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Heat and Thermodynamics question

2023 · 25 Jan · Shift 2 · Q61
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  5. /2023 · 25 Jan · Shift 2 · Q61

Heat and Thermodynamics question

2023 · 25 Jan · Shift 2 · Q61

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-
  1. A
    92R\frac{9}{2}R29​R
  2. B
    52R\frac{5}{2}R25​R
  3. C
    32R\frac{3}{2}R23​R
  4. D
    72R\frac{7}{2}R27​R
View written solutionFree

Correct answer: D

  1. Degrees of freedom of a diatomic molecule

For a diatomic gas:

  • Translational degrees of freedom =3= 3=3
  • Rotational degrees of freedom =2= 2=2 (rotation about two axes perpendicular to the bond axis)

So without vibration, total active degrees of freedom are: f=3+2=5f = 3+2 = 5f=3+2=5

  1. Effect of one vibrational mode

A single vibrational mode contributes:

  • 111 kinetic term
  • 111 potential term

Hence, one vibrational mode contributes 2 degrees of freedom in equipartition.

Therefore total degrees of freedom become: f=3+2+2=7f = 3+2+2 = 7f=3+2+2=7

  1. Molar specific heat at constant volume

According to equipartition theorem, for an ideal gas: CV=f2RC_V = \frac{f}{2}RCV​=2f​R

Substituting f=7f=7f=7: CV=72RC_V = \frac{7}{2}RCV​=27​R

  1. Matching with options
  • A: 92R\frac{9}{2}R29​R ✗
  • B: 52R\frac{5}{2}R25​R ✗
  • C: 32R\frac{3}{2}R23​R ✗
  • D: 72R\frac{7}{2}R27​R ✓

Hence, the correct option is D.

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