Question

A student tries to determine the volume of a bulb like the one shown in Figure \(10.12 .\) These are her results: mass of the bulb filled with dry air at \(23^{\circ} \mathrm{C}\) and \(744 \mathrm{mmHg}=91.6843 \mathrm{~g} ;\) mass of evacuated bulb \(=\) \(91.4715 \mathrm{~g}\). Assume the composition of air is 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2},\) and 1 percent argon by volume. What is the volume (in mL) of the bulb? (Hint: First calculate the average molar mass of air, as shown in Problem \(3.129 .)\)

Short answer

The volume of the bulb is 183.7 mL.

Step-by-step solution

Calculate the Mass of Air

To find the mass of the air inside the bulb, subtract the mass of the evacuated bulb from the mass of the bulb filled with air: \[\text{Mass of air} = 91.6843\, \text{g} - 91.4715\, \text{g} = 0.2128\, \text{g}\]

Calculate the Average Molar Mass of Air

Use the percentages and molar masses of each component to find the average molar mass of air. - Molar mass of \(\text{N}_2 = 28.02\, \text{g/mol}\)- Molar mass of \(\text{O}_2 = 32.00\, \text{g/mol}\)- Molar mass of \(\text{Ar} = 39.95\, \text{g/mol}\)Calculate the weighted average:\[\text{Molar mass of air} = (0.78 \times 28.02) + (0.21 \times 32.00) + (0.01 \times 39.95) \]\[= 21.8556 + 6.72 + 0.3995 = 28.9751\, \text{g/mol}\]

Use the Ideal Gas Law to Find Volume

Use the ideal gas law to solve for the volume of air:\[PV = nRT \]Given: - \(P = 744\, \text{mmHg} = 744 / 760\, \text{atm} = 0.9789\, \text{atm}\)- \(T = 23 + 273.15 = 296.15\, \text{K}\)First calculate the number of moles \(n\) from the mass and the molar mass.\[n = \frac{\text{mass of air}}{\text{molar mass of air}} = \frac{0.2128}{28.9751} = 0.007346\, \text{mol} \]Next, substitute the known values into the ideal gas law:\[V = \frac{nRT}{P} = \frac{0.007346 \times 0.0821 \times 296.15}{0.9789}\, \text{L}\]\[= \frac{0.1798}{0.9789} = 0.1837\, \text{L}\]Convert the volume from liters to milliliters:\[0.1837\, \text{L} = 183.7\, \text{mL}\]

State the Final Volume of the Bulb

The calculated volume of the bulb is 183.7 mL.

Key concepts

Average Molar Mass

The average molar mass of a mixture of gases, like air, is determined by the composition and molar masses of its components. Air is primarily composed of nitrogen (\(\text{N}_2\)), oxygen (\(\text{O}_2\)), and a small amount of argon (\(\text{Ar}\)). To calculate the average molar mass, you combine the molar masses of the individual gases based on their percentages in air.

• Nitrogen (\(\text{N}_2\)): Constitutes 78% of air with a molar mass of 28.02 g/mol.
• Oxygen (\(\text{O}_2\)): Makes up 21% of air and has a molar mass of 32.00 g/mol.
• Argon (\(\text{Ar}\)): Comprises 1% of air and has a molar mass of 39.95 g/mol.

To find the average molar mass, you calculate the weighted sum:\[\text{Molar mass of air} = (0.78 \times 28.02) + (0.21 \times 32.00) + (0.01 \times 39.95) = 28.9751\, \text{g/mol}\]
This average molar mass is crucial when using the ideal gas law to calculate quantities like volume.

Air Composition

Understanding the composition of air helps us calculate various properties, such as average molar mass and density. Air is mostly made of nitrogen and oxygen with a hint of argon. Each component contributes to air's overall behavior based on its volume percentage.
- Nitrogen (\(\text{N}_2\)): Dominates the air at 78%, playing a major role in its properties.- Oxygen (\(\text{O}_2\)): Comes next with 21%, important for combustion and respiration.- Argon: Makes up roughly 1%, being chemically inert, it doesn't contribute much to chemical reactions.
These constituents remain consistent under standard conditions, shaping air's predictable characteristics. Remember, while argon's percentage is small, collectively with nitrogen and oxygen, it impacts atmospheric calculations.

Volume Calculation

When dealing with gases, the Ideal Gas Law is a vital tool for calculating volume. For the bulb in our exercise, we use this law to find the volume of air inside once we know other variables. The Ideal Gas Law is given by \(PV = nRT\), where:
- \(P\) is the pressure, which is given as 744 mmHg and converted to atmospheres (atm).- \(n\) represents the number of moles, calculated from the mass of air over its molar mass.- \(R\) is the gas constant, typically 0.0821 L atm/mol K.- \(T\) is the temperature in Kelvin.
To find the volume \(V\), rearrange the equation:\[V = \frac{nRT}{P}\]Given the mass of air, its molar mass, and conditions in our problem, you can substitute values into this formula to solve for \(V\) in liters, which we convert to milliliters for final use. Understanding each component of the law and conversion is key to deriving the volume accurately.