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

Concept: Molecular Volume Fraction — at STP-like conditions, every gas at the same $T$ and $P$ contains the same number of molecules per unit volume, so we only need the total number of molecules, not the composition.

Step 1: Use the ideal gas law to find the number of moles $n$ in the room.

$PV = nRT$

$P = 1\ \text{atm} = 1.013 \times 10^5\ \text{Pa}$, $V = 25.0\ \text{m}^3$, $T = 27^\circ\text{C} = 300\ \text{K}$, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

Step 2:

$n = \frac{PV}{RT} = \frac{(1.013 \times 10^5)(25.0)}{(8.314)(300)}$

$n \approx \frac{2.5325 \times 10^6}{2494.2} \approx 1015\ \text{mol}$. …

The key insight here is that all air molecules — oxygen, nitrogen, water vapour, argon, everything — behave essentially the same way under the ideal gas law. You don't need to know the composition. At a given temperature and pressure, every gas molecule contributes equally to the pressure, and the total number of molecules depends only on the volume, temperature, and pressure. This is Avogadro's principle in action: equal volumes of gas at the same temperature and pressure contain equal numbers of molecules.

So the problem reduces to: how many molecules are in $25.0\ \text{m}^3$ of an ideal gas at $27\ ^\circ\text{C}$ and $1\ \text{atm}$?

Let's work it through.

1. Convert everything to consistent SI units.

   The ideal gas constant $R$ is most conveniently used as $8.314\ \text{J mol}^{-1}\text{K}^{-1}$ when volume is in $\text{m}^3$ and pressure in $\text{Pa}$.

   • Pressure: $1\ \text{atm} = 1.013 \times 10^5\ \text{Pa}$.
   • Temperature: $T = 27\ ^\circ\text{C} = 27 + 273 = 300\ \text{K}$.
   • Volume: $V = 25.0\ \text{m}^3$ (already given).

2. Use the ideal gas law to find the number of moles $n$.

Substitute:

Compute step by step:

• Numerator: $1.013 \times 10^5 \times 25.0 = 2.5325 \times 10^6\ \text{Pa m}^3$ (and $1\ \text{Pa m}^3 = 1\ \text{J}$).
• Denominator: $8.314 \times 300 = 2494.2\ \text{J mol}^{-1}$.
• So $n = \frac{2.5325 \times 10^6}{2494.2} \approx 1015.5\ \text{mol}$. …

Shortcut via molar volume at STP. Rather than solving $PV = nRT$ from scratch, recall that 1 mole of any ideal gas occupies $22.4\ \text{L}$ at STP ($0^\circ\text{C}$, $1\ \text{atm}$). Scale this reference to the room's actual temperature using $V \propto T$ at constant pressure: at $300\ \text{K}$, 1 mole occupies $22.4 \times \frac{300}{273} \approx 24.6\ \text{L}$. The room's $25{,}000\ \text{L}$ then holds $n \approx 25000/24.6 \approx 101 …