Physics · Kinetic Theory · NEET
A diatomic molecule (dumbbell shape) can move in 3 directions AND spin about 2 axes. That is 5 degrees of freedom. By the law of equipartition, each degree of freedom stores (1/2)RT of energy per mole, so internal energy U = (5/2)RT. Since Cv = dU/dT, we get Cv = (5/2)R. A monatomic gas has only 3 (translational), giving Cv = (3/2)R.
Use Mayer's relation: Cp = Cv + R. So Cp = (5/2)R + R = (7/2)R. Then gamma = Cp/Cv = (7/2R)/(5/2R) = 7/5 = 1.4. A quick shortcut: gamma = 1 + 2/f. For f = 5, gamma = 1 + 2/5 = 7/5.
No. The two atoms lie on the bond axis, so their moment of inertia about that axis is nearly zero. Rotation about that third axis stores almost no energy. That is why we count only 2 rotational degrees of freedom, not 3, giving f = 5 for a rigid diatomic gas.
At high temperature the two atoms can vibrate. One vibrational mode adds 2 more degrees (kinetic + potential energy), so f becomes 7. Then Cv = (7/2)R, Cp = (9/2)R and gamma = 9/7. NCERT uses f = 5 for O2, N2, H2 at ordinary temperature unless the question clearly mentions vibration.
Yes. Molar specific heats (Cv, Cp in J per mole per K) depend only on degrees of freedom, not on the mass of the molecule. So O2, N2 and H2 all have Cv = (5/2)R and Cp = (7/2)R. Only the specific heat per unit mass (per kg) differs, because their molar masses differ.
Match Column-I with Column-II. (A) RMS speed of gas molecules (B) Pressure exerted by ideal gas (C) Average kinetic energy of a molecule (D) Total internal energy of 1 mole of a diatomic gas. Column-II: (i) 1/3 n m v_rms^2 (ii) sqrt(3RT/M) (iii) 5/2 RT (iv) 3/2 kB T
A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
The values of Cp/Cv for hydrogen, helium and another ideal diatomic gas X (whose molecules are not rigid but have an additional vibrational mode) are respectively equal to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a rigid diatomic gas like O2, N2 or H2, the molar specific heat at constant volume is Cv = (5/2)R, which is about 20.8 J per mole per K.
For rigid diatomic gases at ordinary temperature, gamma = Cp/Cv = 7/5 = 1.4. All three (O2, N2, H2) have the same gamma because it depends only on degrees of freedom.
It has 3 translational (motion along x, y, z) and 2 rotational degrees. Rotation about the bond axis is ignored because its moment of inertia is almost zero. Total = 5.
Using Mayer's relation Cp = Cv + R, we get Cp = (5/2)R + R = (7/2)R, which is about 29.1 J per mole per K.
No. Their molar Cv and Cp are the same because these depend only on degrees of freedom (f = 5), not on molecular mass. Only their per-kg specific heats differ.
A monatomic gas has f = 3, so Cv = 3/2 R and gamma = 5/3. A diatomic gas has f = 5, so Cv = 5/2 R and gamma = 7/5. The extra 2 comes from rotation.
