JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Heat & Thermodynamics

The root mean square speed of molecules of nitrogen gas at

2 7^{\circ} C

is approximately:
(Given mass of a nitrogen molecule

= 4.6 \times 1 0^{- 26} kg

and take Boltzmann constant

k_{B} = 1.4 \times 1 0^{- 23} JK^{- 1}

)

(A) $523 m / s$ ✓ Correct
(B) $1260 m / s$
(C) $91 m / s$
(D) $27.4 m / s$

Answer: (A)

Solution

The root mean square speed is given by

v_{rms} = \frac{3 k _{B} T}{m}

. Converting temperature:

T = 27 + 273 = 300

K. Substituting values:

v_{rms} = \frac{3 \times 1.4 \times 1 0 ^{- 23} \times 300}{4.6 \times 1 0 ^{- 26}} = \frac{1.26 \times 1 0 ^{- 20}}{4.6 \times 1 0 ^{- 26}} = 2.74 \times 1 0^{5} \approx 523

m/s.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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