Heat and Thermodynamics — JEE Main Questions

1. A real gas is compressed at constant temperature. Its pressure is found to be less than that predicted by the ideal gas law. Which factor is primarily responsible for this deviation?

1. Intermolecular attractive forces
2. Finite size of gas molecules
3. High molecular speed
4. Large molecular mass

Answer: Intermolecular attractive forces

At constant temperature, compression reduces volume, bringing molecules closer. Intermolecular attractive forces become significant, pulling molecules inward and reducing the force on the container walls, so pressure is less than ideal. Finite size (option a) would cause pressure to be higher, not lower.

2. The van der Waals equation for one mole of a real gas is (P + a/V^2)(V - b) = RT. Which of the following correctly describes the constant 'b'?

1. It is the total volume occupied by one mole of gas molecules.
2. It is the volume excluded due to finite molecular size, approximately four times the molecular volume per mole.
3. It is the volume correction factor equal to the molecular volume per mole.
4. It is the volume at which intermolecular attractions become negligible.

Answer: It is the volume excluded due to finite molecular size, approximately four times the molecular volume per mole.

In the van der Waals equation, 'b' is the excluded volume per mole, accounting for the finite size of molecules. It is approximately four times the actual volume of one mole of molecules, because the excluded volume around each molecule is a sphere of radius equal to the molecular diameter.

3. According to the law of equipartition, the molar specific heat at constant volume of a solid at room temperature is approximately:

1. R
2. 3R
3. 2R
4. 4R

Answer: 3R

Each atom in a solid has 3 vibrational modes, each contributing 2 quadratic terms (KE and PE), so 6 degrees of freedom per atom. Average energy per atom = 3k_B T. For one mole, U = 3RT, so C_v = dU/dT = 3R ≈ 25 J/mol·K (Dulong–Petit law).

4. At a temperature high enough to excite vibration, the total number of degrees of freedom of a diatomic molecule is:

Answer: 7.0

According to equipartition theorem, a diatomic molecule has 3 translational, 2 rotational, and 1 vibrational mode. The vibrational mode contributes 2 quadratic terms (kinetic + potential), so total degrees of freedom = 3 + 2 + 2 = 7.

5. The mean free path λ of a gas is given by λ = 1/(√2 π d² n). If the number density n is 2.5 × 10^25 m⁻³ and molecular diameter d is 2.0 × 10⁻¹⁰ m, what is λ? (Take √2 = 1.414, π = 3.14)

1. 2.25 × 10⁻⁷ m
2. 4.50 × 10⁻⁷ m
3. 9.00 × 10⁻⁷ m
4. 1.80 × 10⁻⁶ m

Answer: 2.25 × 10⁻⁷ m

Using λ = 1/(√2 π d² n). d² = (2.0×10⁻¹⁰)² = 4.0×10⁻²⁰ m². π d² = 3.14×4.0×10⁻²⁰ = 1.256×10⁻¹⁹ m². √2 π d² = 1.414×1.256×10⁻¹⁹ = 1.776×10⁻¹⁹ m². Multiply by n: 1.776×10⁻¹⁹ × 2.5×10²⁵ = 4.44×10⁶ m⁻¹. λ = 1/(4.44×10⁶) = 2.25×10⁻⁷ m.

6. At high temperature, the molar specific heat at constant volume for a diatomic gas is:

1. 5R/2
2. 3R
3. 9R/2
4. 7R/2

Answer: 7R/2

At high temperature, vibrational modes are activated, adding 2 degrees of freedom (one for kinetic, one for potential) to the existing 5 (3 translational + 2 rotational), giving f = 7. Using equipartition, C_v = (f/2)R = 7R/2.

7. A rigid diatomic molecule like N₂ has how many rotational degrees of freedom at room temperature?

Answer: 2.0

A rigid diatomic molecule is modelled as a dumbbell. It can rotate about two axes perpendicular to the bond, giving 2 rotational degrees of freedom. Rotation about the bond axis is ignored due to negligible moment of inertia.

8. Which of the following correctly relates the ideal gas equation in terms of moles and molecules?

1. PV = nRT and PV = NkT, where R = N_A / k
2. PV = nRT and PV = NkT, where R = k / N_A
3. PV = nRT and PV = NkT, where R = N_A k
4. PV = nRT and PV = NkT, where R = N_A k^2

Answer: PV = nRT and PV = NkT, where R = N_A k

The ideal gas law is PV = nRT for moles and PV = NkT for molecules. Since n = N/N_A, substituting gives PV = (N/N_A)RT = N(R/N_A)T. Comparing with PV = NkT yields k = R/N_A, so R = N_A k. Option c correctly states R = N_A k.

9. When the temperature of an ideal gas is increased, which of the following correctly describes the change in its Maxwell–Boltzmann speed distribution curve?

1. The peak shifts to lower speeds and the curve becomes narrower.
2. The peak shifts to higher speeds and the curve becomes narrower.
3. The peak shifts to lower speeds and the curve becomes flatter.
4. The peak shifts to higher speeds and the curve becomes flatter.

Answer: The peak shifts to higher speeds and the curve becomes flatter.

According to the Maxwell–Boltzmann distribution, the most probable speed v_mp = √(2RT/M) increases with temperature. The curve broadens to keep the area constant, so the peak height decreases, making the curve flatter.

10. A real gas is most likely to obey the ideal gas law PV = nRT under which of the following conditions?

1. High pressure and low temperature
2. Low pressure and low temperature
3. High pressure and high temperature
4. Low pressure and high temperature

Answer: Low pressure and high temperature

At low pressure, molecules are far apart so their own volume is negligible. At high temperature, kinetic energy dominates intermolecular attraction. Both conditions minimise deviations from ideal behaviour.

11. According to the law of equipartition, the molar specific heat at constant volume of water is approximately:

1. 3R
2. 9R
3. 6R
4. 12R

Answer: 6R

Water is a nonlinear triatomic molecule. It has 3 translational and 3 rotational degrees of freedom, each contributing (1/2)R to Cv. Vibrational modes are frozen at room temperature. Thus Cv = (3+3)R = 6R.

12. How many translational degrees of freedom does a molecule of oxygen (O₂) have?

Answer: 3.0

Every molecule, regardless of its atomicity, has three translational degrees of freedom corresponding to motion along x, y, and z axes. Oxygen is diatomic but still has 3 translational degrees of freedom.