The average kinetic energy of a monatomic molecule is 0.414 eV at temperature : (Use $\mathrm{K}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}$)

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Heat & Thermodynamics

The average kinetic energy of a monatomic molecule is 0.414 eV at temperature :
(Use

K_{B} = 1.38 \times 1 0^{- 23} J / mol - K

)

(A)3000 K
(B)3200 K✓ Correct
(C)1600 K
(D)1500 K

Answer: (B)

Solution

For a monatomic gas,

K_{a v g} = \frac{3}{2} k_{B} T

. Converting

K_{a v g} = 0.414

eV to joules (

\times 1.6 \times 1 0^{- 19}

T = \frac{2 K _{a v g}}{3 k _{B}} = \frac{2 \times 0.414 \times 1.6 \times 1 0 ^{- 19}}{3 \times 1.38 \times 1 0 ^{- 23}} \approx 3200 K

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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