Thermodynamics & Kinetic Theory of Gases

Thermodynamics governs energy exchange between systems and their surroundings. The Zeroth Law establishes the concept of temperature: if system A is in thermal equilibrium with B, and B with C, then A is in thermal equilibrium with C.
The First Law is the energy conservation statement: Q = $\Delta$-U + W, where Q is the heat absorbed by the system (J), $\Delta$-U is the change in internal energy (J), and W is the work done by the system (J). All have dimensional formula [$M^{1} \; L^{2} \; T^{-2}$]. Sign convention (NEET standard): Q > 0 (heat absorbed), Q < 0 (heat released); W > 0 (expansion), W < 0 (compression); $\Delta$-U > 0 (temperature rise).

Four thermodynamic processes define special conditions.
In an isothermal process (T = constant), $\Delta$-U = 0 for an ideal gas, so Q = W = nRT ln($\frac{V_{2}}{V_{1}}$), where n is moles, R = 8.314 J $mol^{-1} \; K^{-1}$, and PV = constant.
In an adiabatic process (Q = 0), W = -$\Delta$-U = nC_v $\Delta$-T = ($P_{1} \; V_{1}$ - $P_{2} \; V_{2}$)/($\gamma$ - 1); the governing relations are PV^$\gamma$ = constant, TV^($\gamma$-1) = constant. Adiabatic expansion cools the gas; compression heats it.
In an isochoric process (V = constant), W = 0, so Q = $\Delta$-U = nC_v $\Delta$-T; P/T = constant.
In an isobaric process (P = constant), W = P $\Delta$-V = nR $\Delta$-T, Q = nC_p $\Delta$-T; V/T = constant. Work done in any process equals the area under the PV curve.

The Second Law has two equivalent statements. Kelvin-Planck: no engine can convert all absorbed heat into work (100% efficiency is impossible). Clausius: heat cannot spontaneously flow from cold to hot without external work.
The Carnot engine operates between a hot reservoir at $T_{1}$ (K) and a cold reservoir at $T_{2}$ (K) with maximum efficiency: $\eta$ = 1 - $\frac{T_{2}}{T_{1}}$ = W/$Q_{1}$ = ($Q_{1}$ - $Q_{2}$)/$Q_{1}$. Temperatures must be in kelvin; using Celsius gives wrong answers.
A refrigerator has COP = $Q_{2}$/W = $T_{2}$/($T_{1}$ - $T_{2}$).

The kinetic theory of gases connects microscopic molecular motion to macroscopic properties.
The ideal gas equation PV = nRT = Nk_B T, where $k_{B}$ = 1.38 x $10^{-23}$ J/K is the Boltzmann constant.
Kinetic pressure: P = $\frac{1}{3}$ $\rho \; v_{rms}^{2}$ [$M^{1} \; L^{-1} \; T^{-2}$] (Pa).
Molecular speeds: $v_{rms}$ = $\sqrt{3RT/M}$ [$M^{0} \; L^{1} \; T^{-1}$] (m/s), $v_{avg}$ = $\sqrt{8RT/(\pi M}$), $v_{mp}$ = $\sqrt{2RT/M}$.
The ratio $v_{mp}$ : $v_{avg}$ : $v_{rms}$ = $\sqrt{2}$ : $\sqrt{8/\pi}$ : $\sqrt{3}$ = 1 : 1.128 : 1.224. The ordering $v_{mp}$ < $v_{avg}$ < $v_{rms}$ is always maintained — most probable speed is the smallest.

Degrees of freedom (f): monoatomic gases (He, Ne, Ar) have f = 3 (translational only), diatomic ($H_{2}$, $O_{2}$, $N_{2}$) have f = 5 (3 translational + 2 rotational), polyatomic ($\text{CO}_{2}$, $H_{2O}$) have f = 6 (3 translational + 3 rotational).
By the law of equipartition, energy per degree of freedom = $\frac{1}{2}$ $k_{B}$ T per molecule = $\frac{1}{2}$ RT per mole.
Internal energy U = (f/2)nRT.
Specific heats: $C_{v}$ = (f/2)R, $C_{p}$ = $C_{v}$ + R = ((f+2)/2)R, and $\gamma$ = $\frac{C_{p}}{C_{v}}$ = (f+2)/f.
For monoatomic: $C_{v}$ = 3R/2, $C_{p}$ = 5R/2, $\gamma$ = $\frac{5}{3}$ = 1.67.
For diatomic: $C_{v}$ = 5R/2, $C_{p}$ = 7R/2, $\gamma$ = $\frac{7}{5}$ = 1.4.
For polyatomic: $C_{v}$ = 3R, $C_{p}$ = 4R, $\gamma$ = $\frac{4}{3}$ = 1.33.

Solved Numerical 1: Isothermal expansion at T = 300 K, n = 2 mol, $V_{1}$ = 2 L, $V_{2}$ = 6 L (R = 8.314 J $mol^{-1} \; K^{-1}$).
W = nRT ln($\frac{V_{2}}{V_{1}}$) = 2 x 8.314 x 300 x ln($\frac{6}{2}$) = 2 x 8.314 x 300 x ln(3) = 4988.4 x 1.0986 = 5480 J.
Since isothermal: $\Delta$-U = 0, Q = W = 5480 J.

Solved Numerical 2: Carnot engine: $T_{hot}$ = 527 degrees C = 527 + 273 = 800 K, $T_{cold}$ = 127 degrees C = 127 + 273 = 400 K.
Efficiency: $\eta$ = 1 - $\frac{T_{2}}{T_{1}}$ = 1 - $\frac{400}{800}$ = 0.5 = 50%.
If $Q_{1}$ = 2000 J: W = $\eta \; Q_{1}$ = 0.5 x 2000 = 1000 J.
$Q_{2}$ = $Q_{1}$ - W = 2000 - 1000 = 1000 J.
Dimensional check: all quantities in [J] = [$M^{1} \; L^{2} \; T^{-2}$].

Solved Numerical 3: Oxygen (M = 32 x $10^{-3}$ kg/mol) at T = 300 K (R = 8.314 J $mol^{-1} \; K^{-1}$).
$v_{rms}$ = $\sqrt{3RT/M}$ = $\sqrt{3 x 8.314 x 300 / 0.032}$ = $\sqrt{7481.3 / 0.032}$ = $\sqrt{233,790}$ = 483.5 m/s.
$v_{avg}$ = $\sqrt{8RT/(\pi M}$) = $\sqrt{8 x 8.314 x 300 / (\pi x 0.032}$) = $\sqrt{19,954 / 0.10053}$ = $\sqrt{198,480}$ = 445.5 m/s.
$v_{mp}$ = $\sqrt{2RT/M}$ = $\sqrt{2 x 8.314 x 300 / 0.032}$ = $\sqrt{155,888}$ = 394.8 m/s. Ratio check: 394.8 : 445.5 : 483.5 = 1 : 1.128 : 1.225.

The key testable concept is the distinction between thermodynamic processes (especially that Q = 0 in adiabatic does NOT mean $\Delta$-U = 0 or T = constant) and the correct application of Carnot efficiency using kelvin temperatures.