A brief note on Kinetic Theory and Real Gases . Read it before a test. Try each question at the end before you open its answer.
Kinetic theory Assumptions: tiny molecules, random motion, no forces except in collisions, elastic collisions, average kinetic energy proportional to T T T . P V = 1 3 N m u rms 2 PV = \tfrac{1}{3}Nmu_{\text{rms}}^2 P V = 3 1 N m u rms 2 . For one mole, u rms = 3 R T / M u_{\text{rms}} = \sqrt{3RT/M} u rms = 3 R T / M with M M M in kg/mol. Average kinetic energy: 3 2 k T \tfrac{3}{2}kT 2 3 k T for a molecule and 3 2 R T \tfrac{3}{2}RT 2 3 R T for a mole, the same for every gas. At the same T T T , the lighter gas is faster: u ∝ 1 / M u \propto 1/\sqrt{M} u ∝ 1/ M . The Maxwell distribution Fraction of molecules between u u u and u + d u u + du u + d u is f ( u ) d u f(u)\,du f ( u ) d u . The area under the curve is 1 1 1 . u mp = 2 R T / M u_{\text{mp}} = \sqrt{2RT/M} u mp = 2 R T / M , u ˉ = 8 R T / π M \bar{u} = \sqrt{8RT/\pi M} u ˉ = 8 R T / π M , u rms = 3 R T / M u_{\text{rms}} = \sqrt{3RT/M} u rms = 3 R T / M . The ratio is 1 : 1.128 : 1.225 1 : 1.128 : 1.225 1 : 1.128 : 1.225 . A higher T T T moves the peak right and lowers it. The fast tail grows a lot. A lighter gas has a wider curve. Real gases Z = P V m R T Z = \dfrac{PV_m}{RT} Z = R T P V m . Z < 1 Z < 1 Z < 1 : attraction wins. Z > 1 Z > 1 Z > 1 : size wins. Z → 1 Z \to 1 Z → 1 as P → 0 P \to 0 P → 0 . van der Waals: ( P + a n 2 V 2 ) ( V − n b ) = n R T \left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT ( P + V 2 a n 2 ) ( V − nb ) = n R T . a a a is the attraction, b b b the excluded volume. T c = 8 a 27 R b T_c = \dfrac{8a}{27Rb} T c = 27 R b 8 a , P c = a 27 b 2 P_c = \dfrac{a}{27b^2} P c = 27 b 2 a , V c = 3 b V_c = 3b V c = 3 b , Z c = 3 8 Z_c = \tfrac{3}{8} Z c = 8 3 , and T B = a R b T_B = \dfrac{a}{Rb} T B = R b a . A gas below T c T_c T c can be liquefied by pressure. Above T c T_c T c it cannot. A worked example Find the rms speed of argon, M = 39.95 g/mol M = 39.95\ \text{g/mol} M = 39.95 g/mol , at 350 K 350\ \text{K} 350 K . Then find the temperature at which helium, M = 4.003 g/mol M = 4.003\ \text{g/mol} M = 4.003 g/mol , has the same rms speed.
Argon. u rms = 3 × 8.314 × 350 0.03995 = 467 m/s u_{\text{rms}} = \sqrt{\dfrac{3 \times 8.314 \times 350}{0.03995}} = 467\ \text{m/s} u rms = 0.03995 3 × 8.314 × 350 = 467 m/s . Same speed. The speed depends on T / M T/M T / M , so T He = 350 × 4.003 39.95 T_{\text{He}} = 350 \times \dfrac{4.003}{39.95} T He = 350 × 39.95 4.003 . The temperature. T He = 35.1 K T_{\text{He}} = 35.1\ \text{K} T He = 35.1 K . Practice with new numbers
A program to try The program finds the critical temperature, the critical pressure and the Boyle temperature from the constants a a a and b b b .
Run it. Which gas has the highest critical temperature, and what does that say about how easily it liquefies?
1. Add "CH4": (2.303, 0.0431) to the dictionary. What is its critical temperature?
Mistakes to avoid The molar mass in g/mol in u = 3 R T / M u = \sqrt{3RT/M} u = 3 R T / M . Use kg/mol. One speed for all molecules. The molecules have a spread of speeds. A wrong sign in Z Z Z . Z < 1 Z < 1 Z < 1 means attraction, and Z > 1 Z > 1 Z > 1 means that the size of the molecules wins. Forgetting n b nb nb in the volume. The free volume is V − n b V - nb V − nb . Trying to liquefy a gas above T c T_c T c by pressure. Check yourself How does the rms speed change when the temperature rises from 200 K 200\ \text{K} 200 K to 800 K 800\ \text{K} 800 K ? Stuck? Show a hint Show the answerWhich is smaller, the most probable speed or the average speed? Stuck? Show a hint Show the answerCan nitrogen, T c = 126.2 K T_c = 126.2\ \text{K} T c = 126.2 K , be liquefied at room temperature by pressure alone? Stuck? Show a hint Show the answerSources OpenStax, Chemistry 2e , Chapter 9, sections 9.5 and 9.6. CRC Handbook of Chemistry and Physics, for the van der Waals and critical constants.