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Heat and Thermodynamics question

2024 · 27 Jan · Shift 1 · Q80
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  5. /2024 · 27 Jan · Shift 1 · Q80

Heat and Thermodynamics question

2024 · 27 Jan · Shift 1 · Q80

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The average kinetic energy of a monatomic molecule is 0.414 eV0.414 \mathrm{~eV}0.414 eV at temperature : (Use KB=1.38×10−23 J/mol−KK_B=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}KB​=1.38×10−23 J/mol−K)
  1. A
    3000 K
  2. B
    3200 K
  3. C
    1600 K
  4. D
    1500 K
View written solutionFree

Correct answer: B

  1. Use the formula for average kinetic energy of a monatomic gas molecule

For a monatomic molecule, the average kinetic energy is

⟨K⟩=32kBT\langle K \rangle = \frac{3}{2} k_B T⟨K⟩=23​kB​T

Given:

⟨K⟩=0.414 eV\langle K \rangle = 0.414\,\text{eV}⟨K⟩=0.414eV
  1. Convert electron volt to joule

We use

1 eV=1.6×10−19 J1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}1eV=1.6×10−19J

So,

0.414 eV=0.414×1.6×10−19 J0.414\,\text{eV} = 0.414 \times 1.6 \times 10^{-19}\,\text{J}0.414eV=0.414×1.6×10−19J =6.624×10−20 J= 6.624 \times 10^{-20}\,\text{J}=6.624×10−20J
  1. Substitute into the kinetic energy formula
32kBT=6.624×10−20\frac{3}{2} k_B T = 6.624 \times 10^{-20}23​kB​T=6.624×10−20

Given

kB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23}\,\text{J/K}kB​=1.38×10−23J/K

Therefore,

T=23⋅6.624×10−201.38×10−23T = \frac{2}{3} \cdot \frac{6.624 \times 10^{-20}}{1.38 \times 10^{-23}}T=32​⋅1.38×10−236.624×10−20​
  1. Calculate

First,

6.6241.38=4.8\frac{6.624}{1.38} = 4.81.386.624​=4.8

and

10−2010−23=103\frac{10^{-20}}{10^{-23}} = 10^310−2310−20​=103

So,

T=23×4.8×103T = \frac{2}{3} \times 4.8 \times 10^3T=32​×4.8×103 T=3.2×103T = 3.2 \times 10^3T=3.2×103 T=3200 KT = 3200\,\text{K}T=3200K
  1. Evaluate options
  • A: 3000 K3000\,\text{K}3000K ❌
  • B: 3200 K3200\,\text{K}3200K ✅
  • C: 1600 K1600\,\text{K}1600K ❌
  • D: 1500 K1500\,\text{K}1500K ❌

Hence, the correct option is B.

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