NEET Thermodynamics Previous Year Questions with Solutions

In an adiabatic process, there is no heat exchange with the surroundings (Q = 0):

• • Internal energy change of an ideal gas depends solely on temperature: Delta U = n * C_v * (T_f - T_i).

For an ideal monatomic gas with 3 translational degrees of freedom:

• • Given gamma = 5/3, C_v = R / (gamma - 1) = R / (5/3 - 1) = (3/2) R.

Substitute known values into the work equation:

• • W = \frac{3}{2} \times 8.3 \times 10 = 15 \times 8.3 = 124.5\text{ J}.
• • Matches Option (3) exactly.

Because internal energy U is a state function that depends solely on the thermodynamic state variables (P, V, T):

• • First Law of Thermodynamics: Q_{net} = Delta U_{cycle} + W_{net} = 0 + W_{net} = W_{net}.
• • Total net heat supplied over the cycle equals the net work done by the gas.

For a clockwise cyclic process, net work done is the positive enclosed area of the rectangle on the P-V diagram:

• • P_{high} = 300 N/m^2, P_{low} = 100 N/m^2 ==> Delta P = 300 - 100 = 200 N/m^2.
• • V_{high} = 5 m^3, V_{low} = 2 m^3 ==> Delta V = 5 - 2 = 3 m^3.

Multiply the dimensions of the rectangular cycle:

• • Matches Option (3) exactly.

According to the law of equipartition of energy, each degree of freedom contributes to the internal energy of one mole of gas:

• • 3 translational degrees of freedom contribute 3 * (1/2)R = (3/2)R.
• • 3 rotational degrees of freedom contribute 3 * (1/2)R = (3/2)R.
• • f vibrational modes contribute f * R (since each vibrational mode possesses both kinetic and potential energy states).

For any ideal gas, C_p and C_v are related by Mayer's equation C_p = C_v + R:

• • Universal relation C_p - C_v = R holds for all ideal gases.

Set up the ratio of molar heat capacities:

• • Cross-multiply: 7(4 + f) = 8(3 + f).
• • 28 + 7f = 24 + 8f.
• • 8f - 7f = 28 - 24 \implies f = 4.
• • Matches Option (1) exactly.

Free expansion is expansion of a gas into a vacuum where the opposing external pressure is zero (P_ext = 0).

• • Work done w = 0.
• • For an ideal gas undergoing an isothermal process, internal energy depends only on temperature: ΔU = n C_v ΔT = 0.
• • From the First Law (ΔU = q + w): 0 = q + 0 ⇒ q = 0.

The change in entropy of the surroundings depends on the heat exchanged with the surroundings reversibly: ΔS_surroundings = -q_system / T_surroundings Since q_system = 0, no heat is absorbed or released by the surroundings.

• • ΔS_surroundings = 0.

Entropy is a state function. Even though the process is irreversible, ΔS_system equals the entropy change along a reversible path between the same initial and final states.

• • Substitute n = 2 mol, V₁ = 10 L, V₂ = 100 L:
• • ΔS_system = 2.303 × 2 × R × log₁₀(100 / 10) = 4.606 R × log₁₀(10) = 4.606 R.
• • Therefore: ΔS_system = 4.606 R and ΔS_surroundings = 0.

In chemistry, work done during a reversible isothermal expansion or compression of an ideal gas is defined by: w = - ∫ P_ext dV = - ∫ (nRT / V) dV = - nRT ln(V_f / V_i)

• • For n = 1.0 mol and R = 2 cal K⁻¹ mol⁻¹, the prefactor nR = 1.0 × 2 = 2 cal K⁻¹.

Process 1 is an isothermal expansion at temperature T₁ from initial volume V₁ to final volume V₂.

• • Since V₂ > V₁, ln(V₂/V₁) > 0, so w₁ < 0 (work is done by the system).

Process 3 is an isothermal compression at temperature T₂ from initial volume V₃ to final volume V₄.

• • Since V₄ < V₃, ln(V₄/V₃) < 0, so w₃ > 0 (work is done on the system).

Add the two expressions together to obtain the sum of the work done in the two isothermal stages.

• • Matches Option (1).

At the midpoint of thermal denaturation (T = 60 °C), the concentrations of native protein (N) and denatured protein (D) are equal: [N] = [D].

• • Equilibrium constant K = 1.0.

The standard Gibbs free energy change of the reaction is related to the equilibrium constant by: ΔG° = -RT ln K

• • At the transition temperature, the native and denatured states are in thermodynamic equilibrium (ΔG° = 0).

From the Gibbs-Helmholtz relation: ΔG° = ΔH° - T ΔS° = 0 ⇒ ΔS° = ΔH° / T Convert temperature to Kelvin: T = 60 + 273 = 333 K. Substitute ΔH° = 666 kJ mol⁻¹ and T = 333 K:

• • ΔS° = 2.0 kJ K⁻¹ mol⁻¹.
• • Matches Option (1).

The two chambers are enclosed in a container made of a thermal insulator, so no heat is exchanged with the surroundings (Q = 0). The walls of the container are rigid, so no external work is done (W = 0). According to the First Law of Thermodynamics:

• • The total internal energy of the system remains conserved upon removing the partition.

For an ideal gas, internal energy is proportional to the product of pressure and volume: U = (f/2) nRT = (f/2) PV, where f is the degrees of freedom:

• • Degrees of freedom f cancels out.

Substitute the given values P₁ = 1 atm, V₁ = 2 L, P₂ = 2 atm, V₂ = 3 L:

• • Note: The given mole numbers (n₁ = 5, n₂ = 4) are redundant data; equilibrium pressure is completely determined by P₁V₁ + P₂V₂.

For any thermodynamic system, heat added relates to internal energy change and work done by:

• • Given ΔQ_A = ΔQ_B and ΔU_A = ΔU_B, it follows that the work done by both gases must be strictly identical: W_A = W_B.

The cylinders are maintained under constant pressure P. The work done in displacing a cylindrical piston of radius r by a distance Δx is:

• • A = π r² is the cross-sectional area of the piston.

Equate the work expressions for gas A and gas B:

• • Rearranging for the ratio of radii:
• • \frac{r_A}{r_B} = \sqrt{\frac{\Delta x_B}{\Delta x_A}} = \sqrt{\frac{9\text{ cm}}{16\text{ cm}}} = \frac{3}{4}.
• • Corresponds to Option (2).

The given enthalpy change for the formation of methane is negative:

• • Because ΔH is negative, the reaction is strictly exothermic.
• • Reactants (R) have higher potential energy than Products (P).

In diagrams (2) and (4), the product level P is higher than the reactant level R (H_P > H_R), which represents an endothermic process (ΔH > 0):

• • Diagrams (2) and (4) are ruled out immediately.

Examine the labeling of the 74.8 kJ/mol magnitude in diagrams (1) and (3):

• • In diagram (3), the arrow 74.8 is drawn between the reactant baseline and the peak of the activation barrier, which incorrectly labels activation energy as ΔH.
• • In diagram (1), the double-headed arrow 74.8 correctly spans the energy gap between the reactant plateau (R) and the product plateau (P).
• • Thus diagram (1) is the only accurate representation, matching Option (1).

The heat of crystallisation of barium sulfate represents the enthalpy change when aqueous barium and sulfate ions precipitate to form one mole of solid crystal:

• • This is the reverse of the enthalpy of solution.

The enthalpy of reaction equals the enthalpy of formation of products minus the enthalpies of formation of reactants:

• • Given values: ΔH_{cryst}° = -4.5 kcal/mol, ΔH_f°(BaSO₄, s) = -349 kcal/mol, ΔH_f°(SO₄²⁻, aq) = -216 kcal/mol.

Substitute numerical values and solve the linear equation:

• • -4.5 = -133 - \Delta H_f^\circ(\text{Ba}^{2+}, aq)
• • \Delta H_f^\circ(\text{Ba}^{2+}, aq) = -133 + 4.5 = -128.5\text{ kcal/mol}.
• • Matches Option (1) with absolute precision.

In any quasi-static thermodynamic process, the mechanical work done by a gas during transition from an initial state to a final state is represented by the area under the process curve on a P-V diagram:

From the given P-V diagram: - Volume at state b: V_b = 400 cm³ - Volume at state c: V_c = 400 cm³ The process bc is represented by a vertical straight line perpendicular to the volume axis. The volume remains strictly constant throughout the entire transformation.

Substituting dV = 0 into the work integral:

An isothermal process is one in which the temperature of the system remains constant throughout the entire thermodynamic change:

An isochoric process is one in which the total volume of the system remains strictly constant. As a result, no mechanical boundary work is performed by the system:

An isobaric process is carried out at constant pressure. The heat exchanged at constant pressure equals the enthalpy change:

An adiabatic process is thermally insulated such that no heat flows into or out of the system. According to the First Law of Thermodynamics, all work done is at the expense of internal energy:

Entropy S is a state function that quantifies the number of accessible microstates (disorder/randomness) of a system. For any process: - Transition from ordered to disordered phase increases entropy (ΔS > 0). - An increase in the number of gaseous molecules increases entropy (Δng > 0 ⟹ ΔS > 0). - Cooling a substance decreases thermal energy and accessible states (ΔS < 0).

Analyzing each process individually: - A: Evaporation: Molecules in vapour move freely in 3D space with much higher entropy than in the condensed liquid: ΔS_vap > 0. - B: Cooling crystalline solid: As T drops from 130 K to 0 K, atomic vibrations cease. By the Third Law, S approaches 0 at absolute zero: ΔS < 0. - C: Decomposition: 2 NaHCO3(s) ⟶ Na2CO3(s) + CO2(g) + H2O(g). Formation of 2 moles of gaseous products from a solid creates enormous spatial disorder: ΔS > 0. - D: Homolytic dissociation: Cl2(g) ⟶ 2 Cl(g). A single bound diatomic molecule dissociates into two independent free particles: ΔS > 0.

Processes in which entropy increases are A, C, and D. This corresponds precisely to Option (1).

In chemistry, work done on the system is positive, and work done by the system (such as during expansion against external pressure) is negative. For a reversible isothermal expansion of an ideal gas:

Expressing natural logarithm in terms of base-10 logarithm using the factor 2.303:

Given: - n = 1 mole - R = 2.0 cal K⁻¹ mol⁻¹ - T = 25 + 273.15 = 298.15 K (or 298 K) - P1 = 20 atm, P2 = 10 atm ⟹ P1/P2 = 2.0 - log10(2) = 0.30103

Multiplying the terms step by step: 2.303 × 2.0 = 4.606 4.606 × 298.15 = 1373.28 1373.28 × 0.30103 = 413.14 calories. With the negative sign for expansion work:

Thermodynamic efficiency formulas are strictly valid only on the absolute temperature scale (Kelvin). Convert the source temperature T₁ from Celsius to Kelvin:

The efficiency η of a reversible Carnot engine operating between a source at temperature T₁ and a sink at temperature T₂ is given by:

Given efficiency η = 50% = 0.5 and T₁ = 600 K:

Convert T₂ back into degrees Celsius:

For the reversible homogeneous reaction A + B ⇌ C + D, substitute given concentrations:

The standard Gibbs free energy change of reaction is related to the equilibrium constant by:

Given R = 2 cal mol⁻¹ K⁻¹, T = 300 K, and log₁₀(10) = 1:

Enthalpy (H) is defined as the sum of internal energy (U) and pressure-volume energy (PV):

For a process involving an initial state (1) and final state (2):

For reactions involving ideal gases at constant temperature T, let n₁ be moles of gaseous reactants and n₂ be moles of gaseous products. Since P₁V₁ = n₁RT and P₂V₂ = n₂RT:

Substituting Δ(PV) = Δn_g RT into the enthalpy change expression yields:

Heat capacity at constant volume is defined as the rate of change of internal energy with temperature: $C_V = \left(\frac{\partial U}{\partial T}\right)_V$. Heat capacity at constant pressure is defined as the rate of change of enthalpy with temperature: $C_P = \left(\frac{\partial H}{\partial T}\right)_P$.

Enthalpy is defined as $H = U + PV$. For one mole ($n = 1$) of an ideal gas, applying the ideal gas equation of state $PV = RT$:

Differentiating both sides with respect to temperature $T$:

For an ideal gas, intermolecular forces of attraction are zero. Consequently, internal energy $U$ depends solely on temperature (Joule's Law of ideal gases): $$U = f(T) \implies dU = n C_V dT$$ In any isothermal process, temperature remains strictly constant ($\Delta T = 0$): $$\Delta U = n C_V \Delta T = 0$$

The total entropy change of the universe (system plus surroundings) is defined by the Second Law of Thermodynamics: $$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$$ - For an idealized reversible process: $\Delta S_{\text{total}} = 0$. - For any real, irreversible (spontaneous) process: entropy generation occurs due to dissipation, requiring: $$\Delta S_{\text{total}} > 0 \implies \Delta S_{\text{total}} \neq 0$$

Combining the isothermal criterion ($\Delta U = 0$) with the irreversibility criterion ($\Delta S_{\text{total}} \neq 0$):

At chemical equilibrium, the actual Gibbs free energy change $$\Delta_r G = 0$$. Therefore, the standard Gibbs free energy change relates to the equilibrium constant by: $$\Delta_r G = \Delta_r G^\circ + RT \ln K_c = 0$$ $$\Delta_r G^\circ = -RT \ln K_c$$

Given: - $$R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$$ - $$T = 300\text{ K}$$ - $$K_c = 2 \times 10^{13}$$ Substituting directly into the equation: $$\Delta_r G^\circ = -8.314\text{ J mol}^{-1}\text{K}^{-1} \times 300\text{ K} \times \ln(2 \times 10^{13})$$

Because $$K_c = 2 \times 10^{13} \gg 1$$, $$\ln K_c > 0$$. Hence, $$\Delta_r G^\circ$$ must be a large negative value, indicating that the hydrolysis of sucrose to glucose and fructose is strongly spontaneous and exergonic under standard conditions.

Option (2) has a positive sign (violates the negative sign in $$-RT \ln K$$); Options (3) and (4) have incorrect equilibrium constant arguments ($$3 \times 10^{13}$$ and $$4 \times 10^{13}$$).