A brief note on Pressure, Temperature and Molecular Speeds . Read it before a test. Try each question at the end before you open its answer.
The model A gas is a very large number of tiny molecules in random motion. They exert no forces on each other except in collisions, and every collision is elastic. Pressure One molecule gives a wall 2 m v x 2mv_x 2 m v x in each hit and hits it v x 2 L \dfrac{v_x}{2L} 2 L v x times each second. Pressure: p = 1 3 n m v 2 ˉ = 1 3 ρ v 2 ˉ p = \tfrac13 nm\bar{v^2} = \tfrac13\rho\bar{v^2} p = 3 1 nm v 2 ˉ = 3 1 ρ v 2 ˉ , with n n n molecules in each cubic metre. In terms of energy: p = 2 3 E V p = \tfrac23\,\dfrac{E}{V} p = 3 2 V E , where E E E is the total translational kinetic energy. Temperature Total energy: E = 3 2 n R T E = \tfrac32 nRT E = 2 3 n R T for n n n moles. One molecule: K ˉ = 3 2 k B T \bar{K} = \tfrac32 k_BT K ˉ = 2 3 k B T , with k B = 1.38 × 10 − 23 k_B = 1.38 \times 10^{-23} k B = 1.38 × 1 0 − 23 J/K. Temperature measures the average kinetic energy. At one temperature it is the same for every gas. Speeds rms speed: v r m s = 3 R T / M v_{rms} = \sqrt{3RT/M} v r m s = 3 R T / M , with M M M in kg/mol. Average speed: v a v g = 8 R T / ( π M ) v_{avg} = \sqrt{8RT/(\pi M)} v a v g = 8 R T / ( π M ) . Most probable speed: v m p = 2 R T / M v_{mp} = \sqrt{2RT/M} v m p = 2 R T / M . v m p : v a v g : v r m s = 1 : 1.128 : 1.225 v_{mp} : v_{avg} : v_{rms} = 1 : 1.128 : 1.225 v m p : v a v g : v r m s = 1 : 1.128 : 1.225 . A hotter gas has a speed curve that is further right, lower and wider. A worked example Nitrogen ( M = 28 M = 28 M = 28 g/mol) is at 290 290 290 K. Find the average kinetic energy of a molecule and the rms speed.
Energy. K ˉ = 3 2 × 1.38 × 10 − 23 J/K × 290 K = 6.00 × 10 − 21 \bar{K} = \tfrac32 \times 1.38 \times 10^{-23}\ \text{J/K} \times 290\ \text{K} = 6.00 \times 10^{-21} K ˉ = 2 3 × 1.38 × 1 0 − 23 J/K × 290 K = 6.00 × 1 0 − 21 J. Speed. First 3 R T = 3 × 8.314 J/(mol K) × 290 K = 7233 J/mol 3RT = 3 \times 8.314\ \text{J/(mol K)} \times 290\ \text{K} = 7233\ \text{J/mol} 3 R T = 3 × 8.314 J/(mol K) × 290 K = 7233 J/mol . Then v r m s = 7233 J/mol 0.028 kg/mol = 508 v_{rms} = \sqrt{\dfrac{7233\ \text{J/mol}}{0.028\ \text{kg/mol}}} = 508 v r m s = 0.028 kg/mol 7233 J/mol = 508 m/s. Practice with new numbers
A program to try The program works through the example above, and adds the other two speeds.
Run it. Compare the first two lines with the worked example.
1. Change M = 0.028 to M = 0.002, for hydrogen. Which numbers change, and which stays the same?
Mistakes to avoid Using °C in 3 2 k B T \tfrac32 k_BT 2 3 k B T or 3 R T / M \sqrt{3RT/M} 3 R T / M . Using M M M in g/mol in the speed formulas. Leaving out the one third in p = 1 3 ρ v 2 ˉ p = \tfrac13\rho\bar{v^2} p = 3 1 ρ v 2 ˉ . Doubling T T T to double the speed. The temperature must be four times as large. Check yourself The rms speed of a gas is 400 400 400 m/s at 200 200 200 K. What is it at 800 800 800 K? Stuck? Show a hint Show the answerWhich has the larger average kinetic energy at 300 300 300 K, a nitrogen molecule or a helium atom? Stuck? Show a hint Show the answerA gas has a density of 2.0 kg/m 3 2.0\ \text{kg/m}^3 2.0 kg/m 3 and an rms speed of 300 300 300 m/s. What is its pressure? Stuck? Show a hint Show the answerWhy is the average speed of gas molecules larger than the most probable speed? Stuck? Show a hint Show the answer