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Physics · Kinetic Theory of Gases

JEE Main 2024 — 6 April, Shift 2 — Question 41

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

  1. Option A:

    72RT\frac{7}{2} \mathrm{RT}

  2. Option B:

    70 KBT70 \mathrm{~K}_{\mathrm{B}} \mathrm{T}

  3. Option C:

    35 RT

  4. Option D:

    35 KBT35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}

    Correct

Answer: D

Step-by-step solution

Degree of freedom (f)=5+2(3 N−5)(\mathrm{f})=5+2(3 \mathrm{~N}-5)

f=5+2(3×2−1)=7\mathrm{f}=5+2(3 \times 2-1)=7 e

nergy of one molecule =f2 KBT=\frac{f}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}

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}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 2
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
Energy of 10 non rigid diatomic molecules at temperature T is : | JEE Main 2024 PYQ with Solution · DhiX AI