Thermodynamics — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Thermodynamics — ideal gas processes, first law, adiabatic work, specific heats and calorimetry — with step-by-step solutions.
Solved JEE Main 2026 questions from the Thermodynamics chapter, covering ideal gas processes, the first law, adiabatic work, specific heats, degrees of freedom and calorimetry, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
For $n$ moles, $T = \dfrac{PV}{nR}$ with $n = 2$, so $nR\,T = PV$.
Sample A at $V = V_0$:
$$P_A = \frac{P_0}{1 + (V_0/V_0)^2} = \frac{P_0}{2}, \qquad P_A V_0 = \frac{P_0 V_0}{2}$$Sample B at $V = 3V_0$:
$$P_B = \frac{P_0}{1 + (V_0/3V_0)^2} = \frac{P_0}{1 + \tfrac{1}{9}} = \frac{9P_0}{10}, \qquad P_B (3V_0) = \frac{27 P_0 V_0}{10}$$Temperature difference ($n = 2$):
$$T_B - T_A = \frac{P_B V_B - P_A V_A}{2R} = \frac{1}{2R}\left(\frac{27P_0 V_0}{10} - \frac{P_0 V_0}{2}\right) = \frac{1}{2R}\cdot\frac{22P_0 V_0}{10} = \frac{11 P_0 V_0}{10R}$$Answer: B ($\frac{11P_0 V_0}{10R}$)
Solution
For a diatomic gas, $C_V = \dfrac{5}{2}R$ and $C_P = \dfrac{7}{2}R$. At constant pressure, for $n$ moles over $\Delta T$:
$$\Delta Q = nC_P\,\Delta T = \frac{7}{2}nR\,\Delta T$$$$\Delta U = nC_V\,\Delta T = \frac{5}{2}nR\,\Delta T$$$$\Delta W = P\,\Delta V = nR\,\Delta T$$Dividing each by $nR\,\Delta T$:
$$\Delta Q : \Delta U : \Delta W = \frac{7}{2} : \frac{5}{2} : 1 = 7 : 5 : 2$$Answer: D ($7 : 5 : 2$)
Solution
$\gamma = \dfrac{c_p}{c_v} = \dfrac{3R}{2R} = \dfrac{3}{2}$.
Given $P_1 = 8 \times 10^5\ \text{Pa}$, $V_1 = 0.15\ \text{m}^3$, $P_2 = 1 \times 10^5\ \text{Pa}$.
Final volume (adiabatic, $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$):
$$V_2 = V_1\left(\frac{P_1}{P_2}\right)^{1/\gamma} = 0.15 \times 8^{2/3} = 0.15 \times 4 = 0.60\ \text{m}^3$$Work done in adiabatic process:
$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{(8\times10^5)(0.15) - (1\times10^5)(0.60)}{1.5 - 1}$$$$W = \frac{1.2\times10^5 - 0.6\times10^5}{0.5} = \frac{0.6\times10^5}{0.5} = 1.2\times10^5\ \text{J} = 120\ \text{kJ}$$Answer: 120
Solution
Convert $R$ to calories ($1\ \text{cal} = 4.18\ \text{J}$):
$$R = \frac{8.36}{4.18} = 2\ \text{cal/mol·}°C$$Molar specific heat at constant volume (Mayer’s relation):
$$c_v = c_p - R = 8 - 2 = 6\ \text{cal/mol·}°C$$Change in internal energy ($n = 5$, $\Delta T = 20 - 10 = 10\,°C$):
$$\Delta U = n\,c_v\,\Delta T = 5 \times 6 \times 10 = 300\ \text{cal}$$Answer: 300
Solution
The piston is frictionless, so pressure equalises across it; the common final pressure is $P_f = \dfrac{27P}{8}$.
The right compartment is compressed adiabatically (its walls and the piston are adiabatic), so $PV^{\gamma}$ is conserved with $\gamma = 1.5$:
$$P\,V_i^{\gamma} = P_f\,V_f^{\gamma} \implies V_f = V_i\left(\frac{P}{P_f}\right)^{1/\gamma}$$$$V_f = 9\left(\frac{P}{27P/8}\right)^{1/1.5} = 9\left(\frac{8}{27}\right)^{2/3} = 9 \times \frac{4}{9} = 4\ \text{litres}$$since $\left(\dfrac{8}{27}\right)^{2/3} = \dfrac{8^{2/3}}{27^{2/3}} = \dfrac{4}{9}$.
Answer: B ($4$)
Solution
Statement I: For an ideal gas, $C_P - C_V = R$ and $\gamma = C_P/C_V$, so
$$C_V = \frac{R}{\gamma - 1} \implies \Delta U = nC_V\,\Delta T = \frac{nR}{\gamma-1}\,\Delta T$$This is true.
Statement II: With $f$ degrees of freedom, $C_V = \dfrac{f}{2}R$ and $C_P = \dfrac{f+2}{2}R$, giving
$$\gamma = \frac{C_P}{C_V} = \frac{f+2}{f} = 1 + \frac{2}{f}$$This is also true.
However, the second equality in Statement I follows from $C_V = R/(\gamma-1)$ (Mayer’s relation), which does not require the degrees-of-freedom relation of Statement II. Hence R is true but is not the correct explanation of A.
Answer: B (Both A and R are true but R is NOT the correct explanation of A)
Solution
A — True. The zeroth law establishes thermal equilibrium and hence the concept of temperature.
B — True. The first law, $\Delta Q = \Delta U + \Delta W$, introduces internal energy $U$ as a state function.
C — False. For isothermal expansion of an ideal gas, $\Delta U = 0$, so $\Delta Q = \Delta W$ (not $\neq$).
D — True. Intensive $\times$ extensive is extensive (e.g. density $\times$ volume $=$ mass).
E — False. Extensive $\div$ mass gives a specific (intensive) quantity, not extensive.
Correct statements: A, B and D.
Answer: C (A, B and D Only)
Solution
Let $c = 4200\ \text{J/kg·K}$ and $L = 2256\times10^3\ \text{J/kg}$.
Heat released cooling $x$ g from $50\,°C$ to $0\,°C$ ($\Delta T = 50\ \text{K}$):
$$Q_1 = \frac{x}{1000}\,c\,(50)$$Heat needed to evaporate $(1000 - x)$ g of water:
$$Q_2 = \frac{1000 - x}{1000}\,L$$Setting $Q_1 = Q_2$ (the mass factor $1/1000$ cancels):
$$x\,c\,(50) = (1000 - x)\,L$$$$x(4200 \times 50) = (1000 - x)(2256\times10^3)$$$$210000\,x = 2.256\times10^9 - 2256000\,x$$$$x(210000 + 2256000) = 2.256\times10^9 \implies x = \frac{2.256\times10^9}{2466000} \approx 914.8$$Answer: 915
Solution
The piston is massless and moves quasi-statically against atmospheric pressure, so the gas pressure stays equal to $P_{atm}$ and the work done by the gas is:
$$\Delta W = P_{atm}\,\Delta V = P_{atm}\,(A\cdot d)$$$$\Delta W = (100\times10^3)\,(4\times10^{-4})\,(25\times10^{-3}) = 1.0\ \text{J}$$A diatomic molecule with translational and rotational modes (no vibration) has $f = 5$, so $C_V = \dfrac{5}{2}R$. The change in internal energy for $n = 1$ mole is:
$$\Delta U = nC_V\,\Delta T = 1 \times \frac{5}{2}(8.3)(1.2) = 24.9\ \text{J}$$Heat supplied (first law, $\Delta Q = \Delta U + \Delta W$):
$$\Delta Q = 24.9 + 1.0 = 25.9\ \text{J} \approx 25\ \text{J}$$Rounding to the given options, the heat supplied is $\approx 25$ J.
Answer: B (25)
Solution
For a monoatomic gas, $\gamma = \dfrac{5}{3}$ and $C_V = \dfrac{3}{2}R$.
Final pressure (adiabatic, $PV^{\gamma} = $ const):
$$P_f = P\left(\frac{V}{27V}\right)^{5/3} = P\,(27)^{-5/3} = P\,(3^3)^{-5/3} = P\cdot 3^{-5} = \frac{P}{243}$$Product $P_f V_f$:
$$P_f V_f = \frac{P}{243}\times 27V = \frac{27}{243}PV = \frac{PV}{9}$$Change in internal energy (using $\Delta U = nC_V\Delta T = \frac{C_V}{R}(P_f V_f - PV) = \frac{3}{2}(P_f V_f - PV)$):
$$\Delta U = \frac{3}{2}\left(\frac{PV}{9} - PV\right) = \frac{3}{2}\left(-\frac{8}{9}PV\right) = -\frac{4}{3}PV$$Answer: C ($-\frac{4}{3}PV$)