A spherical bulb with a volume of 500 cm3 is evacuated to a negligibly small res

Question

A spherical bulb with a volume of 500 cm3 is evacuated to a negligibly small residual gas pressure and then close off. One hour later, the pressure in the vessel is found to be 2.00 × 10-7 atm because the bulb has a tiny hole in it. Assume that the surroundings are at atmospheric pressure, T = 300 K, and the average molar mass of molecules in the atmosphere is 28.8 g mol-1.

a) Please calculate the radius of the hole on the bulb surface, assuming the tiny hole to be circular

b) assume that the surroundings contain only N2 (molar mass 28 g/mol) and O2 (molar mass 32.0g/mol)and the mole fraction ratio between the two species is X(N2):X(O2) = 4:1. when the pressure in the bulb is found to be 1.00 x10^-7 atm, the mole fraction ratio X(N2):X(O2) inside the bulb should be higher , equal to, or smaller than 4:1? Why? Explain your answer by calculating the enrichment factor for N2

Explanation / Answer

(b): Given the initial mole fraction ratio X(N2):X(O2) = 4:1

The mole fraction ratio X(N2):X(O2) inside the bulb will be higher than the outside 4:1 ratio.

This can be explained from the following calculation.

The rate at which gas molecules enters inside the bulb can be calculated from the following formulae

Reff = (1/4) x (X/V) x Vavg

where Vavg = average velocity.

For N2 gas:

R1 = (1/4) x [X(N2)/V)]x underroot(8RT / PixM(N2)) ------- (1)

For O2 gas:

R2 = (1/4) x [X(O2)/V)]x underroot(8RT / PixM(O2)) ------- (2)

Dividing eqn(1) by (2) we get

R1/R2 = [X(N2) / X(O2)]x underroot[M(O2) / M(N2)]

=> R1/R2 = 4 x underroot(32/28) = 4.276 which is higher than 4.

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