Consider a classroom 10.0m long by 6.0m wide by 4.0m high at "room temperature" 20.C and 1.0 atmosphere.

A) How many moles of gas are contained in the room?

B) Assuming an average molecular weight of 28, what is the mass of the room air?

To find the number of moles of gas contained in the room, we need to use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in m³)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)

First, let's convert the dimensions of the room to volume in cubic meters (m³):
Volume (V) = Length × Width × Height
V = 10.0 m × 6.0 m × 4.0 m
V = 240 m³

Now, let's convert the temperature from Celsius to Kelvin:
Temperature (T) = 20°C + 273.15
T = 293.15 K

Assuming the pressure is 1.0 atmosphere, we can substitute the values into the ideal gas law equation:

(1.0 atm) × (240 m³) = n × (0.0821 L·atm/(mol·K)) × (293.15 K)

Simplifying the equation, we have:

240 = n × 24.070615

Solving for n:
n = 240 / 24.070615
n ≈ 9.974 moles

Therefore, the room contains approximately 9.974 moles of gas.

To calculate the mass of the room air, we can use the average molecular weight formula:

Mass = Number of moles × Molecular weight

Given that the average molecular weight is 28 g/mol, we can substitute the values into the equation:

Mass = 9.974 moles × 28 g/mol
Mass ≈ 279.472 g

Therefore, the mass of the room air is approximately 279.472 grams.