Question

Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m×5 m× 6 m at 100 kPa and 25 C

Answer 1

Answer : The density, specific gravity, and mass of the air in a room is, 1.16825 g/L, 0.916 and 140.19 kg respectively.

Explanation :

First we have to calculate the volume of air.

$Volume=Length\times Breadth\times Height$

$Volume=4m\times 5m\times 6m$

$Volume=120m^3=120000L$      $(1m^3=1000L)$

Now we have to calculate the mole of air.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of air = 100 kPa =  0.987 atm      (1 atm = 101.3 kPa)

V = Volume of air = 120000 L

n = number of moles of air = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of air = $25^oC=273+25=298K$

Putting values in above equation, we get:

$0.987atm\times 120000L=n\times (0.0821L.atm/mol.K)\times 298K$

$n=4841.04mol$

Now we have to calculate the mass of air.

$\text{Mass of air}=\text{Moles of air}\times \text{Molar mass of air}$

As we know that the molar mass of air is, 28.96 g/mol

$\text{Mass of air}=4841.04mol\times 28.96g/mol=140196.5184g=140.19kg$

Now we have to calculate the density of air.

$\text{Density of air}=\frac{\text{Mass of air}}{\text{Volume of air}}$

$\text{Density of air}=\frac{140.19kg}{120000L}=1.16825\times 10^{-3}kg/L=1.16825g/L$

Now we have to calculate the specific gravity of air.

$\text{Specific gravity of air}=\frac{\text{Air density at given condition}}{\text{Air density at STP}}$

As we know that air density at STP is, 1.2754 g/L

$\text{Specific gravity of air}=\frac{1.16825g/L}{1.2754g/L}=0.916$