Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^{3} at a temperature of 27 °C and 1 atm pressure

#### Solution 1

Volume of the room, *V* = 25.0 m^{3}

Temperature of the room, *T* = 27°C = 300 K

Pressure in the room, *P* = 1 atm = 1 × 1.013 × 10^{5} Pa

The ideal gas equation relating pressure (*P*), Volume (*V*), and absolute temperature (*T*) can be written as:

*PV *= *k*_{B}*NT*

Where,

*K*_{B} is Boltzmann constant = 1.38 × 10^{–23} m^{2} kg s^{–2} K^{–1}

*N* is the number of air molecules in the room

`:. N = (PV)/(k_BT)`

`= (1.013xx10^5xx25)/(1.38xx10^(-23)xx300) = 6.11 xx 10^(26)` molecules

Therefore, the total number of air molecules in the given room is 6.11 × 10^{26}.

#### Solution 2

Here, Volume of room, `V= 25.0 m^3`, temperature, `T = 27 ^@C = 300 K` and

Pressure, `P = 1 "atm" = 1.01 xx 10^5 Pa`

According to gas equation, `PV = muRT = muN_A.k_BT`

Hence, total number of air molecules in the volume of given gas

`N = mu.N_A = PV/K_BT`

`:. N = (1.01 xx 10^5 xx25.0)/((1.38 xx 10^(-23))xx300) = 6.1 xx 10^(26)`