Exercises
1: A geyser heats water at 3.0 litres/min from 27 °C to 77 °C. Heat of combustion of gas = 4.0 × 10⁴ J/g. Find rate of fuel consumption.
Answer
Given: Flow rate = 3.0 L/min = 3000 g/min, ΔT = 77 − 27 = 50 °C, cwater = 4.2 J/g/°C
Heat needed per minute = m·c·ΔT = 3000 × 4.2 × 50 = 6.3 × 10⁵ J/min
Rate of fuel consumption = (6.3 × 10⁵) / (4.0 × 10⁴) = 15.75 g/min ≈ 15.8 g/min
2: Heat to raise 2.0 × 10⁻² kg of N₂ by 45 °C at constant pressure. M = 28, R = 8.3 J mol⁻¹ K⁻¹.
Answer
Number of moles: n = 0.020/28 = 7.14 × 10⁻⁴ mol... wait, n = 20 g / 28 g/mol = 0.714 mol
For diatomic gas, CP = (7/2)R = 3.5 × 8.3 = 29.05 J mol⁻¹ K⁻¹
Q = n·CP·ΔT = 0.714 × 29.05 × 45 = 0.714 × 1307.25 = 933 J ≈ 933.4 J
3: Explain why
Answer
(a) Two bodies at T₁ and T₂ in thermal contact don't necessarily reach (T₁ + T₂)/2: They reach thermal equilibrium at a temperature determined by heat balance: m₁c₁(T − T₁) = m₂c₂(T₂ − T). Unless m₁c₁ = m₂c₂, the final temperature is not the mean.
(b) Coolant should have high specific heat: High c means it can absorb large amounts of heat with minimal temperature rise, so it maintains effective cooling over a wide temperature range without frequently needing replacement.
(c) Tyre pressure increases during driving: Friction between tyre and road generates heat, raising the temperature of air in the tyre. By Gay-Lussac's law (P ∝ T at constant V), the pressure increases.
(d) Harbour towns have more temperate climate: Water has a very high specific heat capacity. The sea absorbs large amounts of heat in summer (cooling coastal areas) and releases it in winter (warming them). Deserts have sand/rock with low specific heat — they heat up and cool down rapidly, giving extreme temperatures.
4: Cylinder with 3 moles of H₂ at STP, insulated walls. Compressed to half volume adiabatically. By what factor does pressure increase?
Answer
For adiabatic process: PVγ = constant
For H₂ (diatomic): γ = CP/CV = (7/2)R / (5/2)R = 7/5 = 1.4
P₁V₁γ = P₂V₂γ
P₂/P₁ = (V₁/V₂)γ = (2)1.4 = 21.4
21.4 = 2¹ × 20.4 = 2 × 1.3195 = 2.639
Pressure increases by a factor of 21.4 ≈ 2.64
5: Gas taken adiabatically A → B with 22.3 J work done on it. Via another path, net heat absorbed = 9.35 cal. Find net work done by gas in second case. (1 cal = 4.19 J)
Answer
First path (adiabatic A
⇒ B): Q = 0, Won = 22.3 J
⇒ ΔU = Won = 22.3 J
Internal energy change ΔU = 22.3 J (same for any path between same states)
Second path: Qabsorbed = 9.35 cal = 9.35 × 4.19 = 39.18 J
First law: ΔU = Q − W
⇒ W = Q − ΔU = 39.18 − 22.3 = 16.88 J ≈ 16.9 J
6: Cylinder A (gas at STP) connected to evacuated cylinder B via stopcock. Entire system insulated. Stopcock suddenly opened.
Answer
(a) Final pressure: Gas expands to fill both cylinders (equal capacity), so volume doubles. Since the system is insulated and no work is done against external pressure (free expansion), by ideal gas law: P₂ = P₁/2 = 0.5 atm (standard/2)
(b) Change in internal energy: No heat exchange (insulated), no work done (free expansion into vacuum). By first law: ΔU = Q − W = 0 − 0 = ΔU = 0
(c) Change in temperature: For an ideal gas, U depends only on T. Since ΔU = 0, ΔT = 0 (temperature unchanged).
(d) Intermediate states: The process is irreversible — the gas is not in equilibrium during expansion. The intermediate states are not on the P-V-T surface (which represents equilibrium states only).
7: Electric heater supplies 100 W to a system. System performs work at 75 J/s. Rate of increase of internal energy?
Answer
First law: dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 J/s = 25 W
8: Thermodynamic system taken D → E (linear process on P-V diagram), then E → F (isobaric, volume reduced to original). Calculate total work done by gas from D to E to F.
Answer
Work done in a thermodynamic process = area under P-V graph.
D
⇒ E (linear process): This forms a trapezoid/triangle on P-V diagram. Work = area of the trapezoid = (1/2)(PD + PE)(VE − VD)
E
⇒ F (isobaric compression): Work = PE(VF − VE) = PE(VD − VE) [negative, since volume decreases]
Based on the figure (Fig. 11.13): The D
⇒ E process goes from lower-left to upper-right on a P-V diagram. Taking typical values from the figure where the linear process goes from state D (PD, VD) to E (PE, VE):
WDE = area under line DE = (1/2)(PD + PE)(VE − VD)
WEF = PE(VD − VE) (compression, negative)
Total W = WDE + WEF = area of triangle DEF = (1/2)(PE − PD)(VE − VD)
Note: The exact numerical answer depends on the values read from Fig. 11.13 in the textbook.
Important Extra Questions
Extra Q1: State and explain the First Law of Thermodynamics. How does it relate to conservation of energy?
Answer:
First Law: ΔU = Q − W
The change in internal energy of a system equals the heat added to the system minus the work done by the system.
This is a statement of conservation of energy: heat (Q) and work (W) are both forms of energy exchange. Energy cannot be created or destroyed — it only converts between heat, work, and internal energy.
Sign convention: Q > 0 when heat flows into system; W > 0 when system does work on surroundings.
Applications: Isothermal: ΔU = 0, Q = W. Adiabatic: Q = 0, ΔU = −W. Isochoric: W = 0, ΔU = Q.
Extra Q2: What is a Carnot engine? State the efficiency formula and explain why no real engine can achieve 100% efficiency.
Answer:
A Carnot engine is an ideal heat engine operating between two temperatures TH (hot source) and TC (cold sink) in a reversible cycle: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.
Carnot efficiency: η = 1 − TC/TH
Why not 100%? η = 1 only if TC = 0 K (absolute zero) or TH = ∞, both physically impossible. The Second Law prohibits complete conversion of heat to work in a cyclic process — some heat must always be rejected to a cold reservoir.
Extra Q3: Distinguish between isothermal and adiabatic processes. Compare the P-V curves for each.
Answer:
Isothermal process: Temperature constant (T = const). Heat can exchange with surroundings. For ideal gas: PV = nRT = const → P ∝ 1/V (rectangular hyperbola on P-V graph).
Adiabatic process: No heat exchange with surroundings (Q = 0). Temperature changes. For ideal gas: PVγ = const. Since γ > 1, the adiabatic curve is steeper than the isothermal curve on a P-V diagram at the same state point.
In both cases, as volume increases, pressure decreases — but the adiabatic process has a greater drop in pressure (because temperature also falls, reducing additional pressure).
Extra Q4: State the Second Law of Thermodynamics in two equivalent forms.
Answer:
Kelvin-Planck statement: No heat engine operating in a cycle can absorb heat from a single reservoir and convert it entirely into work. Some heat must always be rejected to a colder reservoir.
Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body without external work being done on the system.
Both statements are equivalent — violating one implies violating the other. They both imply the existence of entropy as a state function that always increases for irreversible processes in an isolated system.
Extra Q5: What is a heat pump? How does its coefficient of performance (COP) compare with a refrigerator?
Answer:
A heat pump extracts heat from a cold source (e.g., outdoor air) and delivers it to a warm space (e.g., a room). It is essentially a refrigerator running in reverse for the purpose of heating.
COP of refrigerator: COPR = QC/W = TC/(TH − TC)
COP of heat pump: COPHP = QH/W = TH/(TH − TC)
Since QH = QC + W: COPHP = COPR + 1
The heat pump always has a higher COP than the refrigerator by exactly 1. For typical conditions, COPHP ≫ 1, meaning a heat pump delivers more heat energy than the electrical energy consumed — making it more efficient than direct electrical heating.