Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
At the transition temperature T, $A \rightleftharpoons B$ and $\Delta G^0 = 105 - 35 \log T$ where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is __________. (Nearest integer)
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Answer: 727
Key idea: Gibbs Free Energy at Phase Transition
At transition temperature, the two states A and B are in equilibrium, so $\Delta G^0 = 0$
Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',
(1) 18 J of the work is done by the gas 'A'.
(2) 2 J of the work is done by the gas 'A'.
(3) 12 J of the work is done on the gas 'A' by the surrounding.
(4) 14 J of the work is done on the gas 'A' by the surrounding.
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Answer: Option (4)
Key idea: First Law of Thermodynamics and State Functions
For process $X \to Y$: Apply first law $\Delta U = Q - W = 10 - 18 = -8$ J
Since internal energy is a state function: $\Delta U_{Y \to X} = -\Delta U_{X \to Y} = +8$ J
For reverse process $Y \to X$: Heat evolved $Q = -6$ J (negative because heat is released)
Apply first law for $Y \to X$: $\Delta U = Q - W$ gives $8 = -6 - W$
Solve for work: $W = -6 - 8 = -14$ J
Negative work by gas means positive work on gas: 14 J of work is done ON the gas by surroundings
Given below are two statements:
**Statement I:** For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure.
**Statement II:** In a constant volume process, no work is produced and all the heat withdrawn goes into the chaotic motion and is reflected by a temperature increase of the ideal gas.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
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Answer: Option (4)
Key idea: Relationship between heat capacities (Cp and Cv) for ideal gases and first law of thermodynamics
For ideal gases, the Mayer's relation states: $C_p - C_v = R$
Since $R > 0$, we get $C_p > C_v$ (heat capacity at constant pressure is GREATER than at constant volume)
Therefore, Statement I is **FALSE** (it claims $C_v > C_p$)
For constant volume process: $W = \int P\,dV = 0$ (no work done)
By first law: $\Delta U = Q - W = Q$, so all heat goes into internal energy (molecular motion)
Therefore, Statement II is **TRUE**
Answer: Statement I is false but Statement II is true
If 3.365g of ethanol $(l)$ is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The $|\Delta H_f^o|$ of ethanol at 298.15 K is __________ $\times 10^2$ kJ mol$^{-1}$. (Nearest integer)
Given: Standard enthalpy for combustion of graphite = $-393.5$ kJ mol$^{-1}$
Standard enthalpy of formation of water (l) = $-285.8$ kJ mol$^{-1}$
Molar mass in g mol$^{-1}$ of C, H, O are 12, 1 and 16 respectively
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Answer: 3
Key idea: Hess's Law and Enthalpy of Formation from Combustion Data
Calculate molar mass of ethanol $C_2H_5OH$: $M = 2(12) + 6(1) + 16 = 46$ g/mol
Find moles of ethanol: $n = \frac{3.365}{46} = 0.07315$ mol
Calculate heat of combustion per mole: $\Delta H_c = -\frac{99.472}{0.07315} = -1360.0$ kJ/mol
For the following reaction at 50 °C and at 2 atm pressure,
$2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$
$N_2O_5$ is 50% dissociated.
The magnitude of standard free energy change at this temperature is $x$.
$x = $ __________ J mol$^{-1}$ [Nearest integer].
Given : R = 8.314 J mol$^{-1}$ K$^{-1}$, log 2 = 0.30, log 3 = 0.48, ln10 = 2.303,
°C + 273 = K
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Answer: 2474
Key idea: Relationship between equilibrium constant and Gibbs free energy; calculation of Kp from degree of dissociation
Set up ICE table with $\alpha = 0.5$: At equilibrium, moles are $N_2O_5 = 1$, $N_2O_4 = 1$, $O_2 = 0.5$, total = 2.5 moles
Consider the following data for the reaction
$X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$
at 600 K. The $\Delta_r G^\circ$ (in kJ mol$^{-1}$) for the reaction is:
[Table with columns: Compound | $\Delta_f H^\circ_{600K}$ (kJ mol$^{-1}$) | $S^\circ_{600K}$ (J mol$^{-1}$ K$^{-1}$)
Rows:
XY(g) | 42 | 200
$X_2(g)$ | 8 | 140
$Y_2(g)$ | 80 | 250]
(1) $-21000$
(2) $-10$
(3) $-1000$
(4) $-9.012$
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Answer: Option (2)
Key idea: Gibbs Free Energy calculation using ΔG° = ΔH° - TΔS°
Arrange the following isothermal processes in order of the magnitude of the work $(p - V)$ involved between states 1 and 2.
A. Expansion in single stage $w_A$
B. Expansion in multi stages $w_B$
C. Compression in single stage $w_C$
D. Compression in multi stages $w_D$
Choose the correct option.
(1) $|w_B| > |w_A| > |w_C| > |w_D|$
(2) $|w_C| > |w_D| > |w_A| > |w_B|$
(3) $|w_C| > |w_D| > |w_B| > |w_A|$
(4) $|w_B| > |w_A| > |w_D| > |w_C|$
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Answer: Option (3)
Key idea: Isothermal work in reversible vs irreversible processes
For isothermal processes: $w = -nRT \ln(V_2/V_1)$
**Expansion** (V₂ > V₁): Multi-stage (reversible) gives maximum work output, so $|w_B| > |w_A|$
**Compression** (V₂ < V₁): Single-stage requires maximum work input, so $|w_C| > |w_D|$
Compression requires more work than expansion for same volume ratio: $|w_C|, |w_D| > |w_B|, |w_A|$
Combining all comparisons: $|w_C| > |w_D| > |w_B| > |w_A|$
Consider the reaction $X \rightleftharpoons Y$ at 300 K. If $\Delta H^{\theta}$ and K are 28.40 kJ mol$^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^{\theta}$ for the reaction in J K$^{-1}$ mol$^{-1}$ is ______. (Nearest integer)
(Given : R = 8.3 J K$^{-1}$ mol$^{-1}$, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)
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Answer: 34
Key idea: Gibbs Free Energy and Thermodynamic Relations
Match **List - I** with **List - II**.
Given $V_1$ and $V_2$ are initial and final volumes respectively.
**List - I** (Isothermal process)
A. Reversible expansion
B. Free expansion
C. Irreversible Compression
D. Cyclic reversible
**List - II** (Expression)
I. $q=0$
II. $q = nRT \ln \frac{V_2}{V_1}$
III. $w = -P_{ext} (V_1 - V_2)$
IV. $\frac{q_{rev}}{T} = 0$
Choose the correct answer from the options given below:
(1) A-II, B-III, C-I, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-II, B-I, C-III, D-IV
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Answer: Option (3)
Key idea: Isothermal processes and thermodynamic expressions for different types of expansions/compressions
**A. Reversible expansion (Isothermal):** For isothermal reversible process, $q = nRT \ln \frac{V_2}{V_1}$ → Matches with **II**
**B. Free expansion (Isothermal):** $P_{ext} = 0$, so $w = 0$. Since $\Delta U = 0$ (isothermal), $q = 0$ → Matches with **I**
**C. Irreversible Compression (Isothermal):** Work done is $w = -P_{ext}(V_2 - V_1) = -P_{ext}(V_1 - V_2)$ for compression → Matches with **III**
**D. Cyclic reversible (Isothermal):** For a complete cycle, $\Delta S = 0$, so $\frac{q_{rev}}{T} = 0$ → Matches with **IV**
Therefore: A-II, B-I, C-III, D-IV
FAQs
Frequently asked questions
How many questions come from Chemical Thermodynamics in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 11 questions on it, about 1.2 per shift.
Where can I practise Chemical Thermodynamics PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
