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Heat and Thermodynamics question

2020 · 3 Sep · Shift 1 · Q48
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Heat and Thermodynamics question

2020 · 3 Sep · Shift 1 · Q48

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
JEE Main 2020 (Online) 3rd September Morning Slot Physics - Heat and Thermodynamics Question 297 English Consider a gas of triatomic molecules. The molecules are assumed to be triangular and made of massless rigid rods whose vertices are occupied by atoms. The internal energy of a mole of the gas at temperature T is :
  1. A
    32RT{3 \over 2}RT23​RT
  2. B
    92RT{9 \over 2}RT29​RT
  3. C
    52RT{5 \over 2}RT25​RT
  4. D
    3RT
View written solutionFree

Correct answer: D

  1. Use equipartition of energy

For an ideal gas, each independent quadratic degree of freedom contributes 12RT\frac{1}{2}RT21​RT per mole to the internal energy.

So, if a molecule has total degrees of freedom fff, then U=f2RT.U = \frac{f}{2}RT.U=2f​RT.

  1. Count the degrees of freedom for the given molecule

The molecule is described as:

  • triatomic,
  • triangular,
  • made of massless rigid rods,
  • hence it behaves like a rigid non-linear molecule.

A rigid non-linear molecule has:

  • 333 translational degrees of freedom,
  • 333 rotational degrees of freedom.

Thus, f=3+3=6.f = 3+3 = 6.f=3+3=6.

  1. Compute internal energy

Therefore, U=62RT=3RT.U = \frac{6}{2}RT = 3RT.U=26​RT=3RT.

  1. Check options
  • A: 32RT\frac{3}{2}RT23​RT — incorrect
  • B: 92RT\frac{9}{2}RT29​RT — incorrect
  • C: 52RT\frac{5}{2}RT25​RT — incorrect
  • D: 3RT3RT3RT — correct

Hence, the internal energy of one mole of this gas is 3RT.\boxed{3RT}.3RT​.

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