# 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%?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Effusion concept. You can view video lessons to learn Effusion Or if you need more Effusion practice, you can also practice Effusion practice problems .

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