NCERT Exemplar: Thermodynamics | Chemistry Class 11 - NEET PDF Download

Multiple Choice Questions

Q.1. Thermodynamics is not concerned about______.
(1) Energy changes involved in a chemical reaction.
(2) The extent to which a chemical reaction proceeds.
(3) The rate at which a reaction proceeds.
(4) The feasibility of a chemical reaction.

Ans. (3)

Sol.
Thermodynamics deals with energy changes, feasibility and extent of reactions by comparing initial and final equilibrium states. It does not study how fast a process proceeds; that is the subject of chemical kinetics. Laws of thermodynamics apply to equilibrium states or transitions between equilibrium states.

Q.2. Which of the following statements is correct?
(1) The presence of reacting species in a covered beaker is an example of open system.
(2) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(3) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(4) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

Ans. (3)

Sol.
A closed system allows exchange of energy (heat, work) but not matter. A closed vessel made of conducting material such as copper or steel permits energy exchange but prevents mass transfer; hence it is a closed system. A thermos flask that is well insulated approximates an adiabatic system (no heat exchange), not a perfectly closed energy-exchanging system.

Q.3. The state of a gas can be described by quoting the relationship between___.
(1) Pressure, volume, temperature
(2) Temperature, amount, pressure
(3) Amount, volume, temperature
(4) Pressure, volume, temperature, amount

Ans. (4)

Sol.
State of an ideal gas is specified by state variables such as pressure (p), volume (V), temperature (T) and amount of substance (n). The ideal gas equation pV = nRT relates all four variables; therefore all are needed to describe the state completely.

Q.4. The volume of gas is reduced to half from its original volume. The specific heat will be ______.
(1) Reduce to half
(2) Be doubled
(3) Remain constant
(4) Increase four times

Ans. (3)

Sol.
Specific heat capacity is the heat required to raise unit mass (or mole) of a substance by one degree. It is an intensive property and does not depend on the amount or volume of the substance. Therefore, halving the volume (for a given mass) does not change the specific heat.

Q.5. During complete combustion of one mole of butane, 2658 kJ of heat is released.The thermochemical reaction for above change is
(1) 2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l) Δ_cH = -2658.0 kJ mol^-1

Ans. (3)

Sol.
Standard enthalpy of combustion is defined per mole of substance combusted under standard states. The thermochemical equation must be written for one mole of the fuel. If 2658 kJ is evolved on burning one mole of butane, the balanced equation for one mole should show ΔcH = -2658 kJ mol^-1; the given option (3) corresponds to the correctly normalized thermochemical equation.

Q.6. ∆_fU^Θ of formation of CH₄(g) at certain temperature is -393 kJ mol^-1. The value of ∆_fH^Θ is
(1) zero
(2) < ∆_fU^Θ
(3) > ∆_fU^Θ
(4) equal to ∆_fU^Θ

Ans. (2)

Sol.
The relation between enthalpy and internal energy changes is:

Δ_fH = Δ_fU + Δn_gRT

For C(s) + 2H₂(g) → CH₄(g), Δn_g = 1 - 2 = -1.

Therefore, Δ_fH = Δ_fU + (-1)RT = Δ_fU - RT.

Since RT is positive, Δ_fH < Δ_fU.

Q.7. In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.
(1) q = 0, ∆T ≠ 0, w = 0
(2) q ≠ 0, ∆T = 0, w = 0
(3) q = 0, ∆T = 0, w = 0
(4) q = 0, ∆T <0, w ≠ 0

Ans. (3)

Sol.
For free expansion into vacuum, external pressure p_ext = 0, so work w = 0. Under adiabatic condition q = 0. From the first law, ΔU = q + w = 0. For an ideal gas ΔU depends only on temperature, so ΔT = 0.

Q.8. The pressure-volume work for an ideal gas can be calculated by using the expression

The work can also be calculated from the pV- plot by using the area under the curve within the specified limits. When an ideal gas is compressed
(a) reversibly or (b) irreversibly from volume V_i to V_f. choose the correct option.
(1) w(reversible) = w (irreversible)
(2) w(reversible) < w (irreversible)
(3) w(reversible) > w (irreversible)
(4) w(reversible) = w (irreversible) + p_ex.∆V

Ans. (2)

Sol.
For compression from V_i to V_f, the reversible path has the smallest magnitude of work done on the surroundings when comparing areas under p-V curves for the same end states. In irreversible (sudden) compression against a larger external pressure, the area (and hence work) is greater. Therefore w(reversible) < w(irreversible).

Q.9. The entropy change can be calculated by using the expression

When water freezes in a glass beaker, choose the correct statement amongst the following :
(1) ∆S (system) decreases but ∆S (surroundings) remains the same.
(2) ∆S (system) increases but ∆S (surroundings) decreases.
(3) ∆S (system) decreases but ∆S (surroundings) increases.
(4) ∆S (system) decreases and ∆S (surroundings) also decreases.

Ans. (3)

Sol.
When water freezes, the system (water → ice) becomes more ordered so ∆S(system) < 0. Freezing releases heat to the surroundings, increasing entropy of surroundings (∆S(surroundings) > 0). The net entropy change depends on temperature and heat released, but the directions are as stated.

∴

Q.10. On the basis of thermochemical equations (a), (b) and (c), find out which of the algebraic relationships given in options (i) to (iv) is correct.
(a) C (graphite) + O₂(g) → CO₂ (g) ;Δ_rH = x kJ mol^-1
(b)

(c)

(1) z = x + y
(2) x = y - z
(3) x = y + z
(4) y = 2z - x

Ans. (3)

Sol.
Use algebraic combination of the given thermochemical equations. By appropriate addition/subtraction of equations (i), (ii) and (iii) one obtains the relation x = y + z. (Detailed algebraic manipulation is shown using the three given equations and their ΔH values.)

Q.11. Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct?
(a) C(g) + 4H(g) → CH₄(g); ∆_rH = x kJ mol^-1
(b) C(graphite,s) + 2H₂(g) → CH₄(g); ∆_rH = y kJ mol^-1
(1) x = y
(2) x = 2y
(3) x > y
(4) x < y

Ans. (3)

Sol.
Reaction (a) atomises all reactants to gaseous atoms before bond formation; reaction (b) starts from molecular hydrogen and graphite. Breaking bonds (or atomisation) costs energy. Thus enthalpy of atomisation plus bond formation in (a) makes x greater than y.

Q.12. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(1) Is always negative
(2) Is always positive
(3) May be positive or negative
(4) Is never negative

Ans. (3)

Sol.
Standard enthalpy of formation may be negative (exothermic formation) or positive (endothermic formation). Example: C(s) + O₂(g) → CO₂(g) has ΔH° = -393.5 kJ mol^-1; formation of some substances (e.g., certain oxides or peroxides under specific conditions) can be endothermic.

Q.13. Enthalpy of sublimation of a substance is equal to
(1) Enthalpy of fusion + enthalpy of vapourisation
(2) Enthalpy of fusion
(3) Enthalpy of vapourisation
(4) Twice the enthalpy of vapourisation

Ans. (1)

Sol.
Sublimation (solid → gas) can be considered as solid → liquid (fusion) followed by liquid → vapour (vapourisation). Therefore ΔH_sub = ΔH_fus + ΔH_vap.

Q.14. Which of the following is not correct?
(1) ∆G is zero for a reversible reaction
(2) ∆G is positive for a spontaneous reaction
(3) ∆G is negative for a spontaneous reaction
(4) ∆G is positive for a non-spontaneous reaction

Ans. (2)

Sol.
Criterion for spontaneity at constant T and p: ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 equilibrium (reversible). Statement (2) is incorrect because for a spontaneous reaction ΔG is negative, not positive.

Q.15. Thermodynamics mainly deals with
(1) Interrelation of various forms of energy and their transformation from one form to another.
(2) Energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.
(3) How and at what rate these energy transformations are carried out.
(4) The system in equilibrium state or moving from one equilibrium state to another equilibrium state.

Ans. (1, 4)

Sol.
Thermodynamics studies interconversion of energy forms and properties of systems in or between equilibrium states. It does not address microscopic kinetics or the rates of processes (choice 3) nor restricted to microscopic few-molecule systems (choice 2).

Q.16. In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system
(1) q_p will be negative
(2) ∆_rH will be negative
(3) q_p will be positive
(4) ∆_rH will be positive

Ans. (1, 2)

Sol.
For exothermic reactions heat is released by the system to surroundings, so q_p < 0 and Δ_rH < 0.

Q.17. The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
(1) Flow of heat from colder to warmer body.
(2) Gas in a container contracting into one corner.
(3) Gas expanding to fill the available volume.
(4) Burning carbon in oxygen to give carbon dioxide.

Ans. (3, 4)

Sol.
Spontaneous processes include diffusion/expansion of gases into available volume (3) and combustion of carbon to CO₂ (4). Flow of heat from colder to warmer body (1) and spontaneous contraction of gas into a corner (2) are non-spontaneous.

Q.18. For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression

A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
(1) Work done at 600 K is 20 times the work done at 300 K.
(2) Work done at 300 K is twice the work done at 600 K.
(3) Work done at 600 K is twice the work done at 300 K.
(4) ∆U = 0 in both cases.

Ans. (3, 4)

Sol.
For isothermal reversible expansion of ideal gas, w = -nRT ln(V_f/V_i). For same n and V_f/V_i=10, w ∝ T. Thus work at 600 K is twice that at 300 K. For an ideal gas in isothermal process ΔU = 0.

Q.19. Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
2Zn(s) + O₂ (g) → 2ZnO (s) ; ∆H = - 693.8 kJ mol^-1
(1) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(2) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(3) 693.8 kJ mol^-1 energy is evolved in the reaction.
(4) 693.8 kJ mol^-1 energy is absorbed in the reaction.

Ans. (1, 3)

Sol.
Δ_rH = ΣΔ_fH(products) - ΣΔ_fH(reactants). Given Δ_rH is -693.8 kJ, so 693.8 kJ heat is evolved; enthalpy of products (2 mol ZnO) is lower than that of reactants by 693.8 kJ.

Short Answer Type Questions

Q.20. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol^-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

Ans.

Sol.
18 g H₂O = 1 mol, ΔH_vap = 40.79 kJ mol^-1.
Enthalpy change for vapourising 2 moles = 2 × 40.79 = 81.58 kJ.
Standard enthalpy of vapourisation ΔH°_vap = 40.79 kJ mol^-1.

Q.21. One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?

Ans.

Sol.
Water has stronger intermolecular hydrogen bonding than acetone, therefore water has higher enthalpy of vapourisation.

Q.22. Standard molar enthalpy of formation, ∆_fH^Θ is just a special case of enthalpy of reaction, ∆_rH^Θ. Is the ∆_rH^Θ for the following reaction same as ∆_fH^Θ? Give reason for your answer.
CaO(s) + CO₂(g) → CaCO₃(s); ∆_fH^Θ = -178.3 kJ mol^-1

Ans.

Sol.
Standard enthalpy of formation is the enthalpy change when one mole of compound forms from its constituent elements in their standard states. The given reaction forms CaCO₃ from CaO and CO₂, not from elemental Ca, C and O₂. Therefore Δ_rH^Θ for this reaction is not the same as Δ_fH^Θ of CaCO₃.

Q.23. The value of ∆_fH^Θ for NH₃ is -91.8 kJ mol^-1. Calculate enthalpy change for the following reaction:
2NH₃(g) → N₂(g) + 3H₂(g)

Ans.

Sol.
Given Δ_fH°(NH₃) = -91.8 kJ mol^-1 per mole of NH₃ formed:
N₂ + 3H₂ → 2NH₃; ΔH° = 2 × (-91.8) = -183.6 kJ.
Reverse reaction 2NH₃ → N₂ + 3H₂ has ΔH° = +183.6 kJ mol^-1.

Q.24. Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆_rH and ∆_rH₁, ∆_rH₂, ∆_rH₃ ..... represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆_rH for overall reaction and ∆_rH₁ , ∆_rH₂ ..... etc. for intermediate reactions.

Ans.

Sol.
Hess's law: Δ_rH (overall) = Σ Δ_rH_i for the intermediate steps. The total enthalpy change is the sum of enthalpy changes of steps, independent of pathway.

Q.25. The enthalpy of atomisation for the reaction CH₄(g) → C(g) + 4H (g) is 1665 kJ mol^-1. What is the bond energy of C-H bond?

Ans.

Sol.
Total atomisation energy = 1665 kJ for breaking 4 C-H bonds.
Bond energy (C-H) = 1665 / 4 = 416.25 kJ mol^-1 (approx. 416.2 kJ mol^-1).

Q.26. Use the following data to calculate ∆_latticeH^Θ for NaBr.
∆_subH^Θ for sodium metal = 108.4 kJ mol^-1
Ionization enthalpy of sodium = 496 kJ mol^-1
Electron gain enthalpy of bromine = - 325 kJ mol^-1
Bond dissociation enthalpy of bromine = 192 kJ mol^-1
∆_fH^Θ for NaBr (s) = - 360.1 kJ mol^-1

Ans.

Sol.
Steps to form NaBr from elements (Born-Haber cycle):

1. Na(s) → Na(g), Δ_subH° = 108.4 kJ mol^-1.

2. Na(g) → Na⁺ + e^-, Δ_iH° = 496 kJ mol^-1.

3. 1/2 Br₂(g) → Br(g), 1/2 Δ_dissH° = 192/2 = 96 kJ mol^-1.

4. Br(g) + e^- → Br^-, Δ_egH° = -325 kJ mol^-1.

Sum with lattice enthalpy gives formation enthalpy:

Δ_fH° = Δ_subH° + Δ_iH° + 1/2Δ_dissH° + Δ_egH° + Δ_latticeH°.

So Δ_latticeH° = Δ_fH° - (108.4 + 496 + 96 - 325) kJ mol^-1.

Compute:

108.4 + 496 + 96 - 325 = 375.4 kJ mol^-1.

Δ_latticeH° = -360.1 - 375.4 = -735.5 kJ mol^-1.

Q.27. Given that ∆H = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?

Ans.

Sol.
Although ΔH = 0, mixing increases disorder (ΔS > 0). Therefore ΔG = ΔH - TΔS = -TΔS < 0, so mixing is spontaneous.

Q.28. Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.

Ans.

Sol.
The thermodynamic relation is ΔS = q_rev / T, where ΔS is change in entropy, q_rev is heat exchanged reversibly, and T is absolute temperature.

Q.29. Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?

Ans.

Sol.
Yes. Thermal equilibrium implies temperatures are equal. Energy lost by system equals energy gained by surroundings (neglecting work), so enthalpy decrease of system equals enthalpy increase of surroundings, and temperatures equalise at equilibrium.

Q.30. At 298 K. K_p for the reaction N₂O₄ (g) ⇌ 2NO₂ (g) is 0.98. Predict whether the reaction is spontaneous or not.

Ans.

Sol.
Δ_rG° = -RT ln K_p. Since ln(0.98) < 0, -RT ln(0.98) > 0. Thus Δ_rG° is positive and the reaction under standard conditions is non-spontaneous.

Q.31. A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Figure. What will be the value of ∆H for the cycle as a whole?

Ans.

Sol.
For any cyclic process, state functions return to the initial value; therefore ΔH = 0 for the whole cycle.

Q.32. The standard molar entropy of H₂O (l) is 70 J K^-1 mol^-1. Will the standard molar entropy of H₂O(s) be more, or less than 70 J K^-1 mol^-1?

Ans.

Sol.
Ice (solid water) is more ordered than liquid water, so its entropy is lower. Therefore standard molar entropy of H₂O(s) < 70 J K^-1 mol^-1.

Q.33. Identify the state functions and path functions out of the following: enthalpy, entropy, heat, temperature, work, free energy.

Ans.

Sol.
State functions: enthalpy (H), entropy (S), temperature (T), free energy (G).
Path functions: heat (q), work (w).

Short Answer Type Questions (continued)

Q.34. The molar enthalpy of vapourisation of acetone is less than that of water. Why?

Ans.

Sol.
Water molecules form strong hydrogen bonds, requiring more energy to vaporise. Acetone has weaker intermolecular forces (dipole-dipole, van der Waals), so its molar enthalpy of vaporisation is smaller.

Q.35. Which quantity out of ∆_rG and ∆_rG^Θ will be zero at equilibrium?

Ans.

Sol.
At equilibrium Δ_rG = 0. The standard Gibbs energy change Δ_rG° is related by Δ_rG = Δ_rG° + RT ln Q. At equilibrium Q = K and Δ_rG = 0, so Δ_rG° = -RT ln K. Δ_rG° equals zero only if K = 1.

Q.36. Predict the change in internal energy for an isolated system at constant volume.

Ans.

Sol.
For an isolated system q = 0 and w = 0. From first law ΔU = q + w = 0. Therefore internal energy does not change.

Q.37. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

Ans.

Sol.
At constant volume, q_v = ΔU, so heat at constant volume equals change in internal energy (a state function). At constant pressure, q_p = ΔH, so heat at constant pressure equals enthalpy change (a state function). Thus under constant-V or constant-p conditions the measured heat is path-independent.

Q.38. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

Ans.

Sol.
External pressure p_ext = 0, so work w = -p_extΔV = 0.
Isothermal free expansion for an ideal gas has q = 0 (no heat exchange in adiabatic vacuum expansion), so ΔU = q + w = 0 + 0 = 0.

Q.39. Heat capacity (C_p) is an extensive property but specific heat (c) is an intensive property. What will be the relation between C_p and c for 1 mol of water?

Ans.

Sol.
Molar mass of water ≈ 18 g mol^-1. Specific heat c ≈ 4.18 J g^-1 K^-1.
Molar heat capacity C_p = mass (g per mol) × c = 18 × 4.18 J K^-1 = 75.24 J K^-1 mol^-1.

Q.40. The difference between C_P and C_V can be derived using the empirical relation H = U + pV. Calculate the difference between C_P and C_V for 10 moles of an ideal gas.

Ans.

Sol.
For ideal gases, C_p - C_v = nR.
For n = 10 mol and R = 8.314 J mol^-1 K^-1, C_p - C_v = 10 × 8.314 = 83.14 J K^-1.

Q.41. If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

Ans.

Sol.
Heat released per 1 g = 20.7 kJ; molar mass of C ≈ 12 g mol^-1.
Molar ΔH = -20.7 × 12 = -248.4 kJ mol^-1 (negative sign denotes exothermic reaction).

Q.42. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
H₂(g) + Br₂(g) → 2HBr(g)
Given that Bond energy of H₂, Br₂ and HBr is 435 kJ mol^-1, 192 kJ mol^-1 and 368 kJ mol^-1 respectively.

Ans.

Sol.
Δ_rH ≈ Σ B.E.(bonds broken) - Σ B.E.(bonds formed).
= B.E.(H₂) + B.E.(Br₂) - 2 × B.E.(HBr).
= 435 + 192 - 2 × 368 = 627 - 736 = -109 kJ mol^-1. (Exothermic.)

Q.43. The enthalpy of vapourisation of CCl₄ is 30.5 kJ mol^-1. Calculate the heat required for the vapourisation of 284 g of CCl₄ at constant pressure. (Molar mass of CCl₄ = 154 g mol^-1).

Ans.

Sol.
1 mol (154 g) requires 30.5 kJ. For 284 g:

Number of moles = 284 / 154 ≈ 1.844 mol.

Heat required = 1.844 × 30.5 ≈ 56.25 kJ.

Q.44. The enthalpy of reaction for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l) is ∆_rH^Θ = - 572 kJ mol^-1.
What will be standard enthalpy of formation of H₂O(l) ?

Ans.

Sol.
The given reaction forms 2 mol H₂O(l) from elements. Standard enthalpy of formation Δ_fH° for one mole of H₂O(l) is half the reaction enthalpy:
Δ_fH°(H₂O,l) = -572 / 2 = -286 kJ mol^-1.

Q.45. What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, p_ext in a single step as shown in Figure. Explain graphically.

Ans.

Sol.
Work done w = -p_ext (V_f - V_i). Graphically, in a p-V diagram the work equals the area under the horizontal line at p_ext between V_i and V_f (the shaded rectangle ABV_IV_II).

Q.46. How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?

Ans.

Sol.
When compression is performed reversibly (infinitesimal steps), the external pressure equals internal pressure at each step. The work is the integral w = -∫_Vi^V_f p dV. Graphically this equals the area under the reversible p-V curve between V_i and V_f.

Q.47. Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof

In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

Ans.

Sol.
(a) Throwing a stone to a higher point requires input of energy; enthalpy (or potential energy) increases. Graphically the energy increases from initial to final state.

(b) For an exothermic reaction where heat is evolved, enthalpy decreases; enthalpy change contributes to spontaneity when ΔH is favourable (negative). Graphically:

Q.48. Enthalpy diagram for a particular reaction is given in Figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain.

Ans.

Sol.
No. Enthalpy change alone does not completely determine spontaneity. Both enthalpy and entropy changes must be considered via ΔG = ΔH - TΔS to decide spontaneity.

Q.49. 1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

Ans.

Sol.
The process shown is reversible (carried out in infinite small steps). For isothermal reversible expansion:

w = -nRT ln(V_f/V_i).

With n = 1 mol, R = 8.314 J mol^-1 K^-1, T = 298 K, and V_f/V_i = 2:

w = -(1)(8.314)(298) ln 2 ≈ -1717.46 J.

Q.50. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

Ans.

Sol.
Work w = -p_ext(V_f - V_i) = -2 bar × (50 - 10) L = -2 × 40 L·bar = -80 L·bar = -80 × 100 J = -8000 J = -8.0 kJ. Negative sign indicates work done by the gas. Reversible isothermal expansion does more work (larger magnitude) because p_ext is infinitesimally smaller than internal pressure at every step; thus |w| (reversible) > |w| (irreversible constant p_ext).

Matching Answer Type Questions

Q.51. In the following questions more than one correlation is possible between options of both columns. Match the following :

Ans. (i) → (e); (ii) → (d); (iii) → (f); (iv) → (a); (v) → (g), (k), (l); (vi) → (b); (vii) → (c); (viii) → (j); (ix) → (h); (x) → (i); (xi) → (a), (l), (m); (xii) → (g), (k)

Q.52. Match the following processes with entropy change:

Ans.

Q.53. Match the following parameters with description for spontaneity:

Ans. (i) → (c); (ii) → (a); (iii) → (b)

Solution.

Q.54. Match the following :

Ans. (i) → (b), (d); (ii) → (b); (iii) → (c); (iv) → (a)

Solution.

Assertion and Reason Answer Type Questions

Q.55. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R): The enthalpies of all elements in their standard state are zero.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

Ans. (2)

Sol.
Assertion is generally true: combustion of many organic compounds is exothermic because products (CO₂, H₂O) are at lower enthalpy than reactants. Reason is true as a convention (standard enthalpy of elements in their most stable forms is taken as zero) but it is not the direct explanation for why combustion is exothermic.

Q.56. Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R): Decrease in enthalpy is a contributory factor for spontaneity.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

Ans. (2)

Sol.
Assertion: Spontaneous processes are typically irreversible; they can be reversed only by external work or by changing conditions; this is true. Reason: decrease in enthalpy (ΔH < 0) often favours spontaneity but is not the sole criterion (ΔS also matters). So R is true but not the full explanation for A.

Q.57. Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R): In crystals, molecules organise in an ordered manner.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

Ans. (1)

Sol.
When a liquid crystallises, the system becomes more ordered, thus entropy decreases. The reason correctly explains the assertion.

Long Answer Type Questions

Q.58. Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.

Ans.

Sol.
Heat absorbed at constant pressure q_p is related to changes in internal energy and expansion work.

From first law: ΔU = q + w.

At constant pressure, work w = -pΔV, so q_p = ΔU + pΔV.

Define enthalpy H = U + pV.

Therefore q_p = (U₂ + pV₂) - (U₁ + pV₁) = ΔH.

So ΔH = ΔU + Δ(pV). For constant pressure, ΔH = ΔU + pΔV.

For reactions involving gases and ideal gas behaviour, pΔV = Δn_gRT where Δn_g = moles of gaseous products - moles of gaseous reactants. Hence:

ΔH = ΔU + Δn_gRT.

Each term meaning:

ΔH: enthalpy change (heat at constant pressure); ΔU: change in internal energy (sum of kinetic and potential energies of molecules); pΔV: pressure-volume work; Δn_gRT: correction for change in moles of gas under ideal behaviour.

Q.59. Extensive properties depend on the quantity of matter but intensive properties do not. Explain whether the following properties are extensive or intensive.
Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.

Ans.

Sol.
Extensive properties (depend on system size): mass, internal energy (U), enthalpy (H), heat capacity (C), volume, total entropy (S).
Intensive properties (independent of system size): pressure (p), temperature (T), density (ρ), mole fraction (x), molarity (c), specific heat (per unit mass), molar heat capacity (per mole).
Molar or specific quantities are intensive because they are normalized per amount of substance or mass.

Q.60. Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (p_i, V_i) to (p_f , V_f). With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure p_f.

Ans.

Sol.
(i) For reversible isothermal expansion, work is the area under the curved p-V isotherm from V_i to V_f. This area can be partitioned into segments represented by IMG_38 and IMG_39; their sum gives the reversible work magnitude.

(ii) For expansion against constant external pressure p_f, work equals the rectangle area p_f(V_f - V_i) shown in IMG_40. The reversible isothermal work (integral under the curve) has greater magnitude (more negative w, i.e., more work done by the system) than the irreversible expansion against constant p_f when going between the same end states. Thus |w|_reversible > |w|_constant-p.