JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Heat & Thermodynamics

Number of molecules in a volume of

4 cm^{3}

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 1 0^{- 14} erg, g = 980 cm / s^{2}

, density of mercury

= 13.6 g / cm^{3}

)

(A) $5.8 \times 1 0^{18}$
(B) $4.0 \times 1 0^{16}$
(C) $5.8 \times 1 0^{16}$
(D) $4.0 \times 1 0^{18}$ ✓ Correct

Answer: (D)

Solution

First convert the pressure to CGS units:

P = h ρ g = 2 \times 13.6 \times 980 \approx 2.67 \times 1 0^{4}

dyn/cm

^{2}

. Kinetic theory gives

P = \frac{2}{3} \frac{N}{V} \overset{ˉ}{E}

, where

\overset{ˉ}{E} = 4 \times 1 0^{- 14}

erg is the mean kinetic energy of one molecule. Hence

N = \frac{3 P V}{2 E ˉ} = \frac{3 \times 2.67 \times 1 0 ^{4} \times 4}{2 \times 4 \times 1 0 ^{- 14}} \approx 4.0 \times 1 0^{18}

molecules.

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Heat & Thermodynamics

Solution

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