JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
Match the C_P/C_V ratio for ideal gases with different type of molecules :
| Molecule Type | C P /C V |
|---|---|
| (A) Monatomic | (I) 7/5 |
| (B) Diatomic rigid molecules | (II) 9/7 |
| (C) Diatomic non-rigid molecules | (III) 4/3 |
| (D) Triatomic rigid molecules | (IV) 5/3 |
- (a) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- (b) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- (c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- (d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Answer and solution
Answer (c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Solution
Physics sir · 2 steps
Count the degrees of freedom f for each molecule, then γ = 1 + 2/f does all the matching.
- Gamma from f. By equipartition each degree of freedom carries 1/2RT per mole, so C_V = f/2R. For an ideal gas C_P = C_V + R.
γ = C_P/C_V = (f/2R + R)/(f/2R) = 1 + 2/f
Molecule Degrees of freedom f γ = 1 + 2/f (A) Monatomic 3 translational, so f = 3 5/3, that is (IV) (B) Diatomic rigid 3 translational + 2 rotational, so f = 5 7/5, that is (I) (C) Diatomic non-rigid 5 + 2 vibrational (kinetic and potential), so f = 7 9/7, that is (II) (D) Triatomic rigid (non-linear) 3 translational + 3 rotational, so f = 6 8/6 = 4/3, that is (III) - So. The matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III), which is option (c).
The idea to keep. More degrees of freedom means γ moves closer to 1. Remember that one vibrational mode adds 2 to f, one for kinetic and one for potential energy.
Answer from the NTA final answer key. Solution by PrepWiser.