Verified Solution • 2025–26 Session moderate

QJEEM20Ph9678: JEE Main 2020 Physics - Class 11 JEEM20Ph9678 Solution: Number of molecules in a volume of 4 cm 3 of a perfect…

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)

Multiple Choice Options

  • A 5.8 ×\times 1018
  • B 4.0 ×\times 1016
  • C 5.8 ×\times 1016
  • D 4.0 ×\times 1018

Step-by-Step Verified Solution

E=32kT⇒(T=2E3k),E = {3 \over 2}kT \Rightarrow \left( {T = {{2E} \over {3k}}} \right),

Also  PV=NkT\,PV = NkT

P=ρgh, V=4cm3P = \rho gh,\,V = 4c{m^3}

∴\therefore ( ρgh\rho gh )V = Nk×2E3kNk \times {{2E} \over {3k}}

⇒\Rightarrow 13.6×103×9.8×2×10−2×4×10−613.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}

=Nk×2E3k=N×23×4×10−14×10−7= Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}

⇒\Rightarrow N=13.6×19.6×4×10−5×3×108N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}

⇒\Rightarrow N=399.84×1016N = 399.84 \times {10^{16}}

=3.99×1018= 3.99 \times {10^{18}}

⇒\Rightarrow N=4×1018N = 4 \times {10^{18}}

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