# How many air molecules are in a room whose dimensions are 13.5*ftxx12.0*ftxx10.0*ft at a temperature of 20 ""^@C?

Dec 17, 2016

There are approx. $1.14 \times {10}^{27}$ $\text{air molecules}$.

$\text{Pretty vast but not infinite!}$

#### Explanation:

So we need $\text{(i) the volume of the room..........}$

And $\text{(ii) the number of air molecules..........}$.

$\text{(i) Volume}$ $=$ $13.5 \times 12.0 \times 10.0 \cdot f {t}^{3} = 1620 \cdot f {t}^{3}$. (I never thought I would use $\text{feet and inches}$; I'll be bidding $\text{guineas}$ at an auction next!)

$\text{Volume}$ $=$ $1620 \cdot f {t}^{3} \times 28.2 \cdot L \cdot f {t}^{-} 3 \times {10}^{-} 3 \cdot {m}^{3} \cdot {L}^{-} 1 = 45.68 \cdot {m}^{3}$.

And now we need the molar volume at $20$ ""^@C, i.e. $293 \cdot K$

$V = \frac{n R T}{P}$

$= \frac{1 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 293 \cdot K}{1 \cdot a t m} =$

$= 24.05 \cdot L$

And so if we divide the $\text{volume}$ by the $\text{molar volume}$, and then multiply this number by the $\text{Avocado number}$ (yes, I know but I am partial to avocadoes), we should reasonably get the number of air molecules in the room. It is a fact that ${10}^{3} L \equiv 1 \cdot {m}^{3}$.

So $\frac{45.68 \cdot {m}^{3} \times {10}^{3} \cdot L \cdot {m}^{-} 3}{24.05 \cdot L \cdot m o {l}^{-} 1} \cong 1900 \cdot m o l$

And thus $\left(i i\right)$ there are $1900 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 1.14 \times {10}^{27}$ $\text{air molecules}$

$\text{all care taken but no responsibility admitted}$.