Determine the mass of air in a room with dimensions 4 m × 5 m × 6 m at p = 1 atm and T = 25°C

Answer

The room volume is $V = 4\times5\times6 = 120\,\text{m}^3$. Treating air as an ideal gas, the mass is $m = \dfrac{pV}{RT} = \dfrac{(101325\,\text{Pa})(120\,\text{m}^3)}{(287\,\text{J/(kg·K)})(298.15\,\text{K})} \approx 1.42\times10^2\,\text{kg}$. So the mass of air in the room is about $142\,\text{kg}$.

Explanation

What we are trying to find

You are given the room size (so you can get the volume) and the air conditions ($p$ and $T$). With those, you can find the mass of air using the ideal-gas relation in a form that uses the specific gas constant for air.

Step 1: Compute the room volume

$$ V = (4\,\text{m})(5\,\text{m})(6\,\text{m}) = 120\,\text{m}^3 $$

Step 2: Convert pressure and temperature to SI units

Use absolute temperature in kelvin and pressure in pascals:

$p = 1\,\text{atm} = 101325\,\text{Pa}$
$T = 25^\circ\text{C} = 25 + 273.15 = 298.15\,\text{K}$

Step 3: Use the ideal gas law to solve for mass

For an ideal gas, $$ pV = mRT $$ so $$ m = \frac{pV}{RT} $$ For (dry) air, $R \approx 287\,\text{J/(kg·K)}$.

Substitute: $$ m = \frac{(101325)(120)}{(287)(298.15)} \approx 142\,\text{kg} $$

Quick reasonableness check

A typical air density near room conditions is about $1.18\,\text{kg/m}^3$. Then $m \approx \rho V \approx (1.18)(120) \approx 142\,\text{kg}$, which matches.