Kinetic Theory of Gases JEE Main Questions 2025:
The JEE Main 2025 aspirants must be aware of the JEE Main Kinetic Theory of Gases questions since understanding the correct approaches and techniques are necessary for excelling in the JEE Main Physics paper. The Kinetic Theory of Gases chapter covers molecular explanations for the behavior of gases and candidates can find a minimum of 2 questions from the Kinetic Theory of Gases chapter in the
JEE Main 2025
exam. Hence, it is essential that candidates familiarize themselves with the important practice questions of the Kinetic Theory of Gases chapter for JEE Main 2025. Candidates are advised to take a look at this article for the JEE Main Kinetic Theory of Gases questions along with their solutions.
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JEE Main 2025 Kinetic Theory of Gases Important Questions
Candidates can check the important practice questions for JEE Main Kinetic Theory of Gases 2025 along with their solutions as provided in the table below.
S.No. | Question | Solution |
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1. | The specific heats, CP and CV of gas of diatomic molecules, A, is given (in units of J mol–1 K–1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then (a) A has a vibrational mode but B has none. (b) A has one vibrational mode and B has two. (c) A is rigid but B has a vibrational mode. (d) Both A and B have a vibrational mode each. | Solution Here Cp and Cv of A are 29 and 22 and Cp and Cv of B are 30 and 21. γ = Cp/Cv = 1 + 2/f For A, Cp/Cv = 1 + 2/f ⇒ f=6 Molecule A has 3 translational, 2 rotational and 1 vibrational degree of freedom For B, Cp/Cv = 1 + 2/f ⇒ f=5 i.e., B has 3 translational and 2 rotational degrees of freedom. Answer: (a) A has a vibrational mode but B has none. |
2. | N moles of a diatomic gas in a cylinder is at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas? (a) 0 (b) (5/2)nRT (c) (½)nRT (d) (3/2)nRT | The initial kinetic energy of the system Ki = (5/2)RTN Final kinetic energy of the system Kf = (5/2)RT(N-n) + (3/2)RT(2n) Kf – Ki =ΔK = nRT(3 – 5/2) = (½)nRT Answer: (c) (½)nRT |
3. | The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in m s–1 is [kB = 1.4 × 10–23 J/K; mHe = 7 × 10–27 kg] (a) 1.3 × 103 (b) 1.3 × 105 (c) 1.3 × 102 (d) 1.3 × 104 | (3/2)kBT = ½mv2 v=3kBTm=3×1.4×10−23×3007×10−27 = 1.3 x103 ms-1 Answer:(a) 1.3 × 103 |
4. | An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (CP) and at constant volume (CV) is (a) 6 (b) 7/2 (c) 5/2 (d) 7/5 | An ideal gas has molecules with 5 degrees of freedom, then Cv = (5/2)R and Cp = (7/2)R Cp/Cv = (7/2)R/(5/2)R Cp/Cv = 7/5 Answer: (d) 7/5 |
5. | A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is (a) 20 RT (b) 12 RT (c) 4 RT (d) 15 RT | U = (f1/2)n1RT + (f2/2)n2RT U = (5/2)(3RT) + (3/2)(5RT) U = 15RT Answer: (d) 15 RT |
6. | In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is (a) 2 (b) 1 (c) 4/5 (d) 4/7 | Pressure, P = ⅓(mN/V)V2rms P = (mN)T/V If the gas mass and temperature are constant then P ∝ (Vrms)2 ∝ T So, force ∝ (Vrms)2 ∝ T I.e., Value of q = 1 |
7. | Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will (a) increase (b) decrease (c) remain the same d) decrease for some, while the increase for others. | It is the relative velocities between molecules that are important. Root mean square velocities are different from lateral translation. Answer: (c) remain the same |
8. | One kg of a diatomic gas is at a pressure of 8 × 104 N/m2. The density of the gas is 4 kg/m3. What is the energy of the gas due to its thermal motion? (a) 3 × 104 J (b) 5 × 104 J (c) 6 × 104 J (d) 7 × 104 J | The thermal energy or internal energy is 5 2 U RT for diatomic gases. (degree of freedom for diatomic gas = 5) But PV = RT V = mass/density = 1kg/(4 kg/m3) = (¼)m3 P = 8 x 104 N/m2 U = (5/2) x 8 x 104 x ¼ = 5 x 104 J Answer: (b) 5 × 104 J |
9. | In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to a)kinetic energy of molecules and collision of molecules b)size of molecules and expansion of the container c)expansion of the container and intermolecular attraction d)intermolecular attraction and the size of molecules Correct answer is 'D'. Can you explain this answer? | The equation of state for real gases, PV = RT, is an ideal gas equation that assumes that the gas molecules have no volume and do not interact with each other. However, in reality, gas molecules do occupy space and experience intermolecular attractions. Therefore, corrections need to be applied to account for these deviations from ideal behavior. Corrections for Pressure (P): The measured pressure of a real gas needs to be corrected to account for the size of the gas molecules. The pressure measured by a pressure gauge is the sum of both the pressure exerted by the gas molecules hitting the walls of the container and the pressure exerted by the gas molecules colliding with the gauge. This collision pressure is due to the size of the gas molecules. Corrections for Volume (V): The measured volume of a real gas needs to be corrected to account for the intermolecular attractions between the gas molecules. The volume measured by a measuring device is the sum of both the actual volume occupied by the gas molecules and the additional space that is occupied due to intermolecular attractions. This additional space is due to the size of the gas molecules and the attractive forces between them. Explanation of Option D: Option D states that the corrections for P and V arise due to intermolecular attraction and the size of molecules. This is the correct answer because it correctly identifies the two main factors that need to be considered for the corrections. |
10. |
The kinetic energy of one mole of an ideal gas is E=3/2 RT. Then Cρ will be
| The kinetic energy of one mole of an ideal gas is given by E = 3/2 RT, where R is the gas constant and T is the temperature in Kelvin. - The specific heat capacity of a gas at constant volume (Cv) is defined as the amount of heat required to raise the temperature of one mole of the gas by one degree while keeping the volume constant. The specific heat capacity of a gas at constant pressure (Cp) is defined as the amount of heat required to raise the temperature of one mole of the gas by one degree while keeping the pressure constant. - The relationship between Cp and Cv is given by Cp - Cv = R, where R is the gas constant. Using the above equation, we can calculate the value of Cp in terms of Cv and R as Cp = Cv + R. - Now, we can substitute the value of Cv from the ideal gas law as Cv = (3/2) R, and the value of R from the same law as R = (Nk)/V, where N is the number of moles, k is the Boltzmann constant, and V is the volume. Substituting these values in the equation for Cp, we get Cp = (5/2) R = 2.5 R. - Therefore, the correct answer is option C, 2.5 R. |
11. |
Oxygen and nitrogen in two enclosures have the same mass, volume and pressure. The ratio of the temperature of oxygen to that of nitrogen is:
| For the same mass the ratio of moles of oxygen to that of nitrogen is 14:16 = 7:8 And we know that PV = nRT Hence as V and P are also the same, ratio of the temperature of oxygen to that of nitrogen is the inverse of the ratio of moles that is 8:7 |
12. |
Value of gas constant, R for one mole of a gas is independent of the
| We know that PV=nRT also PM=dRT So in the equation The value of R depends on P, V , n, T , d, M except atomicity so the answer is A |
13. |
A region of the earth’s atmosphere contains n molecules (treated as ideal gas molecules) per unit volume. The temperature of air in the region is T. If k represents Boltzmann’s constant and R represents a universal gas constant, the pressure of air in the region is
| PV = nRT Where n = number of moles = m/NA So, P = (m/V)(R/NA)T Also, we know that R/NA = k So, P = nkT |
14. |
One mole of any substance at any temperature, pressure or volume always contains ________ molecules.
| One mole of any substance at any temperature, pressure or volume always contains 6.02×10^23 molecules. |
15. |
Which one of the following quantities can be zero on an average for the molecules of an ideal gas in equilibrium?
| In the case of ideal gases, the average velocity is always zero. Hence the average momentum is zero. Whereas average speed is non-zero so the kinetic energy is also non-zero, as these two are scalar quantities. |
16. |
The degree of freedom for tri-atomic gas is:
| Degrees of freedom are the ways in which a molecule of the gas can execute motion. So in the case of triatomic gas molecule: 1. It can translate (move) in all 3 dimensions, which accounts for 3 degrees of freedom (since there are 3 dimensions in which it could translate (move)). 2. This molecule can also revolve with Moment of Inertia ≠ 0 around all three axes, x, y, and z, which accounts for another 3 degrees of freedom (since there are 3 axes of rotation). |
17. |
Kinetic theory explains the behaviour
| Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules. This is possible as the inter-atomic forces, which are short-range forces that are important for solids and liquids, can be neglected for gases. |
18. |
Mean free path is the
| The mean free path is the average distance travelled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties. |
19. |
Four moles of an ideal diatomic gas is heated at constant volume from 20° C to 30° C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol-1K-1 and the universal gas constant (R) is 8.3 Jmol-1K-1. The increase in internal energy of the gas is
| The value of Cp is 30.3 and as Cp-Cv = R(8.3) hence Cv = 30.3-8.3 Cv is 22 change in internal energy = no of moles × Cv × change in temperature hence change in internal energy = 22 × 4 × 10 = 880j Hence Option D is correct. |
20. |
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
| Mass, m = 14.5 kg Length of the steel wire, l = 1.0 m Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2 Let Δl be the elongation of the wire when the mass is at the lowest point of its path. When the mass is placed at the position of the vertical circle, the total force on the mass is: F = mg + mlω2 = 14.5 × 9.8 + 14.5 × 1 × (12.56)2 = 2429.53 N Young’s modulus = Strss / Strain Y = (F/A) / (∆l/l) ∴ ∆l = Fl / AY Young’s modulus for steel = 2 × 1011 Pa ∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m Hence, the elongation of the wire is 1.87 × 10–3 m Hence 1.87 × 10–3 m |
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