$C_p - C_v = R$$

The difference between molar specific heats at constant pressure and constant volume equals the universal gas constant.

$C_p$=Molar specific heat at constant pressure(J/(mol·K))

$C_v$=Molar specific heat at constant volume(J/(mol·K))

$R$=Universal gas constant (8.314 J/(mol·K))(J/(mol·K))

Monatomic ideal gas

→

$C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$, $\gamma = C_p/C_v = 5/3$

Diatomic ideal gas (moderate T)

→

$C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, $\gamma = 7/5 = 1.4$

Why $C_p > C_v$: At constant pressure, some heat goes into expansion work ($P\Delta V = R\Delta T$ for 1 mol), so more heat is needed for the same $\Delta T$.

At Constant Volume: $\Delta U = C_v \Delta T$ -- all heat goes to internal energy, no work done.

Derivation: From first law, $C_p \Delta T = C_v \Delta T + P\Delta V = C_v \Delta T + R\Delta T$, giving $C_p = C_v + R$.