Chapter 5.7, Problem 5.6CC

(a)

Interpretation Introduction

Interpretation:

Among the $\text{Ar}$ gas and $\text{He}$ gas present in the separate given experimental apparatus containing a round flask, valve and long tube, the gas that first reaches the end of the long tube after opening of valves has to be explained

Concept Introduction:

Graham’s law of effusion:

Graham’s law of effusion states that “at constant temperature and pressure, the rate of effusion of gas molecule from a particular hole is inversely proportional to the square root  of the molecular weight of the gas” and is given as,

$\text{Rate of effusion of molecules }\propto \frac{\text{1}}{\sqrt{{\text{M}}_{\text{m}}}}$

Where,

${\text{M}}_{\text{m}}$ is the molar mass of the gas

(b)

Interpretation Introduction

Interpretation:

Among the $\text{Ar}$ gas and $\text{He}$ gas present in the separate given experimental apparatus containing a round flask, valve and long tube.

The experimental change needed to be done for reaching the end of long tube by $\text{Ar}$ gas at the same time as the $\text{He}$ gas has to be explained.

Concept Introduction:

Graham’s law of effusion:

Graham’s law of effusion states that “at constant temperature and pressure, the rate of effusion of gas molecule from a particular hole is inversely proportional to the square root  of the molecular weight of the gas” and is given as,

$\text{Rate of effusion of molecules }\propto \frac{\text{1}}{\sqrt{{\text{M}}_{\text{m}}}}$

Where,

${\text{M}}_{\text{m}}$ is the molar mass of the gas