Question

The pressure of a gas in a container is $10^{-11}$ Pascal at $27^{\circ}C$. The number of molecules per unit volume of the vessel will be.
\[\begin{align}
  & A.6\times {{10}^{23}}c{{m}^{-3}} \\
 & B.2.68\times {{10}^{19}}c{{m}^{-3}} \\
 & C.2.5\times {{10}^{6}}c{{m}^{-3}} \\
 & D.2400c{{m}^{-3}} \\
\end{align}\]

Answer 1

Hint: Ideal gas law or the general gas equation is the combination of Boyles Law, Charles’s law, Avogadro’s law and Gay Lussac’s law. It gives the relationship between the $P$ pressure, $V$ volume, $T$ temperature and the $n$ numbered moles of an ideal gas.

Formula Used:
Ideal gas law: $PV=nRT$ where $R$ is the gas constant.

Complete step-by-step answer:
Let us consider that the given gas is ideal in nature in some vessel. Then it is given that at $27^{\circ}C$, or $T=273+27=300K$, the gas experiences a pressure $P=10^{-11}$.
Let us consider the volume of the gas present in the vessel to be $V$ and the number of moles present in the gas to be $n$. then from the ideal gas law, we can say that $PV=nRT$
Then we can write, $\dfrac{n}{V}=\dfrac{P}{RT}$ given the number of moles per volume of the gas.
But we need, number of molecules per volume.
Then we also know that we can rewrite the equation as $PV=nN_{A}kT$, where $N_{A}$ is the Avogadro number and $k$ is the Boltzmann constant. Also $nN_{A}$ gives the number of molecules here.
Then we get, $\dfrac{nN_{A}}{V}=\dfrac{P}{kT}$ gives the number of molecules per volume, if $V=1$
Substituting the values, we get, $\dfrac{nN_{A}}{V}=\dfrac{10^{-11}}{1.38\times 10^{-23}\times 300}=2400cm^{-3}$
Thus we get the number of molecules per unit volume of the vessel to be $2400cm^{-3}$
Hence the correct answer is option \[D.2400c{{m}^{-3}}\]

Note: However, the ideal gas law doesn’t give any information of the nature of reaction, i.e. when the gas is expanding or compressing does it absorb heat or release heat. Also such gases don't exist in the real world they are hypothetical in nature.