Chapter 12: Thermodynamics
Physics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master the concepts of Laws of thermodynamics and heat engine.
Question 12.1
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is \(4.0 \times 10^4 \, \text{J/g}\)?
Answer & Explanation:
Given: Water flow rate = 3.0 L/min = 3000 g/min, ΔT = 77 – 27 = 50°C
Specific heat of water, s = 4.186 J/g°C ≈ 4.2 J/g°C
Heat required per minute: Q = m s ΔT = 3000 × 4.2 × 50 = 6.3×10⁵ J/min
Heat of combustion = 4.0×10⁴ J/g
Fuel consumed per minute = Q / (heat of combustion) = 6.3×10⁵ / 4.0×10⁴
= 15.75 g/min
Specific heat of water, s = 4.186 J/g°C ≈ 4.2 J/g°C
Heat required per minute: Q = m s ΔT = 3000 × 4.2 × 50 = 6.3×10⁵ J/min
Heat of combustion = 4.0×10⁴ J/g
Fuel consumed per minute = Q / (heat of combustion) = 6.3×10⁵ / 4.0×10⁴
= 15.75 g/min
Question 12.2
What amount of heat must be supplied to \(2.0 \times 10^{-2} \, \text{kg}\) of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N₂ = 28; R = 8.3 J mol⁻¹ K⁻¹).
Answer & Explanation:
Mass of N₂ = 0.02 kg = 20 g
Number of moles, μ = 20/28 = 0.714 mol
For diatomic gas at constant pressure: Cₚ = (7/2)R
Q = μ Cₚ ΔT = 0.714 × (7/2) × 8.3 × 45
≈ 0.714 × 3.5 × 8.3 × 45
≈ 933 J
Alternatively, using Cₚ ≈ 29 J/mol·K for N₂:
Q = 0.714 × 29 × 45 ≈ 933 J
Number of moles, μ = 20/28 = 0.714 mol
For diatomic gas at constant pressure: Cₚ = (7/2)R
Q = μ Cₚ ΔT = 0.714 × (7/2) × 8.3 × 45
≈ 0.714 × 3.5 × 8.3 × 45
≈ 933 J
Alternatively, using Cₚ ≈ 29 J/mol·K for N₂:
Q = 0.714 × 29 × 45 ≈ 933 J
Question 12.3
Explain why:
(a) Two bodies at different temperatures T₁ and T₂ when brought in thermal contact do not necessarily settle to the mean temperature (T₁+T₂)/2.
(b) Coolant in a plant should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) Climate of a harbour town is more temperate than a desert town at same latitude.
(a) Two bodies at different temperatures T₁ and T₂ when brought in thermal contact do not necessarily settle to the mean temperature (T₁+T₂)/2.
(b) Coolant in a plant should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) Climate of a harbour town is more temperate than a desert town at same latitude.
Answer & Explanation:
(a) Final temperature depends on masses and specific heats of the bodies, not just the mean of initial temperatures.
(b) High specific heat means coolant can absorb more heat per unit mass, making it more efficient.
(c) Driving increases tyre temperature (due to friction), increasing air pressure (by Gay-Lussac's law: P ∝ T at constant V).
(d) Water has high specific heat; harbour areas experience moderating effect of sea breezes, reducing temperature extremes.
(b) High specific heat means coolant can absorb more heat per unit mass, making it more efficient.
(c) Driving increases tyre temperature (due to friction), increasing air pressure (by Gay-Lussac's law: P ∝ T at constant V).
(d) Water has high specific heat; harbour areas experience moderating effect of sea breezes, reducing temperature extremes.
Question 12.4
A cylinder with a movable piston contains 3 moles of hydrogen at STP. The walls and piston are insulated. By what factor does pressure increase if gas is compressed to half its original volume?
Answer & Explanation:
Since the process is adiabatic (insulated) and hydrogen is diatomic (γ = 7/5 = 1.4):
For adiabatic process: P₁V₁ᵞ = P₂V₂ᵞ
Given V₂ = V₁/2
P₂/P₁ = (V₁/V₂)ᵞ = (2)¹·⁴
2¹·⁴ = 2^(7/5) = (2^7)^(1/5) = 128^(0.2)
≈ 2.639
Pressure increases by a factor of about 2.64.
For adiabatic process: P₁V₁ᵞ = P₂V₂ᵞ
Given V₂ = V₁/2
P₂/P₁ = (V₁/V₂)ᵞ = (2)¹·⁴
2¹·⁴ = 2^(7/5) = (2^7)^(1/5) = 128^(0.2)
≈ 2.639
Pressure increases by a factor of about 2.64.
Question 12.5
In changing state of a gas adiabatically from A to B, work done on system is 22.3 J. If gas is taken from A to B via another process where net heat absorbed is 9.35 cal, what is net work done by system in latter case? (1 cal = 4.19 J)
Answer & Explanation:
For adiabatic process: ΔU = –Won = –22.3 J (since Q=0)
So, ΔU = –22.3 J for A→B.
In second process: Q = 9.35 cal = 9.35 × 4.19 ≈ 39.18 J
By First Law: ΔU = Q – Wby
–22.3 = 39.18 – Wby
Wby = 39.18 + 22.3 = 61.48 J
Net work done by system ≈ 61.5 J.
So, ΔU = –22.3 J for A→B.
In second process: Q = 9.35 cal = 9.35 × 4.19 ≈ 39.18 J
By First Law: ΔU = Q – Wby
–22.3 = 39.18 – Wby
Wby = 39.18 + 22.3 = 61.48 J
Net work done by system ≈ 61.5 J.
Question 12.6
Two cylinders A and B of equal capacity connected via stopcock. A has gas at STP; B is evacuated. System is insulated. Stopcock opened suddenly. Find:
(a) Final pressure
(b) Change in internal energy
(c) Change in temperature
(d) Do intermediate states lie on P-V-T surface?
(a) Final pressure
(b) Change in internal energy
(c) Change in temperature
(d) Do intermediate states lie on P-V-T surface?
Answer & Explanation:
(a) Free expansion into vacuum: work done = 0, heat exchange = 0 ⇒ ΔU = 0.
For ideal gas, U depends only on T ⇒ T remains constant.
Initially: P₁V (in A) = nRT
Finally: P₂(2V) = nRT (since volume doubles)
⇒ P₂ = P₁/2 = 1 atm / 2 = 0.5 atm
(b) ΔU = 0 (since T constant for ideal gas)
(c) ΔT = 0
(d) No, intermediate states are non-equilibrium states; gas is not in uniform pressure/temperature, so they do not lie on P-V-T surface.
For ideal gas, U depends only on T ⇒ T remains constant.
Initially: P₁V (in A) = nRT
Finally: P₂(2V) = nRT (since volume doubles)
⇒ P₂ = P₁/2 = 1 atm / 2 = 0.5 atm
(b) ΔU = 0 (since T constant for ideal gas)
(c) ΔT = 0
(d) No, intermediate states are non-equilibrium states; gas is not in uniform pressure/temperature, so they do not lie on P-V-T surface.
Question 12.7
A steam engine delivers 5.4×10⁸ J of work per minute and absorbs 3.6×10⁹ J of heat per minute from boiler. What is efficiency? How much heat wasted per minute?
Answer & Explanation:
Work output per minute, W = 5.4×10⁸ J
Heat absorbed per minute, Q₁ = 3.6×10⁹ J
Efficiency η = W/Q₁ = (5.4×10⁸) / (3.6×10⁹) = 0.15 = 15%
Heat wasted, Q₂ = Q₁ – W = 3.6×10⁹ – 5.4×10⁸ = 3.06×10⁹ J/min
= 3.06×10⁹ J/min
Heat absorbed per minute, Q₁ = 3.6×10⁹ J
Efficiency η = W/Q₁ = (5.4×10⁸) / (3.6×10⁹) = 0.15 = 15%
Heat wasted, Q₂ = Q₁ – W = 3.6×10⁹ – 5.4×10⁸ = 3.06×10⁹ J/min
= 3.06×10⁹ J/min
Question 12.8
An electric heater supplies heat to a system at 100 W. If system performs work at 75 J/s, at what rate is internal energy increasing?
Answer & Explanation:
Heat supplied per second, ΔQ/Δt = 100 J/s
Work done by system per second, ΔW/Δt = 75 J/s
First Law: ΔU = Q – W
Rate form: dU/dt = dQ/dt – dW/dt
= 100 – 75 = 25 J/s
Internal energy increases at 25 W.
Work done by system per second, ΔW/Δt = 75 J/s
First Law: ΔU = Q – W
Rate form: dU/dt = dQ/dt – dW/dt
= 100 – 75 = 25 J/s
Internal energy increases at 25 W.
Question 12.9
A thermodynamic system is taken from D to E by linear process, then from E to F isobarically back to original volume. Calculate total work done by gas from D→E→F. (Refer to Fig. 12.13 P–V diagram.)
Answer & Explanation:
From D→E (linear expansion): Work = area under DE = trapezium area
WDE = ½ × (P_D + P_E) × (V_E – V_D)
From graph: P_D = 600 N/m², P_E = 300 N/m², V_E – V_D = 5 – 2 = 3 m³
WDE = ½ × (600+300) × 3 = 1350 J
From E→F (isobaric compression at P=300 N/m²):
WEF = P × (V_F – V_E) = 300 × (2 – 5) = –900 J
Total work W = WDE + WEF = 1350 – 900 = 450 J
WDE = ½ × (P_D + P_E) × (V_E – V_D)
From graph: P_D = 600 N/m², P_E = 300 N/m², V_E – V_D = 5 – 2 = 3 m³
WDE = ½ × (600+300) × 3 = 1350 J
From E→F (isobaric compression at P=300 N/m²):
WEF = P × (V_F – V_E) = 300 × (2 – 5) = –900 J
Total work W = WDE + WEF = 1350 – 900 = 450 J
Question 12.10
A refrigerator maintains inside temperature at 9°C. Room temperature is 36°C. Calculate coefficient of performance.
Answer & Explanation:
T₂ = 9°C = 282 K (cold reservoir)
T₁ = 36°C = 309 K (hot reservoir)
For Carnot refrigerator:
Coefficient of performance α = T₂/(T₁ – T₂)
= 282 / (309 – 282) = 282 / 27
≈ 10.44
The refrigerator can theoretically transfer about 10.44 J of heat from inside per joule of work input.
T₁ = 36°C = 309 K (hot reservoir)
For Carnot refrigerator:
Coefficient of performance α = T₂/(T₁ – T₂)
= 282 / (309 – 282) = 282 / 27
≈ 10.44
The refrigerator can theoretically transfer about 10.44 J of heat from inside per joule of work input.
📘 Exam Preparation Tip:
These exercise questions cover the core of thermodynamics: First Law, specific heats, adiabatic/isothermal processes, heat engines, refrigerators, Carnot efficiency, and practical applications. Solving them builds conceptual clarity for board exams and competitive tests like NEET/JEE, linking theory to real-world systems like engines and cooling devices.