$\begin{aligned}
& \text { Degree of freedom }(f)=5+2(3 \mathrm{N}-5) \
& \qquad f=5+2(3 \times 2-1)=7 \
& \text { energy of one molecule }=\frac{f}{2} K_B T
\end{aligned}energyof10molecules=10\left(\frac{\mathrm{f}}{2} \mathrm{K}{\mathrm{B}} \mathrm{T}\right)=10\left(\frac{7}{2} \mathrm{~K}{\mathrm{B}} \mathrm{T}\right)=35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$
Energy of 10 non rigid diatomic molecules at temperature T is :
JEE Main 2024 — Physics Thermodynamics
Energy of 10 non rigid diatomic molecules at temperature T is :
Held on 6 Apr 2024 · Verified 6 Jul 2026.
Options
- A
70 KBT
- B
35 KBT
- C
27RT
- D
35RT
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