Kinetic Theory

JEE Main level, single correct, medium

128 of 168

Question 128 of 168

Energy of 10 non rigid diatomic molecules at temperature TT is :

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Answer: (B) 35 KBT35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}

$\begin{aligned}
& \text { Degree of freedom }(f)=5+2(3 \mathrm{~N}-5) \\
& \qquad f=5+2(3 \times 2-1)=7 \\
& \text { energy of one molecule }=\frac{f}{2} K_B T
\end{aligned}$
energy of 10 molecules
=10(f2 KBT)=10(72 KBT)=35 KBT=10\left(\frac{\mathrm{f}}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=10\left(\frac{7}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}

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