Q. Find out the number of ideal gas molecules per $cm^{3}$ that exist at a pressure of $10^{- 11}mm$ of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at $27^{o}C$ .
NTA AbhyasNTA Abhyas 2020
Solution:
Using ideal gas equation in terms of number of molecules :
$PV=NkT$
where $N$ = Number of molecules
$k=$ boltzman constant
Here $P=\left(10\right)^{- 11}mmofHg,T=\left(\right.27+273\left.\right)K$
Putting these values in above equation we get:
$\frac{N}{V}=\frac{P}{k T}$ $\Rightarrow \frac{N}{V}=\frac{10^{- 11} \times 10^{- 3} \times 13 . 6 \times 10^{3} \times 9 . 8}{1 . 38 \times 10^{- 23} \times 300}=3.22\times 10^{5}$
$PV=NkT$
where $N$ = Number of molecules
$k=$ boltzman constant
Here $P=\left(10\right)^{- 11}mmofHg,T=\left(\right.27+273\left.\right)K$
Putting these values in above equation we get:
$\frac{N}{V}=\frac{P}{k T}$ $\Rightarrow \frac{N}{V}=\frac{10^{- 11} \times 10^{- 3} \times 13 . 6 \times 10^{3} \times 9 . 8}{1 . 38 \times 10^{- 23} \times 300}=3.22\times 10^{5}$