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Heat and Thermodynamics question

2021 · 17 Mar · Shift 2 · Q49
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Heat and Thermodynamics question

2021 · 17 Mar · Shift 2 · Q49

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
If one mole of the polyatomic gas is having two vibrational modes and β\betaβ is the ratio of molar specific heats for polyatomic gas (β=CPCV)\left( {\beta = {{{C_P}} \over {{C_V}}}} \right)(β=CV​CP​​) then the value of β\betaβ is :
  1. A
    1.02
  2. B
    1.35
  3. C
    1.2
  4. D
    1.25
View written solutionFree

Correct answer: C

  1. Degrees of freedom for a polyatomic gas

For one mole of an ideal gas, CV=f2R,C_V = \frac{f}{2}R,CV​=2f​R, where fff is the total number of active degrees of freedom.

For a non-linear polyatomic gas:

  • Translational degrees of freedom =3=3=3
  • Rotational degrees of freedom =3=3=3
  • Each vibrational mode contributes 222 degrees of freedom

Given two vibrational modes, vibrational contribution is 2×2=4.2 \times 2 = 4.2×2=4.

Hence total degrees of freedom: f=3+3+4=10.f = 3 + 3 + 4 = 10.f=3+3+4=10.

  1. Calculate CVC_VCV​ and CPC_PCP​

Thus, CV=102R=5R.C_V = \frac{10}{2}R = 5R.CV​=210​R=5R.

Also, CP=CV+R=5R+R=6R.C_P = C_V + R = 5R + R = 6R.CP​=CV​+R=5R+R=6R.

  1. Find the ratio β\betaβ

Given β=CPCV=6R5R=65=1.2.\beta = \frac{C_P}{C_V} = \frac{6R}{5R} = \frac{6}{5} = 1.2.β=CV​CP​​=5R6R​=56​=1.2.

  1. Match with options

β=1.2\beta = 1.2β=1.2 So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

This matches the derived answer.

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