Question

The lowest pressure (the best Vacuum) that can be created in the laboratory at $27$ degree is ${10^{ - 11}}mm$ of Hg. At this pressure, the number of ideal gas molecules per $c{m^3}$ will be
A) $\left( {\text{A}} \right){\text{ }}3.22 \times {10^{12}}$
B) $\left( {\text{B}} \right){\text{ }}1.61 \times {10^{12}}$
C) $\left( {\text{C}} \right){\text{ }}3.21 \times {10^6}$
D) $\left( {\text{D}} \right){\text{ }}3.22 \times {10^5}$

Answer 1

-Use the ideal gas equation in terms of pressure, volume, temperature, molecule number for two cases.
-The first case should be taken according to STP and for the second one use the given information about the Ideal gas.
-We have to find the number of molecules per $c{m^3}$ that means the number of molecules should be divided by the volume.

Formula used:
The ideal gas equation,
$PV = nRT$
Where $P$= the pressure of the gas,
$V$= the volume of the gas,
$n$= the number of gas molecules,
$R$= Universal gas constant,
$T$= the temperature in kelvin.
$\dfrac{{PV}}{{nT}} = R$
$\Rightarrow PV \propto nT$
$\Rightarrow \dfrac{{{P_1}{V_1}}}{{{n_1}{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{n_2}{T_2}}}$ .

Complete step by step answer:
The ideal-gas equation states that, for the $n$ number of gas molecules of pressure $P$, volume $V$, and temperature $T$,
$PV = nRT$
$\Rightarrow \dfrac{{PV}}{{nT}} = R$
Since,$R$= Universal gas constant.
$PV \propto nT$
So we can write the equation for two cases,
$\Rightarrow \dfrac{{{P_1}{V_1}}}{{{n_1}{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{n_2}{T_2}}}$
For this problem, in the laboratory at the lowest pressure that can be created at ${27^0}c$ is ${10^{ - 11}}mm$of Hg. And we have to calculate the number of molecules per unit volume of the ideal gas at this pressure.
So, we take the first case for the ideal gas in the STP condition.
Hence, ${P_1} = 760mm$ $of$ $Hg$
${V_1} = 22400c{m^3}$
${T_1} = 273K$
${n_1} = 6.023 \times {10^{23}}$
And for the second case,
Given, ${P_2} = {10^{ - 11}}mm$ $of$ $Hg$
${T_1} = (27 + 273)K = 300K$
Now,
$\dfrac{{{P_1}{V_1}}}{{{n_1}{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{n_2}{T_2}}}$
$\Rightarrow \dfrac{{{n_2}}}{{{V_2}}} = \dfrac{{{P_2}{n_1}{T_1}}}{{{P_1}{V_1}{T_2}}}$
$\Rightarrow \dfrac{{{n_2}}}{{{V_2}}} = \dfrac{{{{10}^{ - 11}} \times 6.023 \times {{10}^{23}} \times 273}}{{760 \times 22400 \times 300}}$
$\Rightarrow \dfrac{{{n_2}}}{{{V_2}}} = 3.22 \times {10^5}$
So, the number of gas molecules per unit volume is $3.22 \times {10^5}$.
Hence, the right answer is in option $\left( D \right)$

Note: The ideal gas law formula $PV = nRT$ is the combination of Boyle’s law, Charl’s Law, Avogadro’s law, and Gay-Lussac’s law.
For the $n$ number of gas molecules of pressure $P$, volume $V$, and temperature $T$, the mathematical representations of the laws are,
Boyle’s Law $\Rightarrow P \propto \dfrac{1}{V}$
Charl’s Law $\Rightarrow V \propto T$
Avogadro’s law $\Rightarrow V\propto n$
Gay-Lussac’s law$\Rightarrow P \propto T$
The combined form of all these laws is, $PV \propto nT$
$\Rightarrow PV = nRT$, $R$ is the Universal gas-constant.