Question

Number of molecules in a volume of 4 cm3 of a perfect monatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to?
(Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2, density of mercury = 13.6 g/cm3)
1) $4.0 \times {10^{18}}$
2) $4.0 \times {10^{6}}$
3) $5.8 \times {10^{16}}$
4) $5.8 \times {10^{18}}$

Answer 1

Hint: Here, the pressure is in cm of hg, so we need to do unit conversion. The question is related to No.of molecules,volume,pressure, temperature which gives us a hint to use Ideal Gas Equation. Also, mean kinetic energy is given so we will be using kinetic energy formula.

Complete answer:
The number of independent variables or parameters that make up a system is referred to as the degree of freedom. The term "freedom" is used in several scientific disciplines to refer to the boundaries of what is physically achievable in terms of mobility or other physical processes.

For monoatomic gases, degree of freedom(f)= 3.
\[\therefore \]Mean Kinetic Energy(E)= \[
  \dfrac{f}{2}kT \\
    \\
\] (where k=Boltzman constant=\[\dfrac{R}{{{N_a}}}\])
\[\therefore \] \[E = \dfrac{3}{2}\dfrac{R}{{{N_a}}}T\]

By Ideal Gas Equation, \[PV = nRT\]
\[\therefore \]\[\dfrac{{PV}}{{RT}} = n = \dfrac{N}{{{N_a}}}\] (Here, N = No.of molecules and \[{N_a}\]=\[6.022 \times {10^{24}}\])
\[\therefore \]\[\dfrac{{PV{N_a}}}{{RT}} = N\]

Substituting Value of RT from previous equation,
\[\therefore \]\[\dfrac{{3PV{N_a}}}{{2E{N_a}}} = N\]
\[\therefore \]\[\dfrac{{3PV}}{{2E}} = N\]

Now to convert dimension of pressure,
\[\therefore \]P = \[(13.6\dfrac{g}{{c{m^3}}})(980\dfrac{{cm}}{{{s^2}}}).2cm\]
\[\therefore \]PV = \[(13.6\dfrac{g}{{c{m^3}}})(980\dfrac{{cm}}{{{s^2}}}).2cm \times 4c{m^3}\]
\[\therefore \]PV = \[(13.6 \times 980 \times 8)\dfrac{{g.c{m^2}}}{{{s^2}}}\]
\[\therefore \]PV = \[(13.6 \times 980 \times 8)\dfrac{{kg.c{m^2}}}{{1000{s^2}}}\]
\[\therefore \]PV = \[(13.6 \times 980 \times 8)\dfrac{{kg.(\dfrac{{{m^2}}}{{{{10}^4}}})}}{{1000{s^2}}}\]
\[\therefore \]PV = \[(13.6 \times 980 \times 8)\dfrac{{kg.{m^2}}}{{{{10}^7}{s^2}}}\]
E=\[4 \times {10^{ - 14}}erg\]

We know that, (\[1erg = {10^{ - 7}}J\])
\[\therefore \]E= \[4 \times {10^{ - 21}}J\]

Substituting all these in above equation,
\[\therefore \]\[N = \dfrac{{3(13.6 \times 980 \times 8)\dfrac{{kg.{m^2}}}{{{{10}^7}{s^2}}}}}{{2 \times 4 \times {{10}^{ - 21}}J}}\]
\[\therefore \]\[N\]\[ = 3.99 \times {10^{18}}\]molecules

Ttherefore, the correct answer is \[4 \times {10^{18}}\] i.e., option A.

Note: You should be very careful during conversion of units. You should be thorough with the dimensions and conversion constants like \[1erg = {10^{ - 7}}J\]. You should remember the value of various constants like k,Na etc.