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MCQEasyJEE 2024Degrees of Freedom & Law of Equipartition

JEE Physics 2024 Question with Solution

N moles of a polyatomic gas (f=6f = 6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of NN is:

  • A

    66

  • B

    33

  • C

    44

  • D

    22

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Polyatomic gas has degrees of freedom f1=6f_1 = 6 and number of moles n1=Nn_1 = N. Monoatomic gas has degrees of freedom f2=3f_2 = 3 and number of moles n2=2n_2 = 2. The mixture must behave like a diatomic gas, so effective degrees of freedom fmix=5f_{\text{mix}} = 5.

Find: The value of NN.

For a gas mixture, the average degrees of freedom is

fmix=n1f1+n2f2n1+n2f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}

Substituting the given values,

5=N⋅6+2⋅3N+25 = \frac{N \cdot 6 + 2 \cdot 3}{N + 2}

So,

5(N+2)=6N+65(N + 2) = 6N + 6

Expanding and simplifying,

5N+10=6N+65N + 10 = 6N + 6 5N+4=6N5N + 4 = 6N N=4N = 4

Therefore, the value of NN is 44. Hence, the correct option is C.

Average Degrees of Freedom Approach

Given: n1=N,f1=6,n2=2,f2=3n_1 = N, f_1 = 6, n_2 = 2, f_2 = 3 and the equivalent diatomic behavior means feq=5f_{\text{eq}} = 5.

Find: NN.

Using the average degrees of freedom relation,

feq=n1f1+n2f2n1+n2f_{\text{eq}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}

Therefore,

5=N×6+2×3N+25 = \frac{N \times 6 + 2 \times 3}{N + 2}

This gives

5=6N+6N+25 = \frac{6N + 6}{N + 2}

Multiplying both sides by N+2N + 2,

5(N+2)=6N+65(N + 2) = 6N + 6 5N+10=6N+65N + 10 = 6N + 6

Now rearrange,

10−6=6N−5N10 - 6 = 6N - 5N N=4N = 4

Thus, the required number of moles of the polyatomic gas is 44.

Common mistakes

  • Using total degrees of freedom directly instead of average degrees of freedom for the mixture is incorrect. The mixture behaves according to degrees of freedom per mole, so use the weighted average formula.

  • Taking a monoatomic gas to have f=1f = 1 or a diatomic gas to have f=2f = 2 is wrong. For this thermodynamics model, monoatomic, diatomic, and the given polyatomic gases have degrees of freedom 33, 55, and 66 respectively.

  • Making an algebraic mistake while solving 5(N+2)=6N+65(N+2)=6N+6 can lead to a wrong value. Expand both sides carefully and collect like terms before solving for NN.

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