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Physics Question on The Kinetic Theory of Gases

The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use kB=1.38×10−23 J/mol-Kk_B = 1.38 \times 10^{-23} \, \text{J/mol-K})

A

3000 K

B

3200 K

C

1600 K

D

1500 K

Answer

3200 K

Explanation

Solution

For a monatomic gas, the average kinetic energy per molecule is:

KE=32kBTKE = \frac{3}{2} k_B T

Given KE=0.414 eVKE = 0.414 \, \text{eV}, convert this to joules:

0.414 eV=0.414×1.6×10−19 J=6.624×10−20 J0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J} = 6.624 \times 10^{-20} \, \text{J}

Now,

6.624×10−20=32×1.38×10−23×T6.624 \times 10^{-20} = \frac{3}{2} \times 1.38 \times 10^{-23} \times T

Solving for TT:

T=6.624×10−20(32)×1.38×10−23≈3200 KT = \frac{6.624 \times 10^{-20}}{\left(\frac{3}{2}\right) \times 1.38 \times 10^{-23}} \approx 3200 \, \text{K}