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A room is 11.5 m by 7.2 m by 2.8 m is filled with air at 22°C and 1 atm. The average molar mass of air is 28.8 g/mol. Find the number of moles of air to fill the room and the weight of the air in Kg.

User ICantC
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Final answer:

The room contains approximately 10348 moles of air, and the air weighs about 298.1 kilograms. This calculation uses the ideal gas law at 22°C and 1 atm pressure, and the average molar mass of air, which is 28.8 g/mol.

Step-by-step explanation:

Calculating Moles and Weight of Air in a Room

To find the number of moles of air to fill a room with dimensions 11.5 m by 7.2 m by 2.8 m, and to calculate the weight of the air in kilograms, we can use the ideal gas law and the known molar mass of air. The volume of the room in cubic meters is 11.5 m × 7.2 m × 2.8 m. Since air at 22°C and 1 atm has a molar volume of 22.4 L at standard temperature and pressure (STP), we convert the room's volume to liters and then use the molar volume to find the number of moles:

Room Volume: 11.5 m × 7.2 m × 2.8 m = 231.84 m³ = 231840 L

Number of Moles: 231840 L × (1 mol/22.4 L) ≈ 10348 moles

The average molar mass of air is 28.8 g/mol, so to find the weight in kilograms, we multiply the number of moles by the molar mass and convert from grams to kilograms:

Weight of Air: 10348 moles × 28.8 g/mol = 298102.4 g = 298.1 kg

Therefore, the room contains approximately 10348 moles of air, which weighs about 298.1 kilograms.

User TPS
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