Question: A balloon filled with helium gas is found to take6hours to deflate to50%of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?

Graham's law states that a gas's rate of diffusion or effusion is inversely related to its square root of molecular weight.

According to Graham’s Law

Rate of Effusion\( \propto \frac{{\rm{1}}}{{\sqrt {\rm{\mu }} }}\).

where\({\rm{\mu }}\)is the molar mass.

Hence, we have

\(\begin{array}{c}\frac{{{\rm{ rate of effusion of A}}}}{{{\rm{ rate of effusion of B}}}}{\rm{ = }}\frac{{\sqrt {{{\rm{\mu }}_{\rm{B}}}} }}{{\sqrt {{{\rm{\mu }}_{\rm{A}}}} }}\\\frac{{{\rm{ rate of effusion of Hydrogen }}}}{{{\rm{ rate of effusion of Helium }}}}{\rm{ = }}\frac{{\sqrt {{{\rm{\mu }}_{{\rm{Helium }}}}} }}{{\sqrt {{{\rm{\mu }}_{{\rm{Hydrogen }}}}} }}\\{\rm{ = }}\frac{{\sqrt {\rm{4}} }}{{\sqrt {\rm{2}} }}\\{\rm{ = }}\frac{{\rm{2}}}{{\sqrt {\rm{2}} }}\\{\rm{ = }}\sqrt {\rm{2}} \\\frac{{{\rm{ rate of effusion of Hydrogen }}}}{{{\rm{ rate of effusion of Helium }}}}{\rm{ = }}\sqrt {\rm{2}} \\{\rm{ rate of effusion of Hydrogen = }}\sqrt {\rm{2}} {\rm{(rate of effusion ofHelium)}}\end{array}\)

Now, to obtain the time, we have

\({\rm{rate of effusion = }}\frac{{{\rm{ amount of gas transferred }}}}{{{\rm{ time }}}}\)

Rate of effusion \( \propto \frac{{\rm{1}}}{{{\rm{ time }}}}\)

Time for effusion of Hydrogen \({\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{2}} }}\)x Time of effusion of Helium

\(\begin{array}{c}{\rm{ = }}\left( {\frac{{\rm{1}}}{{\sqrt {\rm{2}} }}} \right){\rm{ \times (6) hours}}\\{\rm{ = }}\frac{{\rm{6}}}{{\sqrt {\rm{2}} }}{\rm{ hours}}\\{\rm{ = 4}}{\rm{.2 hours }}\end{array}\)

Therefore, the value for time is \({\rm{4}}{\rm{.2 hours}}\).