Question

In outer space there are 10 molecules per $c{m^3}$ on an average and the temperature there is $3K$ . The average pressure of the light gas is
$A.\,\,{10^5}\,N{m^{ - 2}} \\\ B.\,\,5 \times {10^{ - 14}}\,N{m^{ - 2}} \\\ C.\,\,0.4 \times {10^{ - 16}}\,N{m^{ - 2}} \\\ D.\,\,4.14 \times {10^{ - 16}}\,N{m^{ - 2}} \\\$

Answer 1

Hint
In the question, the number of molecules and temperature of the gas is given. By using the equation of absolute pressure of the ideal gas, we find the value of the average pressure of the gas has been calculated.
Formula Used:
The expression for finding the absolute pressure of the light gas is
$P = \dfrac{{nKT}}{V}$
Where, $P$ be the absolute pressure of the gas, $n$ be the number of atoms and molecules in the gas, $T$ be the absolute temperature, $V$ be the volume it occupies and $K$ be the Boltzmann’s constant.

Complete step by step answer
Given that,
Number of molecules $n = 10\,molecules/c{m^3}$
Temperature of the gas $\,T = 3K$ (In terms of Kelvin).
In the equation of ideal gas formula, the value of $K$ is taken from Boltzmann's constant. Because It relates to the temperature and the energy of the gas.
So that, $K = 1.38 \times {10^{ - 23}}$
$P = \dfrac{{nKT}}{V}........\left( 1 \right)$
Substitute all the known values in the equation $\left( 1 \right)$
$P = \dfrac{{10 \times \left( {1.38 \times {{10}^{ - 23}}} \right) \times 3}}{{{{10}^{ - 6}}}}$
Simplify the above equation we get,
$P = 4.14 \times {10^{ - 16}}\,N{m^{ - 2}}.$
Therefore, the value of the average pressure of the gas is $4.14 \times {10^{ - 16}}$ .
Hence, from the above options, option (D) is correct.

Note
In the question, we have that the dimension of the energy per degree of the temperature is given. Boltzmann’s constant relates the energy and temperature of the gas. So that we find the value of average pressure by substituting the value of Boltzmann’s constant. The value of Boltzmann’s constant is obtained by dividing the gas constant by the Avogadro number.