Question

The average molecular speed in a sample of He gas at a certain temperature is 1.37x10 m/s. The average molecular speed in a sample of Kr gas is m/s at the same temperature.

Rate of Effusion Problems

Answer 1

Answer 1) 0.299 * 10³ m/s

Average molecular speed is inversely proportional to mass according to the formula therefore we can use the relative average velocity ratio which will be equal to square root of masses

Average molecular speed = 18RT TM Varl Var kr - Now He He V Mua

Atomic masses

M_Kr = 83.80 g/mol

M_He = 4 g/mol

Vav Kr - Mhe o Vas the Var kr 1 4 g/mol 1.37X103 m/s 783.80 g/mol - 10.047 Vav kr 1.370 10 3 m/s = 0.218 Vow kr 1.37 X10% m/s Var kr = 0.218 X 1.378 10² m/s. Var kr = 0.299 x 10 m/s.

Answer 2) 2.82 * 10^-4 mol/h

according to Graham law the rate of effusion is proportional to its root mean square velocity and it is further inversely proportional to molecular mass so we can calculate rate of effusion by taking the relative values ratio

Molecular masses

M_CH4 = 16 g/mol , M_{N2 =} 28 g /mol

Rate of effusion of usms = 4/3 RT MCHA pretty M Na 16 g/mol 3.13x10 mol/h rN2 10 571 3.13 31o "mele de TN = (10.571 ) X 3.13 x 10 mol/h r N₂ = 0.756 X 3,73 x 104 mol/h

Date r = 2.819 x 10 mol/h r = 2.82 x 104 mol/h