Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

$rate=\left(\frac{n}{t}\right)$

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

$\frac{RateofeffusionofgasA}{Rateofeffusionofgas\text{ B}}={\left(\frac{MolarmassofgasB}{Molarmassofgas\text{ A}}\right)}^{1/2}$