Question

3. For a sample of CO₂ at 20 ºC, calculate the most probable molecular speed, mean molecular speed and root-mean-square molecular speed. Assume behavior described by the kinetic theory of gases. Also, what would the temperature need to increase to for each of the speeds in part (a) to double? [Hint: Look at the equations! No need for extensive calculations.]

Answer 1

T= 20°c = 293K For (0₂, molemlar mass, M = 44 gmole (a) Caluclation of most brobable speed: 2x 8.314 Imol x 293K 44 x103 kg motel 110. 7274 x103 Kamps-2 1 kg 110727.4 ms - 19 = 332. 76 ms77 J Am (aluation of mean molecular sbeed. The mean molecular speed is given by <c) = TM 18x 8.314 Imitex x 293K 3.14 x 44X103 kg molet

= px 8.314 x 293 x1000 kg m252 3.14 x 44 141053.96 <C) = 375.57 meyl Ans © Calulahon of moot mean square : Krzy's 3x 8.314 Jmoler KT x 2935 ✓ 44X103 kg moles 3x 8.314 x 293 X 1000 km2 44 kg 166091.04 mst <>= 407.5 mx' | Any # Catulation of temperature in part (9): - Let initial & most probable velocity = Ump of T - After increasent temp., most probable = v velouty . Given', at Tz U'mp = a V mp

26 RT RT -= 24 (squaring on both sides) .. 3 T2 = 4x T, = 4 x 293K | T2 = 1172 k r Thus, temperature increases four times.