Question

An effusion cell has a circular hole of diameter 2.50 mm. If the molar mass of the solid in the cell is 260 g mol-1and its vapour pressure is 0.853 Pa at 400K, by how much will the mass of the solid decrease in a period of 2.00 hours?

Answer 1

Solution

$v = \frac{\Delta P \times Na \times A}{(2 \times \pi \times R \times T)^{0.5}}$

$A = \pi \times R^2$

$A = 1.96 \times 10^{-5} m^2$

$v = \frac{0.853 \times 6.02 \times 10^{23} \times 1.96 \times 10^{-5}}{2 \times 3.14 \times 8.31 \times 400} = 6.97 \times 10^{16} particles/sec.$

$N = v \times t$

$N = 6.97 \times 10^{16} \times 7200 = 5.02 \times 10^{20} particles$

$\frac{m}{M} = \frac{N}{Na}$

$m = \frac{M \times N}{Na} = 0.22 \ g$

Answer:

m = 0.22 g