Question

A collapsed balloon is filled with He to a volume of \(12.5 \mathrm{I}\) at a pressure of \(1.00 \mathrm{atm} .\) Oxygen, \(\mathrm{O}_{2}\) is then added so that the final volume of the balloon is 26 L with a total pressure of 1.00 atm. The temperature, which remains constant throughout, is \(21.5^{\circ} \mathrm{C}\) (a) What mass of He does the balloon contain? (b) What is the final partial pressure of He in the balloon? (c) What is the partial pressure of \(\mathrm{O}_{2}\) in the balloon? (d) What is the mole fraction of each gas?

Short answer

(a) 2.072 g He. (b) 0.481 atm. (c) 0.519 atm. (d) Mole fractions: He = 0.481, O2 = 0.519.

Step-by-step solution

Calculate Moles of Helium

Use the ideal gas law, \( PV = nRT \), to find the moles of helium initially in the balloon.For helium, \( P = 1.00 \, \text{atm} \), \( V = 12.5 \, \text{L} \), \( R = 0.0821 \, \text{L atm/mol K} \), and \( T = 21.5^{\circ} \text{C} = 294.65 \, \text{K} \).Rearranging the equation gives: \( n = \frac{PV}{RT} \).Plug the values into the equation: \[ n = \frac{1.00 \, \text{atm} \times 12.5 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 294.65 \, \text{K}} \approx 0.518 \, \text{mol}. \]

Calculate Mass of Helium

The molar mass of helium is approximately \( 4.00 \, \text{g/mol} \). Use the moles calculated previously to determine the mass using the formula: \( \text{mass} = n \times \text{molar mass} \).Substitute the values: \[ \text{mass} = 0.518 \, \text{mol} \times 4.00 \, \text{g/mol} = 2.072 \, \text{g}. \]

Calculate Final Partial Pressure of Helium

Since the total pressure remains 1.00 atm and the volume doubled (from 12.5 L to 26 L), the partial pressure of helium will also change.The initial partial pressure is the same as the initial total pressure (1.00 atm) since it's the only gas initially. By the equation \( P_1V_1 = P_2V_2 \) and given \( V_1 = 12.5 \, \text{L} \), \( V_2 = 26 \, \text{L} \).Calculate \( P_2 \): \[ P_2 = \frac{P_1V_1}{V_2} = \frac{1.00 \, \text{atm} \times 12.5 \, \text{L}}{26 \, \text{L}} \approx 0.481 \, \text{atm}. \]

Calculate Partial Pressure of Oxygen

The total pressure is 1.00 atm and helium contributes 0.481 atm. Therefore, the partial pressure of oxygen can be found as follows: \( P_{\text{O}_2} = P_{\text{total}} - P_{\text{He}} \).Substitute the known values: \[ P_{\text{O}_2} = 1.00 \, \text{atm} - 0.481 \, \text{atm} = 0.519 \, \text{atm}. \]

Calculate Mole Fraction of Each Gas

The mole fraction is the ratio of the moles of a component to the total moles. We already have moles for helium, \( n_{\text{He}} = 0.518 \, \text{mol} \). Now assume total pressure gives total moles for oxygen using ideal gas law.For oxygen, rearrange \( P_{\text{O}_2}V = nRT \) to find moles \( n_{\text{O}_2} \): \[ n_{\text{O}_2} = \frac{0.519 \, \text{atm} \times 26 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 294.65 \, \text{K}} \approx 0.559 \, \text{mol}. \] Calculate mole fractions: \( X_{\text{He}} = \frac{0.518 \, \text{mol}}{0.518 \, \text{mol} + 0.559 \, \text{mol}} \approx 0.481 \).\( X_{\text{O}_2} = \frac{0.559 \, \text{mol}}{0.518 \, \text{mol} + 0.559 \, \text{mol}} \approx 0.519 \).

Key concepts

Partial Pressure

Partial pressure is an important concept in chemistry, especially when dealing with mixtures of gases. It's defined as the pressure that a specific gas in a mixture would exert if it were the only gas present in the volume. In other words, partial pressure is the contribution of a single gas's pressure to the total pressure of the gas mixture.
In the given exercise, we use the initial conditions of helium to determine its final partial pressure. Since the balloon's volume changes from 12.5 L to 26 L, the partial pressure of helium must also adjust accordingly. Using the relationship in Boyle's Law, where the initial pressure and volume product equals the final state, we have:

• Initial partial pressure of helium
• Change in the volume of the balloon
• Total pressure remains constant

Calculations show that the partial pressure of helium inside the balloon drops to 0.481 atm after the addition of oxygen. This is because, despite the increased volume, the same amount of helium occupies a larger space, leading to a reduced pressure per unit volume.

Mole Fraction

The mole fraction of a gas is the ratio of the number of moles of that gas to the total number of moles in the gas mixture. This fraction is a dimensionless number that provides insight into the proportion of each gas in the mixture. It's calculated using the formula:

• \( X_i = \frac{n_i}{n_{\text{total}}} \)

Where \(X_i\) is the mole fraction and \(n_i\) is the moles of each component gas.
In solving this problem, we start by finding the moles of each gas, helium, and oxygen, using the ideal gas law. From the exercise, we know:

• Moles of helium, \(n_{He} = 0.518\)
• Moles of oxygen, \(n_{O_2} = 0.559\)
• Total moles, \(n_{\text{total}} = 0.518 + 0.559 = 1.077\)

By dividing the individual moles of each gas by the total moles, we find that the mole fraction for helium is approximately 0.481, and for oxygen, it is about 0.519.
This simple calculation helps provide insight into how much of each gas is present relative to the total mixture inside the balloon.

Helium

Helium is a noble gas, known for its low density and inert characteristics. It is the second lightest and second most abundant element in the observable universe.
For problems involving gases, helium is often treated as an ideal gas due to its simple non-reactive nature and low molecular weight. In the exercise, helium is initially the only gas in the balloon, making it straightforward to apply the ideal gas law.The calculations begun with:

• Using the ideal gas law, \( PV = nRT \), to find moles
• Finding moles of helium, which is \( n = 0.518 \, \text{mol} \)
• Finding its partial pressure after adding another gas

The chemical inertness of helium keeps it stable even when mixed with other gases, like oxygen, without any risk of unwanted reactions. This makes it especially useful in experiments and settings that require inert atmospheres.

Oxygen

Oxygen, denoted as \(O_2\), is a highly reactive non-metal and the element essential for life on Earth. It makes up a significant portion of the Earth's atmosphere and is vital for respiration in living organisms.
In the context of gas laws, when oxygen is added to a gas mixture like in the exercise, its role is to adjust the total pressure and thus influence the behavior of the existing gas, in this case, helium.In solving the exercise:

• Oxygen is added to reach a total volume of 26 L
• The additional moles of oxygen were calculated to be \( n = 0.559 \, \text{mol} \)
• Partial pressure due to oxygen was found as \(0.519 \, \text{atm}\)

This showcases how the presence of oxygen impacts the overall gas dynamics inside the balloon by adding its pressure to reach a standard total pressure of 1 atm. Furthermore, understanding oxygen's properties is crucial when predicting its behavior and interaction with other gases in a mixture.