# Problem: 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%?

###### FREE Expert Solution
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###### FREE Expert Solution

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.

This means that when comparing two gases:

Since time is given:

Let's designate He as gas 1 and H2 as gas 2. Recall that rate = volume/time. This means:

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

Since the volume of He = volume of H2, the equation simplifies to:

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

Plugging this into Graham’s Law:

91% (287 ratings)
###### Problem Details

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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