Question
In the text, it was shown that $\dfrac{N}{V} = 2.68 \times 10^{25} \textrm{ m}^{-3}$ for gas at STP. (a) Show that this quantity is equivalent to $\dfrac{N}{V} = 2.68 \times 10^{19} \textrm{ cm}^{-3}$, as stated. (b) About how many atoms are there in one $\mu\textrm{m}^3$ (a cubic micrometer) at STP? (c) What does your answer to part (b) imply about the separation of atoms and molecules?
1. $2.68 \times 10^{19} \textrm{ /cm}^3$
2. $2.68 \times 10^7 \textrm{ atoms}$
3. $3.34 \textrm{ nm}$
Solution Video