Previous year questions: Kinetic Theory

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the formulas on the first page of this chapter, and every answer is worked out in full so that the method behind the correct option is clear. The set covers molecular mixtures, kinetic energy, specific heats and gas laws, and every answer keeps the molar mass and temperature units explicit.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Heat is supplied to a diatomic gas at constant pressure. Find the ratio of the heat supplied to the change in internal energy to the work done by the gas.

Answer: For one mole, the heat supplied is Q = C_P delta T = (7R/2)delta T. The internal energy change is delta U = C_V delta T = (5R/2)delta T, and the work done is P delta V = R delta T. The ratio Q : delta U : delta W is therefore 7 : 5 : 2. The same counting works for any gas once its degrees of freedom are known. For a monatomic gas the ratio would be 5 : 3 : 2, which is a useful check on the diatomic result.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A mixture of carbon dioxide and oxygen occupies 8310 cm^3 at 300 K and 100 kPa and has a total mass of 13.2 g. Treating both gases as ideal, find the number of moles of carbon dioxide and of oxygen in the mixture.

Answer: The total number of moles is n = PV/RT = (100 x 10^3 x 8310 x 10^-6)/(8.31 x 300) = 0.333. Let x be the moles of CO2; then the moles of O2 are 0.333 - x. The total mass gives 44x + 32(0.333 - x) = 13.2, so 12x = 13.2 - 10.67 = 2.53 and x = 0.21. The oxygen moles are 0.333 - 0.21 = 0.12. The correct option is 0.21 and 0.12. A quick check: 0.21 mol of CO2 has mass 9.24 g and 0.12 mol of O2 has mass 3.84 g, and the sum is close to the given 13.2 g.

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

One gas of n1 mole at temperature T1, volume V1 and pressure P1 is mixed with another gas of n2 mole at T2, V2 and P2, giving a mixture of pressure P and volume V. Find the temperature of the mixture.

Answer: Internal energy is conserved in mixing if no heat is lost, so n1 C_V T1 + n2 C_V T2 = (n1 + n2) C_V T, which gives T = (n1T1 + n2T2)/(n1 + n2). Now use the ideal gas equation for each gas: n1 = P1V1/(RT1) and n2 = P2V2/(RT2), and for the mixture n1 + n2 = PV/(RT). Substituting and simplifying gives T = T1T2PV/(T2P1V1 + T1P2V2). This matches the option T1T2PV/(T2P1V1 + T1P2V2).

Q4 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Assertion A: if the average kinetic energy of hydrogen and oxygen molecules kept in containers of different sizes is the same, then their temperatures are the same. Reason R: the rms speed of hydrogen and oxygen molecules is the same at the same temperature. Choose the correct option.

Answer: The average translational kinetic energy per molecule is (3/2)k_BT, which depends only on temperature and not on the gas or the container, so A is true. The rms speed is sqrt(3RT/M), which contains the molar mass, so at the same temperature hydrogen and oxygen have different rms speeds. Hence R is false. The correct option is that A is true but R is false. The container size does not enter the kinetic energy argument at all, because the result holds for any volume.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Two moles of a monatomic ideal gas at temperature T are mixed with six moles of another monatomic ideal gas at temperature 2T. Find the temperature of the mixture.

Answer: For monatomic gases the internal energy per mole is (3/2)RT, and the same C_V applies to both, so the final temperature follows from energy conservation: T_mix = (n1T1 + n2T2)/(n1 + n2) = (2T + 6 x 2T)/8 = 14T/8 = 7T/4. The correct option is 7T/4. Since both gases are monatomic, the common C_V cancels in the energy balance, which is why only the mole numbers and temperatures matter.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

One mole of a diatomic gas with rotational modes only is kept in a cylinder with a piston of area 4 cm^2. The gas is heated slowly, raising its temperature by 1.2 degrees C while the piston moves out by 25 mm. Taking atmospheric pressure as 100 kPa and R = 8.3 J mol^-1 K^-1, find the heat supplied to the gas.

Answer: The diatomic gas has five degrees of freedom, so C_V = 5R/2. The internal energy change is delta U = nC_V delta T = 1 x (5 x 8.3/2) x 1.2 = 24.9 J. The work done against the atmosphere is P delta V = 100 x 10^3 x (4 x 10^-4 x 25 x 10^-3) = 1.0 J. The heat supplied is therefore about 26 J, which corresponds to the option closest to 25 J. The small difference from the printed option comes from rounding R and the temperature change, and the method of adding the internal-energy change to the work is the part to remember.

Q7 (NCERT Exemplar, 1 mark)

A cubic vessel containing an ideal gas at normal temperature and pressure is carried by a rocket moving vertically upward at 500 m s^-1. How does the pressure of the gas inside, observed from the ground, compare with the pressure when the vessel is at rest?

Answer: The uniform motion of the vessel as a whole does not change the relative motion of the gas molecules with respect to the walls. Since pressure depends on these relative motions and not on the motion of the container through the laboratory, the pressure stays the same. The correct choice is that it remains the same because the motion of the vessel as a whole does not affect the relative motion of the molecules. The same argument holds inside a train moving at constant speed, where a gas thermometer reads exactly what it reads at rest.

Q8 (NCERT Exemplar, 1 mark)

Boyle's law is applicable for which type of process?

Answer: Boyle's law connects pressure and volume at constant temperature, so it applies to an isothermal process. It is also an approximate law for a real gas near ideal behaviour. The correct choice is the isothermal process.

Q9 (NCERT Exemplar, 1 mark)

A cylinder containing an ideal gas stands vertically with a piston of mass M free to move without friction. If the temperature is increased, what happens to the pressure and the volume of the gas?

Answer: The piston is free, so it moves until the pressure inside the cylinder balances the atmospheric pressure plus the weight of the piston per unit area. That balance fixes the pressure, so the pressure does not change. With the pressure constant, the volume must increase in proportion to the absolute temperature, which is Charles's law. The correct choice is that the volume changes but the pressure does not. This is the standard behaviour of a gas under a freely moving piston: it is a constant-pressure process, not a constant-volume one.

Q10 (NCERT Exemplar, 2 marks)

One mole of hydrogen gas is contained in a box of volume 1.00 m^3 at 300 K. It is heated to 3000 K, and the molecules dissociate completely into hydrogen atoms. Assuming ideal behaviour, find the ratio of the final pressure to the initial pressure.

Answer: The pressure is P = nRT/V. Initially the number of moles is 1 and the temperature is 300 K. After dissociation one mole of H2 becomes two moles of H atoms, so n becomes 2, while the volume is unchanged. The final pressure is therefore (2 x 3000)/(1 x 300) = 20 times the initial pressure. The correct choice is 20 times. Dissociation raises the number of moles, and heating raises the temperature, so both factors combine in the same direction.