Medium MCQ +4 / -1 PYQ · JEE Mains 2020

Number of molecules in a volume of 4 cm3 of a perfect monoatomic 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 $\times$ 10–14 erg, g = 980 cm/s2, density of
mercury = 13.6 g/cm3)

  1. A 5.8 $\times$ 10<sup>18</sup>
  2. B 4.0 $\times$ 10<sup>16</sup>
  3. C 5.8 $\times$ 10<sup>16</sup>
  4. D 4.0 $\times$ 10<sup>18</sup> Correct answer

Solution

$E = {3 \over 2}kT \Rightarrow \left( {T = {{2E} \over {3k}}} \right),$ <br><br>Also $\,PV = NkT$<br><br>$P = \rho gh,\,V = 4c{m^3}$ <br><br>$\therefore$ ($\rho gh$)V = $Nk \times {{2E} \over {3k}}$ <br><br>$\Rightarrow$ $$13.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}$$<br><br>$$ = Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}$$<br><br>$\Rightarrow$ $$N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}$$<br><br>$\Rightarrow$ $N = 399.84 \times {10^{16}}$<br><br>$= 3.99 \times {10^{18}}$<br><br>$\Rightarrow$ $N = 4 \times {10^{18}}$

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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