Answer to Question #322834 in Physical Chemistry for Vikram

Question

Calculate the 1) average speed, 2) root mean square speed and, 3) most probable speed of oxygen molecules at 515k. (Given Mm(O2)=0.016kgmol^-1).

Answer 1

T = 515 K;

M(O₂) = 0.016 kg/mol = 16 g/mol;

R = 83140000 erg/(K * mol);

$\pi$ = 3.14;

1) The average speed: S_average = ((8 * R * T)/($\pi$ * M))^1/2 = ((8 * 83140000 * 515)/(3.14 * 16))^1/2 = 82571 sm/sec = 825.71 m/sec;

2) The root mean square speed: S_{root mean square} = ((3 * R * T)/(M))^1/2 = ((3 * 83140000 * 515)/(16))^1/2 = 89600 sm/sec = 896.00 m/sec;

3) The most probable speed: S_{most probable} = ((2 * R * T)/(M))^1/2 = ((2 * 83140000 * 515)/(16))^1/2 = 73158 sm/sec = 731.58 m/sec.

Answer: 1) 825.71 m/sec;

2) 896.00 m/sec;

3) 731.58 m/sec.