Physics Circle home Student Login

Home›Chapters›Thermodynamics PYQs›Degrees of Freedom & Internal Energy›Question 10

JEE Main 2024 (6 Apr, Shift 2) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2024 (6 Apr, Shift 2)

Q10JEE Main 2024 · 6 APR · SHIFT 2Medium

Degrees of Freedom & Internal Energy

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

Show answer and solution

Answer: (d)

  1. Non-rigid diatomic: f=3+2+2=7f = 3 + 2 + 2 = 7 (vibration gives 2 terms).
  2. Energy per molecule =72kBT= \frac{7}{2}k_BT; for 10 molecules E=10×72kBTE = 10 \times \frac{7}{2}k_BT.
  3. So the answer is (d) 35kBT35k_BT.
Report a mistake

More questions on Degrees of Freedom & Internal Energy