Previous year questions: Thermodynamics

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 the first law, adiabatic processes, cycles, Carnot efficiency and refrigerators, with the signs of heat and work stated at every step.

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, that is the ratio of the three quantities.

Answer: Let the temperature rise be delta T. The heat supplied is Q = nC_P delta T = n(7R/2)delta T. The change in internal energy is delta U = nC_V delta T = n(5R/2)delta T, and the work done is delta W = P delta V = nR delta T. The ratio Q : delta U : delta W is therefore 7 : 5 : 2. The same method works for any gas once C_V and C_P for that gas are known.

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A vessel contains 0.15 m^3 of gas at 8 bar and 140 degrees C, with C_P = 3R and C_V = 2R. The gas expands adiabatically until its pressure falls to 1 bar. Find the work done during the process in kJ. Take R = 8.314 J mol^-1 K^-1.

Answer: First find the number of moles from PV = nRT with T = 413 K: n = (8 x 10^5 x 0.15)/(8.314 x 413) = 34.96 mol. For the adiabatic expansion, T2 = T1 (P2/P1)^((gamma - 1)/gamma) with gamma = 3/2, so T2 = 413 x (1/8)^(1/3) = 413 x 0.5 = 206.5 K. The work done by the gas is nC_V(T1 - T2) = 34.96 x (2 x 8.314) x 206.5 = 1.2 x 10^5 J = 120 kJ. The answer is 120.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Given the two statements: Statement I: the change in internal energy of n moles of an ideal gas can be written as delta U = nC_V(Tf - Ti) = nR(Tf - Ti)/(gamma - 1). Statement II: the relation between the degrees of freedom f and gamma = C_P/C_V is gamma = 1 + 2/f. Choose the correct option.

Answer: Both statements are individually correct. The first is the standard expression for the internal energy change of an ideal gas, with C_V = R/(gamma - 1). The second is the standard relation between degrees of freedom and gamma. The second statement, however, is not the explanation of the first: it is a separate relation, so the correct option is that both A and R are true but R is not the correct explanation of A.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Consider the statements: A. the zeroth law gives the concept of temperature; B. the first law gives the concept of internal energy; C. during an isothermal expansion of an ideal gas, delta Q and delta W are unequal; D. the product of an intensive and an extensive variable is extensive; E. the ratio of any extensive variable to mass is an extensive variable. Choose the correct combination.

Answer: A is true because the zeroth law makes temperature measurable. B is true because the first law defines the change of internal energy. C is false because an isothermal process of an ideal gas has delta U = 0, so delta Q equals delta W. D is true: pressure (intensive) times volume (extensive) is work, which is extensive. E is false because an extensive variable divided by mass gives a specific quantity, which is intensive. The correct combination is A, B and D only.

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

A cylinder with adiabatic walls is divided into two compartments by a frictionless adiabatic piston. Both compartments contain the same ideal gas at the same P, V and T, with V = 9 litres on each side. Heat is supplied to the left side until its pressure becomes 27P/8. Find the final volume of the right-hand compartment. Take C_P/C_V = 1.5 for the gas.

Answer: The right side is compressed adiabatically, so P V^gamma stays constant there. Since the final pressure is 27/8 times the initial pressure, V_right = V_initial (P_initial/P_final)^(1/gamma) = 9 x (8/27)^(2/3) = 9 x (2/3)^2 = 9 x 4/9 = 4 litres. The left side expands and takes up the rest of the volume. The answer is 4 litres.

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

The initial pressure and volume of a monoatomic ideal gas are P and V. It expands adiabatically to a final volume of 27V. Find the change in its internal energy.

Answer: For a monoatomic gas gamma = 5/3. In an adiabatic expansion T V^(gamma - 1) stays constant, so T_f = T(V/27V)^(2/3) = T/9. The internal energy of n moles is (3/2)nRT = (3/2)PV at the initial state, so delta U = (3/2)(P_f V_f - P V). Using P_f V_f = nRT_f = PV/9, delta U = (3/2)(PV/9 - PV) = (3/2)(-8PV/9) = -4PV/3. The answer is -4PV/3.

Q7 (NCERT Exemplar, 3 marks)

A Carnot engine operates between 500 K and 300 K and produces 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer: The Carnot efficiency is 1 - T2/T1 = 1 - 300/500 = 0.4. The heat absorbed from the hot reservoir is Q1 = W/eta = 1000/0.4 = 2500 J = 2.5 kJ. The heat rejected to the cold reservoir is Q2 = Q1 - W = 2500 - 1000 = 1500 J = 1.5 kJ. The reversible heat ratio Q1/T1 = Q2/T2 gives the same values, which is a useful check. If an engine is claimed to be more efficient than this limit, the claim must be wrong.

Q8 (NCERT Exemplar, 2 marks)

The initial state of a gas is (Pi, Vi, Ti). It expands to volume Vf in two ways: (a) at constant temperature, (b) at constant pressure. Plot the processes on a P-V diagram and state in which case the work done by the gas is more.

Answer: The isothermal process follows PV = constant and is a hyperbola from (Pi, Vi) to a lower pressure at Vf. The isobaric process is a horizontal line at Pi from Vi to Vf. For an expansion with Vf greater than Vi, the isobaric work is Pi(Vf - Vi) while the isothermal work is PiVi ln(Vf/Vi). For Vf/Vi greater than 1, (Vf/Vi - 1) is greater than ln(Vf/Vi), so the constant-pressure process does more work.

Q9 (NCERT Exemplar, 2 marks)

The coefficient of performance of a refrigerator is 5 and it operates with its outside at room temperature 27 degrees C. Find the temperature inside the refrigerator.

Answer: For a reversible refrigerator, the coefficient of performance is beta = T2/(T1 - T2), with T1 the room temperature in kelvin, so 5 = T2/(300 - T2). Solving gives 5(300 - T2) = T2, so 1500 = 6 T2 and T2 = 250 K. Converting back to Celsius, the inside temperature is 250 - 273 = -23 degrees C.

Q10 (NCERT Exemplar, 3 marks)

A refrigerator is driven by a 1 kW motor and removes heat from a cold space at -3 degrees C into surroundings at 27 degrees C. The refrigerator works at 50 percent of the performance of a perfect refrigerator. Find the heat removed per second.

Answer: The temperatures are T2 = 270 K and T1 = 300 K. The coefficient of performance of a perfect (reversible) refrigerator is T2/(T1 - T2) = 270/30 = 9. At 50 percent of that performance the actual coefficient is 4.5. With a work input of 1000 J each second, the heat removed is Q2 = beta W = 4.5 x 1000 = 4500 J per second, that is 4.5 kW. The calculation shows why a refrigerator cannot be judged by its motor rating alone: the coefficient of performance decides how much cooling each unit of work produces.