Question

The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft × 11.0 ft and has an 10.0 ft ceiling? 1 in = 2.54 cm (exactly); 1 L = 103 cm3.

Answer 1

Answer: There are 37 kg of air in the room.

Step-by-step explanation:

To calculate the volume of cuboid (room), we use the equation:

`V=lbh`

where,

V = volume of cuboid

l = length of room = 11 ft

b = breadth of room = 10 ft

h = height of room= 10 ft

Putting values in above equation, we get:

`V=10* 11* 10=1100ft^3=1100* 28.3L=31130L` (Conversion factor:
`1ft^3=28.3L`

To calculate mass of a substance, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

We are given:

Density of air = 1.19 g/L

Volume of air = volume of room = 31130 L

Putting values in above equation, we get:

`1.19g/L=\frac{\text{Mass of air}}{31130L}\\\\\text{Mass of air}=37000g=37.0kg` (1kg=1000g)

Hence, the mass of air is 37 kg.

Answer 2

Answer: The mass of air present in the room is 37.068 kg

Explanation : Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

`V=lbh`

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

`V=10.0ft* 11.0ft* 10.0ft=1100ft^3=31148.53L`

Conversion used :
`1ft^3=28.3168L`

Now we have to calculate the mass of air in the room.

`Density=(Mass)/(Volume)`

`1.19g/L=(Mass)/(31148.53L)`

`Mass=37066.7507g=37.068kg`

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg