Questions: Problem 5.63: Using Avagadro's Law: If 5.00 g helium gas is added to a 4.00-L balloon containing 1.00 g of helium gas, what is the new volume of the balloon? Assume no change in temperature or pressure.

Solution Steps

Avogadro's Law states that the volume of a gas is directly proportional to the number of moles of the gas, provided the temperature and pressure remain constant. Mathematically, it can be expressed as: \[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \] where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( n_1 \) and \( n_2 \) are the initial and final moles of gas, respectively.

First, we need to calculate the initial moles of helium in the balloon. Given that the initial mass of helium is 1.00 g and the molar mass of helium (He) is approximately 4.0026 g/mol, we can find the initial moles (\( n_1 \)) as follows: \[ n_1 = \frac{1.00 \, \text{g}}{4.0026 \, \text{g/mol}} = 0.2498 \, \text{mol} \]

Next, we add 5.00 g of helium to the balloon. The total mass of helium now is: \[ 1.00 \, \text{g} + 5.00 \, \text{g} = 6.00 \, \text{g} \] The final moles of helium (\( n_2 \)) are: \[ n_2 = \frac{6.00 \, \text{g}}{4.0026 \, \text{g/mol}} = 1.4993 \, \text{mol} \]

We know the initial volume (\( V_1 \)) is 4.00 L. Using Avogadro's Law, we can find the final volume (\( V_2 \)): \[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \] \[ \frac{4.00 \, \text{L}}{0.2498 \, \text{mol}} = \frac{V_2}{1.4993 \, \text{mol}} \] Solving for \( V_2 \): \[ V_2 = \frac{4.00 \, \text{L} \times 1.4993 \, \text{mol}}{0.2498 \, \text{mol}} = 24.00 \, \text{L} \]

Final Answer