CBSE Class 12 Physics Heat and Thermodynamics Objective Questions

CBSE Class 12 Physics Heat and Thermodynamics Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

OBJECTIVES
CLASS 12 PHYSICS : HEAT AND THERMODYNAMICS

Choose the correct option(s):

1. Two identical rectangular strips, one of copper and the other of steel, are rivetted together to form a bimetallic strip. On heating this strip, it will (α_copper > α_steel)
(a) remain straight
(b) bend with copper on the convex side
(c) bend with steel on the convex side
(d) get twisted

Answer: (b) bend with copper on the convex side

1. On heating, copper expands more than steel because α_copper is greater.
2. The strip bends so that the longer, more expanded metal lies on the outer (convex) side.
3. Hence, the copper side is convex and the correct option is (b).

Teacher's Note:
a) The metal with the larger coefficient of expansion always ends up on the outside of the curve.
b) On cooling, the bending is reversed.

2. If I is the moment of inertia of solid body, the change in I corresponding to a small change in temperature ΔT is
(a) αI ΔT
(b) 1/2 αI ΔT
(c) 2 αI ΔT
(d) 3 αI ΔT

Answer: (c) 2 αI ΔT

1. Moment of inertia I is proportional to (length)², as I = Mk².
2. Every length increases by the factor (1 + αΔT), so I' = I(1 + αΔT)² ≈ I(1 + 2αΔT).
3. Change in I = 2αI ΔT.
4. Hence, the correct option is (c).

Teacher's Note:
a) Each length scale appears squared in I, so the change is twice the linear strain.
b) Neglect (αΔT)² as it is very small.

3. The length of the steel rod which would have the same difference in length with a copper rod of length 16 cm at all temperature (α_copper = 18 × 10^-6 K^-1, α_steel = 12 × 10^-6 K^-1) is
(a) 20 cm
(b) 18 cm
(c) 24 cm
(d) 30 cm

Answer: (c) 24 cm

1. The difference in length stays the same at all temperatures only if both rods expand by the same amount.
2. So α_copperL_copperΔT = α_steelL_steelΔT.
3. L_steel = (18 × 10⁻⁶ × 16)/(12 × 10⁻⁶) = 24 cm.
4. Hence, the correct option is (c).

Teacher's Note:
a) A constant difference needs equal expansions, not equal strains.
b) Check: 18 × 16 = 288 and 12 × 24 = 288.

4. The density of a solid at 0°C and 500°C is in the ratio 1.027 : 1. The coefficient of linear expansion of the solid is
(a) 20 × 10^-6 K^-1
(b) 12 × 10^-6 K^-1
(c) 18 × 10^-6 K^-1
(d) 24 × 10^-6 K^-1

Answer: (c) 18 × 10^-6 K^-1

1. Density is inversely proportional to volume, so ρ₀/ρ₅₀₀ = V₅₀₀/V₀ = 1.027.
2. V₅₀₀ = V₀(1 + γ × 500), so γ × 500 = 0.027 and γ = 5.4 × 10⁻⁵ K⁻¹.
3. α = γ/3 = 1.8 × 10⁻⁵ K⁻¹ = 18 × 10⁻⁶ K⁻¹.
4. Hence, the correct option is (c).

Teacher's Note:
a) The question asks for the linear coefficient, so divide γ by 3.
b) Density falls on heating, so the ratio at 0°C to 500°C is more than 1.

5. At 4°C, 98% of the volume of a body is immersed in water. The temperature at which the entire body gets immersed in water (γ_ω = 3.3 × 10^-4 K^-1) is (neglect the expansion of the body)
(a) 40.8° C
(b) 64.6° C
(c) 60.0°C
(d) 58.8°C

Answer: (b) 64.6° C

1. At 4°C the body floats with 98% of its volume under water, so ρ_body = 0.98 ρ₄.
2. The body is fully immersed when the density of water falls to ρ_body, that is ρ₄/(1 + γΔT) = 0.98 ρ₄.
3. 1 + γΔT = 1/0.98, so γΔT ≈ 0.02 and ΔT = 0.02/(3.3 × 10⁻⁴) ≈ 60.6°C.
4. T = 4 + 60.6 = 64.6°C.
5. Hence, the correct option is (b).

Teacher's Note:
a) Water becomes less dense on heating above 4°C, so the body sinks deeper.
b) The small-expansion approximation 1/(1 - 0.02) ≈ 1.02 is used.

6. If α is the coefficient of linear expansion, the change in the period t of a physical pendulum with a temperature change of ΔT is
(a) 1/2 αtΔT
(b) 1/4 αtΔT
(c) 3/4 αtΔT
(d) 1/3 αtΔT

Answer: (a) 1/2 αtΔT

1. For a physical pendulum t = 2π√(I/(mgd)), where I is proportional to L² and d is proportional to L.
2. So t is proportional to √(L²/L) = √L.
3. Fractional change in t = (1/2) × fractional change in L = (1/2)αΔT.
4. Change in period = (1/2) α t ΔT.
5. Hence, the correct option is (a).

Teacher's Note:
a) The period depends on the square root of a length, so the change is half of the strain.
b) Do not confuse this with I alone, which changes by 2αΔT.

7. Which one of the following expressions (notations having their usual meanings) does not represent ideal gas behaviour?
(a) p = ρrT
(b) pV = (m/M)RT
(c) pV =(m/M)rT
(d) p = nkT

Answer: (a) p = ρrT

1. The ideal gas equation is pV = (m/M)RT, which is option (b), and p = nkT is the same law with n as number density.
2. Dividing pV = (m/M)RT by V gives p = ρRT/M, where ρ is the density. Option (a) lacks the factor 1/M, so it does not represent the ideal gas equation when r stands for the same constant as in option (c).
3. Hence, the correct option is (a).

Teacher's Note:
a) Convert between forms by using m/V = ρ and N/V = n.
b) Check that every form has the molar mass M with the density.

8. The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures p₁ and p₂ are shown in the figure here.
(a) p₁ < p₂
(b) p₁ = p₂
(c) p₁ > p₂
(d) It cannot be ascertained.

Answer: (c) p₁ > p₂

1. For a perfect gas V = (nR/p)T, so the slope of the V-T graph is nR/p.
2. The line labelled p₂ is steeper than the line labelled p₁.
3. A larger slope means a smaller pressure, so p₂ is less than p₁.
4. Hence, the correct option is (c).

Teacher's Note:
a) At the same temperature a steeper V-T line means a larger volume, so a lower pressure.
b) Compare the slopes, not the heights of the lines.

9. The molar heat capacity of an ideal gas at constant volume is given by
(a) \( \frac{γP}{γ-1} \)
(b) \( \frac{R}{γ-1} \)
(c) γR
(d) (γ - 1)R

Answer: (b) \( \frac{R}{γ-1} \)

1. C_p - C_v = R and C_p/C_v = γ.
2. So (γ - 1)C_v = R.
3. C_v = R/(γ - 1).
4. Hence, the correct option is (b).

Teacher's Note:
a) Option (a) contains P, so it cannot be a molar heat capacity.
b) Quick check: γ = 5/3 gives C_v = 3R/2.

10. The work done by a gas in an adiabatic process is
(a) \( \frac{p1V1 - p2V2}{γ-1} \)
(b) \( \frac{p2V2 - p1V1}{γ-1} \)
(c) \( \frac{R(T1 - T2)}{γ-1} \)
(d) \( \frac{nR(T2 - T1)}{γ-1} \)

Answer: (a) \( \frac{p1V1 - p2V2}{γ-1} \) and (c) \( \frac{R(T1 - T2)}{γ-1} \)

1. In an adiabatic process W = -ΔU = nC_v(T₁ - T₂) with C_v = R/(γ - 1).
2. So W = nR(T₁ - T₂)/(γ - 1) = (p₁V₁ - p₂V₂)/(γ - 1).
3. For one mole this is the expression in option (c), and the pressure-volume form is option (a).
4. Options (b) and (d) have the opposite sign, so they give the work done on the gas.
5. Hence, the correct options are (a) and (c).

Teacher's Note:
a) The work done by the gas is positive when the gas expands and cools, so T₁ is greater than T₂.
b) Option (c) is for one mole of gas, as printed.

11. An ideal gas is compressed isothermally. The compression agent does 36 joules of work. The amount of heat that flows out from the gas is
(a) zero
(b) +36 J
(c) - 36 J
(d) infinite

Answer: (b) +36 J

1. For an ideal gas in an isothermal process ΔU = 0.
2. By the first law, Q = ΔU + W, so Q = W = -36 J (work done by the gas is -36 J during compression).
3. So 36 J of heat flows out of the gas, that is, the heat given out is +36 J.
4. Hence, the correct option is (b).

Teacher's Note:
a) The heat taken in by the gas is -36 J, so the heat flowing out is +36 J.
b) The work done on the gas leaves as heat because the internal energy cannot change.

12. In an isobaric (constant pressure) process
(a) Δ Q : Δ U = 1 : 1
(b) Δ Q : Δ U = 1 : γ = 1
(c) Δ Q : Δ U = γ - 1 : 1
(d) Δ Q : Δ U = γ : 1

Answer: (d) Δ Q : Δ U = γ : 1

1. In an isobaric process ΔQ = nC_pΔT and ΔU = nC_vΔT.
2. ΔQ : ΔU = C_p : C_v = γ : 1.
3. Hence, the correct option is (d).

Teacher's Note:
a) The internal energy change depends only on ΔT, whatever the process.
b) Heat at constant pressure is always more than ΔU because of the work done.

13. In an isobaric process
(a) Δ Q : Δ W = 1 : 1
(b) Δ Q : Δ W = γ : γ - 1
(c) Δ Q : Δ W = γ - 1 : γ
(d) Δ Q : Δ W = γ : 1

Answer: (b) Δ Q : Δ W = γ : γ - 1

1. ΔQ = nC_pΔT and ΔW = pΔV = nRΔT.
2. ΔQ : ΔW = C_p : R = C_p : (C_p - C_v).
3. Dividing by C_v gives γ : (γ - 1).
4. Hence, the correct option is (b).

Teacher's Note:
a) Use R = C_p - C_v to link the work with the heat.
b) For a monoatomic gas ΔQ : ΔW = 5 : 2, which matches γ = 5/3.

14. The internal energy of a mole of air (γ = 7/5) in a room of volume V = 40 m³ at 1 atmospheric pressure (= 10⁵ N m^-2) is
(a) 10⁵ J
(b) 10⁴ J
(c) 10⁷ J
(d) 10⁸ J

Answer: (c) 10⁷ J

1. Internal energy U = nC_vT = nRT/(γ - 1) = pV/(γ - 1).
2. U = (10⁵ × 40)/(7/5 - 1) = (4 × 10⁶)/0.4 = 10⁷ J.
3. Hence, the correct option is (c).

Teacher's Note:
a) This is the internal energy of all the air filling the room; the options are based on that reading.
b) Use pV in place of nRT when n is not given.

15. In an isobaric process
(a) the heat added is \( \frac{γ}{γ-1} \) times the work done by the gas
(b) the work done is zero
(c) the heat added is (γ -1) times the work done
(d) the heat added has no relation with the work done

Answer: (a) the heat added is \( \frac{γ}{γ-1} \) times the work done by the gas

1. ΔQ = nC_pΔT and ΔW = nRΔT.
2. ΔQ/ΔW = C_p/(C_p - C_v) = γ/(γ - 1).
3. Hence, the heat added is γ/(γ - 1) times the work done and the correct option is (a).

Teacher's Note:
a) The same ratio appears in the previous question.
b) The work is not zero because the volume changes at constant pressure.

16. For an adiabatic compression, the quantity PV
(a) increases
(b) decreases
(c) remains constant
(d) may either increase or decrease

Answer: (a) increases

1. In an adiabatic process PV^γ = constant, so PV = constant × V^1-γ.
2. Since γ is greater than 1, the power (1 - γ) is negative, so PV rises when V decreases.
3. Also PV = nRT, and the temperature rises during adiabatic compression.
4. Hence, the correct option is (a).

Teacher's Note:
a) PV is proportional to T, and adiabatic compression heats the gas.
b) PV is constant only in an isothermal process.

17. Q calories of heat is required to raise the temperature of 1 mole of a monoatomic gas from 20°C to 30°C. The amount of heat required to raise the temperature of diatomic gas from 20° C to 25° C is
(a) Q
(b) 3Q/2
(c) 5Q/3
(d) 7Q/3

Answer: (d) 7Q/3

1. For the monoatomic gas, Q = C_vΔT = (3/2)R × 10 = 15R.
2. For the diatomic gas heated at constant pressure through the same 10 K, Q' = C_pΔT = (7/2)R × 10 = 35R.
3. Q'/Q = 35/15 = 7/3, so Q' = 7Q/3.
4. Hence, the correct option is (d).

Teacher's Note:
a) The options fit a 10 K rise for the diatomic gas; the printed end temperature of 25°C gives no matching option, so it appears to be a misprint of 30°C.
b) Monoatomic C_v = 3R/2 and diatomic C_p = 7R/2.

18. The heat capacity of an ideal gas in a polytropic process is C = C_V + 0 .1 R. The value of the polytropic exponent is
(a) 11
(b) 10
(c) - 10
(d) -9

Answer: (d) -9

1. For a polytropic process pVⁿ = constant, the molar heat capacity is C = C_V + R/(1 - n).
2. Equating, R/(1 - n) = 0.1R, so 1 - n = 10.
3. n = -9.
4. Hence, the correct option is (d).

Teacher's Note:
a) Take care with the sign: 1 - n = 10 gives a negative n.
b) Check: n = 0 gives C_p and n = 1 gives infinity.

19. An ideal gas expands according to the law pV² = constant. The molar heat capacity C is
(a) C_V + R
(b) C_V - R
(c) C_V + 2R
(d) C_V - 3R

Answer: (b) C_V - R

1. The law pVⁿ = constant with n = 2 is a polytropic process.
2. C = C_V + R/(1 - n) = C_V + R/(1 - 2).
3. C = C_V - R.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use the polytropic formula directly.
b) For n greater than 1 the term R/(1 - n) is negative.

20. If W₁ is the work done in compressing an ideal gas from a given initial state through a certain volume isothermally and W₂ is the work done in compressing the same gas from the same initial state through the same volume adiabatically, then
(a) W₁ = W₂
(b) W₁ < W₂
(c) W₁ > W₂
(d) W₁ = 2 W₂

Answer: (b) W₁ < W₂

1. On a p-V diagram the adiabatic curve is steeper than the isothermal curve from the same initial point.
2. For the same decrease in volume, the pressure in the adiabatic compression is higher at every stage, so the area under it is larger.
3. Hence, W₂ is greater than W₁ and the correct option is (b).

Teacher's Note:
a) In adiabatic compression the temperature rises, which makes the gas harder to compress.
b) Remember: adiabatic slope = γ × isothermal slope.

21. The number of molecules per unit volume at pressure p and temperature T is given by (k = Boltzmann's constant)
(a) \( \frac{T}{kp} \)
(b) \( \frac{p}{kT} \)
(c) \( \frac{kp}{T} \)
(d) \( \frac{kT}{p} \)

Answer: (b) \( \frac{p}{kT} \)

1. The ideal gas law in molecular form is pV = NkT.
2. Number per unit volume N/V = p/(kT).
3. Hence, the correct option is (b).

Teacher's Note:
a) Check the dimensions: p/(kT) has the dimension of 1/volume.
b) This is the same as the relation p = nkT.

22. Three vessels A, B, and C are at the same temperature and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only O₂, B only N₂, and C, a mixture of equal quantities of O₂ and N₂. If the average speed of O₂ molecules in vessel A is v₁, that of the N₂ molecules in vessel B is v₂, the average speed of the O₂ molecules in vessel C is
(a) \( \frac{v1 + v2}{2} \)
(b) v₁
(c) \( \sqrt{v_1 v_2} \)
(d) \( \sqrt{3kT/M} \)

Answer: (b) v₁

1. The average speed of molecules of a gas depends only on the temperature and the mass of one molecule: v = √(8kT/πm).
2. All the vessels are at the same temperature, and in vessel C the O₂ molecules have the same mass as in vessel A.
3. So the average speed of O₂ molecules in C is v₁.
4. Hence, the correct option is (b).

Teacher's Note:
a) The printed key gives (d), which is wrong; √(3kT/M) is the root mean square speed, not the average speed, and does not equal v₁.
b) Mixing gases does not change the speed distribution of each gas at a given temperature.

23. A wall has two layers A and B, each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is 36° C. The temperature difference across the layer A is
(a) 6° C
(b) 12° C
(c) 18° C
(d) 24° C

Answer: (b) 12° C

1. In the steady state the same heat current flows through both layers: K_A(ΔT_A)A/d = K_B(ΔT_B)A/d.
2. With K_A = 2K_B, ΔT_B = 2ΔT_A.
3. ΔT_A + ΔT_B = 36, so 3ΔT_A = 36 and ΔT_A = 12°C.
4. Hence, the correct option is (b).

Teacher's Note:
a) The better conductor shows the smaller temperature drop.
b) This is like two resistors in series, with the drop proportional to the resistance.

24. A slab ABCD consists of two layers of thermal conductivities k₁ = 300 and k₂ = 200 SI units but of the same length and area of cross section. If the faces AD and BC are maintained at different temperature, that is, heat current flows longitudinally along the layers the equivalent thermal conductivity of the slab is
(a) 500
(b) 250
(c) 1000
(d) 120

Answer: (b) 250

1. The layers are side by side, so the heat flows through them in parallel.
2. Total heat current = (k₁A₁ + k₂A₂)ΔT/L, with A₁ = A₂ = A/2 for the whole cross section A.
3. k_eqA = (k₁ + k₂)A/2, so k_eq = (300 + 200)/2 = 250.
4. Hence, the correct option is (b).

Teacher's Note:
a) The two layers have equal area, so the equivalent conductivity is the average of the two.
b) Do not add the conductivities directly; the total area also doubles.

25. The ends of a rod of length l and thermal conductivity k are maintained at temperatures θ₁ and θ₂ and (θ₁ > θ₂). The temperature of the rod at a distance x from the hot end in the steady state is
(a) θ = \( \frac{θ1 + θ2}{2} \) x
(b) θ = θ₁ - \( \frac{θ1 - θ2}{l} \) x
(c) θ = θ₂ + \( \frac{θ1 - θ2}{l} \) x
(d) θ = θ₂ - \( \frac{θ1 - θ2}{l} \) x

Answer: (b) θ = θ₁ - \( \frac{θ1 - θ2}{l} \) x

1. In the steady state the temperature falls uniformly along the rod, with gradient (θ₁ - θ₂)/l.
2. At the hot end x = 0, θ = θ₁.
3. At distance x, θ = θ₁ - [(θ₁ - θ₂)/l] x.
4. Check: at x = l, θ = θ₂.
5. Hence, the correct option is (b).

Teacher's Note:
a) Verify with the two ends, x = 0 and x = l.
b) A linear temperature profile applies because the heat current is constant.

26. The pressure of an ideal gas is written as \( p = \frac{2E}{3V} \). Here E refers to
(a) translational kinetic energy
(b) rotational kinetic energy
(c) vibrational kinetic energy
(d) total kinetic energy.

Answer: (a) translational kinetic energy

1. From kinetic theory, p = (1/3)(N m v²/V) = (2/3)(E/V), where E = (1/2)Nmv².
2. This E is the kinetic energy due to the translational motion of the molecules only.
3. Hence, the correct option is (a).

Teacher's Note:
a) Pressure is caused by the translational motion of molecules hitting the walls.
b) Rotational and vibrational energies do not contribute to pressure.

27. The quantity \( \frac{pV}{kT} \) represents
(a) mass of the gas
(b) kinetic energy of the gas
(c) number of moles of the gas
(d) number of molecules in the gas.

Answer: (d) number of molecules in the gas.

1. For an ideal gas pV = NkT, where N is the number of molecules.
2. So pV/kT = N.
3. Hence, the correct option is (d).

Teacher's Note:
a) If R were used instead of k, the quantity would give the number of moles.
b) Check that k is Boltzmann's constant and not the gas constant.

28. Which of the following quantities is the same for all ideal gases at the same temperature?
(a) the kinetic energy of 1 mole
(b) the kinetic energy of 1 g
(c) the number of molecules in 1 mole
(d) the number of molecules in 1 g

Answer: (a) the kinetic energy of 1 mole and (c) the number of molecules in 1 mole

1. The average kinetic energy per molecule is (3/2)kT, so 1 mole has (3/2)RT, the same for all ideal gases at the same temperature.
2. 1 mole of any gas contains Avogadro's number of molecules, N_A = 6.02 × 10²³.
3. The kinetic energy of 1 g and the number of molecules in 1 g depend on the molar mass.
4. Hence, the correct options are (a) and (c).

Teacher's Note:
a) Per mole quantities are universal; per gram quantities depend on the gas.
b) This question has more than one correct option.

29. Consider the quantity \( \frac{MkT}{pV} \) of an ideal gas where M is the mass of the gas. It depends on the
(a) temperature of the gas
(b) volume of the gas
(c) pressure of the gas
(d) nature of the gas

Answer: (d) nature of the gas

1. pV = NkT, so kT/(pV) = 1/N.
2. MkT/(pV) = M/N, which is the mass of one molecule of the gas.
3. This does not depend on temperature, volume or pressure; it depends only on the nature of the gas.
4. Hence, the correct option is (d).

Teacher's Note:
a) The mass of a molecule is a property of the type of gas.
b) Substitute pV = NkT to simplify the expression.

30. Figure shows two processes A and B on a system. Let ΔQ₁ and ΔQ₂ be the heat given to the system in processes A and B respectively. Then
(a) ΔQ₁ > ΔQ₂
(b) ΔQ₁ = ΔQ₂
(c) ΔQ₁ < ΔQ₂
(d) ΔQ₁ ≤ ΔQ₂

Answer: (a) ΔQ₁ > ΔQ₂

1. Both processes join the same initial and final states, so the change in internal energy ΔU is the same.
2. The area under A is larger than the area under B, so W_A is greater than W_B.
3. ΔQ = ΔU + ΔW, so ΔQ₁ is greater than ΔQ₂.
4. Hence, the correct option is (a).

Teacher's Note:
a) ΔU is a state function, but work depends on the path.
b) The curve with more area under it needs more heat.

31. Let ΔU₁ and ΔU₂ be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B. Then
(a) ΔU₁ + ΔU₂ = 0
(b) ΔU₁ - ΔU₂ = 0
(c) ΔQ - ΔW = 0
(d) ΔQ + ΔW = 0

Answer: (a) ΔU₁ + ΔU₂ = 0 and (c) ΔQ - ΔW = 0

1. Process A takes the system from the initial state to the final state, and process B brings it back, so A + B is a cyclic process.
2. Internal energy is a state function, so ΔU₁ + ΔU₂ = 0 over the cycle.
3. By the first law, ΔQ = ΔU + ΔW = 0 + ΔW, so ΔQ - ΔW = 0.
4. Hence, the correct options are (a) and (c).

Teacher's Note:
a) In a closed cycle the net heat absorbed equals the net work done.
b) ΔU₁ = -ΔU₂, not ΔU₁ = ΔU₂.

32. Consider the process on a system shown in figure. During the process, the work done by the system
(a) continuously increases
(b) continuously decreases
(c) first increases then decreases
(d) first decreases then increases.

Answer: (a) continuously increases

1. The work done by the system is the area under the P-V curve, dW = P dV.
2. In the process the volume increases all the way, and the pressure stays positive, so each small step adds positive work.
3. Hence, the work done continuously increases and the correct option is (a).

Teacher's Note:
a) The pressure first falls and then rises, but the work still grows because dV is always positive.
b) Do not confuse the pressure with the work done.

33. An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process
(a) is positive
(b) is negative
(c) is zero
(d) cannot be obtained from this information

Answer: (c) is zero

1. The graph is a straight line through the origin on a P-T diagram, so P is proportional to T.
2. From pV = nRT, V = nR × (T/P) is constant.
3. The process is isochoric, so ΔV = 0 and W = ∫p dV = 0.
4. Hence, the correct option is (c).

Teacher's Note:
a) A straight line through the origin on a P-T graph means constant volume.
b) Work is zero whenever the volume does not change.

34. Consider two processes on a system as shown in figure. The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW₁ and ΔW₂ be the work done by the system in the processes A and B respectively.
(a) ΔW₁ > ΔW₂
(b) ΔW₁ = ΔW₂
(c) ΔW₁ < ΔW₂
(d) Nothing can be said about the relation between ΔW₁ and ΔW₂.

Answer: (c) ΔW₁ < ΔW₂

1. In the P-T graph both processes are at constant pressure (horizontal lines), so each is isobaric.
2. The volume change is the same in both, but the pressure of B is higher than that of A.
3. W = pΔV, so W_B is greater than W_A, that is, ΔW₁ is less than ΔW₂.
4. Hence, the correct option is (c).

Teacher's Note:
a) For an isobaric process the work is the product of pressure and the change in volume.
b) A higher horizontal line on a P-T graph means a higher pressure.

35. A system can be taken from the initial state p₁, V₁ to the final state p₂, V₂ by two different methods. Let ΔQ and ΔW represent the heat given to the system and the work done by the system. Which of the following must be the same is both the methods?
(a) ΔQ
(b) ΔW
(c) ΔQ + ΔW
(d) ΔQ - ΔW

Answer: (d) ΔQ - ΔW

1. By the first law, ΔU = ΔQ - ΔW.
2. ΔU is a state function, so it depends only on the initial and final states and is the same for both methods.
3. ΔQ and ΔW individually depend on the path.
4. Hence, ΔQ - ΔW is the same and the correct option is (d).

Teacher's Note:
a) Heat and work are path dependent; their difference is not.
b) Watch the sign convention: ΔW is the work done by the system.