A tube with a porous wall allows 0.53 litre of N_(2) to escape per minute from a pressure of 1 atm to an evacuated chamber. What will be the amount escaping under the same conditions for He, C Cl_(4) vapour and UF_(6) ? (He = 4, N = 14, C = 12, Cl = 35.5, F = 19, U = 28)

A tube with a porous wall allows 0.53 litre of N_(2) to escape per min

1.	A tube with a porous wall allows 0.53 litre of N_(2) to escape per minute from a pressure of 1 atm to an evacuated chamber. What will be the amount escaping under the same conditions for He, C Cl_(4) vapour and UF_(6) ? (He = 4, N = 14, C = 12, Cl = 35.5, F = 19, U = 28)
Answer» Solution :for He, `(V_(He))/(V_(N_(2))) = SQRT((M_(N_(2)))/(M_(He))), (V_(He))/(0.53) = sqrt((28)/(4)) = sqrt(7)` `V_(He) = 0.53 xx sqrt(7) = 1.40` litre per minute. Similarly, for `C Cl_(4)` vapour and `UF_(6)`, `(V_(C Cl_(4)))/(V_(N_(2))) = sqrt((M_(N_(2)))/(M_(C Cl_(4)))) = sqrt((28)/(152)), V_(C Cl_(4)) = 0.227` lit. per min. `(V_(UF_(6)))/(V_(N_(2))) = sqrt((M_(N_(2)))/(M_(UF_(6)))) = sqrt((28)/(352)) , V_(UF_(6)) = 0.149` lit. per min.