Kinetic Theory

JEE Main level, single correct, medium

132 of 168

Question 132 of 168

The average kinetic energy of a monatomic molecule is 0.414eV0.414eV at temperature:
(Use KB=1.38×10−23Jmol−1K−1{K}_{B}=1.38\times {10}^{-23}J{mol}^{-1}{K}^{-1})

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Answer: (B) 3200K3200K

For monatomic molecule, degree of freedom =3=3.
The formula for the average kinetic energy is given by
Kavg=12fKBT...(1){K}_{avg}=\dfrac{1}{2}f{K}_{B}T...\left(1\right)
From equation (1), it follows that
0.414eV×1.6×10−19J1eV=32×1.38×10−23Jmol−1K−1×T⇒T=0.414eV×1.6×10−19J1eV×23×1.38×10−23Jmol−1K−1=3200K0.414eV\times \dfrac{1.6\times {10}^{-19}J}{1eV}=\dfrac{3}{2}\times 1.38\times {10}^{-23}J{mol}^{-1}{K}^{-1}\times T \Rightarrow T=\dfrac{0.414eV\times \dfrac{1.6\times {10}^{-19}J}{1eV}\times 2}{3\times 1.38\times {10}^{-23}J{mol}^{-1}{K}^{-1}} =3200K

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