Answer to Question #318785 in Physical Chemistry for Lucky

Question

Calculate the : 1) average speed 2) root mean square speed and 3) most probable speed of oxygen molecule at 515k. {Given Mm(O2)= 0.016Kg mol(–1)

Answer 1

T = 515 K;

R = 83140000 erg * K^-1 * mol^-1;

M(O₂) = 32 g * mol^-1;

$\pi$ = 3.14;

1) The average speed: S_average = ((8 * R * T)/($\pi$ * M))^1/2 = ((8 * 83140000 * 515)/(3.14 * 32))^1/2 = 58387 sm *sec^-1 = 583.87 m *sec^-1;

2) The root mean square speed: S_{root mean square} = ((3 * R * T)/(M))^1/2 = ((3 * 83140000 * 515)/(32))^1/2 = 63357 sm *sec^-1 = 633.57 m *sec^-1;

3) The most probable speed: S_{most probable} = ((2 * R * T)/(M))^1/2 = ((2 * 83140000 * 515)/(32))^1/2 = 51731 sm *sec^-1 = 517.31 m *sec^-1.

Answer: 1) 583.87 m *sec^-1;

2) 633.57 m *sec^-1;

3) 517.31 m *sec^-1.