1.) The average molecular speed in a sample of CO2 gas at a certain temperature

Question

1.) The average molecular speed in a sample of CO2 gas at a certain temperature is 416 m/s. The average molecular speed in a sample of O2 gas is________ m/s at the same temperature.

2.)The rate of effusion of Ne gas through a porous barrier is observed to be mol/h. Under the same conditions, the rate of effusion of Kr gas would be _______mol/h.

3.) A sample of Ne gas is observed to effuse through a pourous barrier in 4.34 minutes. Under the same conditions, the same number of moles of an unknown gas requires 11.3 minutes to effuse through the same barrier. The molar mass of the unknown gas is __________g/mol.

Explanation / Answer

1) at same temp all gases have nearly same kinetic energy , hence

0.5 x m(CO2) x v(CO2)^2 = 0.5 x m ( O2) x v (O2) ^2

0.5 x 44g/mol x (416m/s)^2 = 0.5 x 32g/mol x v(O2)^2

v(O2) = 488 m/s

average molecular speed in sample of O2 gas is 488 m/s

3) rate of effusion of unkown gas / rate of effusion of Ne = ( Ne molar mass / unknown molar mass) ^ 1/2

( 1/11.3 min) / ( 1/4.34 min) = ( 20 / unknown molar mass) ^ 1/2

( 4.34/11.3)^2 = 20 / unknown molar mass

unknown molar mass = 135.6 g/mol

2) for 2nd question rate of effusion of Ne is not given , the formula we use is

rate of effusion of Ne / rate of effusion of Kr = ( Kr molar mass / Ne molar mass)^ 1/2

( 10mol/hr / rate of effusion of Kr) = ( 83.8g/mol / 20g/mol) ^ 1/2

rate of effusin of Kr = 4.885 mol/hr

( here since value is not given random value of 10 mol/hr is taken for Ne)

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