You have a partially filled party balloon with 2.00 g of helium gas. You then add 3.45 g of hydrogen gas to the balloon. Assuming constant temperature and pressure, how many times bigger is the party balloon - comparing before and after the hydrogen gas has been added?

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Quizplus · Accepted Answer

At constant temperature and pressure, the volume of a gas is directly proportional to the amount (in moles) of gas. The molar mass of helium (He) is approximately 4.00 g/mol, and the molar mass of hydrogen (H2) is approximately 2.00 g/mol. Initially, there are $2.00 \, \text{g He} \times \frac{1 \, \text{mol He}}{4.00 \, \text{g He}} = 0.50 \, \text{mol He}$. After adding hydrogen, there are $3.45 \, \text{g H2} \times \frac{1 \, \text{mol H2}}{2.00 \, \text{g H2}} = 1.725 \, \text{mol H2}$. The total moles of gas after adding hydrogen is $0.50 \, \text{mol He} + 1.725 \, \text{mol H2} = 2.225 \, \text{mol}$. The ratio of the final amount of gas to the initial amount is $2.225 \, \text{mol} / 0.50 \, \text{mol} = 4.45$, making the balloon 4.45 times bigger.

You Have a Partially Filled Party Balloon with 2