Recall that * Graham's Law of Effusion* allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

$\overline{){\mathbf{rate}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{\sqrt{\mathbf{MM}}}}$

This means that when comparing two gases:

$\overline{)\frac{{\mathbf{rate}}_{\mathbf{1}}\mathbf{}}{{\mathbf{rate}}_{\mathbf{2}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\sqrt{\frac{{\mathbf{MM}}_{\mathbf{2}}}{{\mathbf{MM}}_{\mathbf{1}}}}}$

Since time is given:

Let's designate He as gas 1 and H_{2} as gas 2. Recall that **rate = volume/time**. This means:

$\frac{{\mathbf{rate}}_{\mathbf{He}}}{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}\mathbf{=}\frac{{\displaystyle \frac{{V}_{\mathrm{He}}}{{t}_{\mathrm{He}}}}}{{\displaystyle \frac{{\mathbf{V}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{t}}_{{\mathbf{H}}_{\mathbf{2}}}}}}$

Since the **volume of He = volume of H_{2}**, the equation simplifies to:

$\frac{{\mathbf{rate}}_{\mathbf{He}}}{{\mathbf{rate}}_{\mathbf{disilane}}}\mathbf{=}\frac{{\displaystyle \frac{\overline{){\mathbf{V}}_{\mathbf{He}}}}{{\mathbf{t}}_{\mathbf{He}}}}}{\frac{\overline{){\mathbf{V}}_{{\mathbf{H}}_{\mathbf{2}}}}}{{\mathbf{t}}_{{\mathbf{H}}_{\mathbf{2}}}}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{rate}}_{\mathbf{He}}}{{\mathbf{rate}}_{{\mathbf{H}}_{\mathbf{2}}}}\mathbf{=}\frac{{\mathbf{t}}_{{\mathbf{H}}_{\mathbf{2}}}}{{\mathbf{t}}_{\mathbf{He}}}$

Plugging this into Graham’s Law:

A balloon filled with helium gas is found to take 6 hours to deflate to 50% 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%?

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