Chapter 11 – Thermodynamics | AAI ATC Exam Prep

Q1

The Zeroth Law of Thermodynamics leads to the concept of: Easy

(A)Internal energy

(B)Entropy

(C)Temperature

(D)Heat capacity

Solution

✓ Correct: (C)

The Zeroth Law states: "Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other." This establishes the concept of temperature as the common thermodynamic variable. Formulated by R.H. Fowler in 1931.

Q2

The First Law of Thermodynamics is written as: Easy

(A)ΔQ = ΔU − ΔW

(B)ΔQ = ΔU + ΔW

(C)ΔU = ΔQ × ΔW

(D)ΔW = ΔU + ΔQ

Solution

✓ Correct: (B)

ΔQ = ΔU + ΔW

ΔQ = heat supplied to system, ΔU = change in internal energy, ΔW = work done by system. It is the general law of conservation of energy applied to thermodynamic systems. Note: ΔU is a state variable; ΔQ and ΔW are path-dependent.

Q3

Internal energy of an ideal gas depends on: Easy

(A)Temperature only

(B)Pressure only

(C)Volume only

(D)Both pressure and volume

Solution

✓ Correct: (A)

For an ideal gas (negligible intermolecular forces), internal energy U = sum of kinetic energies of molecules, which depends only on temperature T. This is why ΔU = 0 in isothermal processes for ideal gases. For real gases, U depends on T and V.

Q4

In an isothermal process for an ideal gas, which is correct? Moderate

(A)ΔU = 0, ΔQ = 0

(B)ΔU = 0, ΔQ = ΔW

(C)ΔW = 0, ΔQ = ΔU

(D)ΔQ = 0, ΔU = −ΔW

Solution

✓ Correct: (B)

Isothermal = T = constant → ΔU = 0 (for ideal gas). From First Law: ΔQ = ΔU + ΔW = 0 + ΔW → ΔQ = ΔW. All heat supplied is converted to work. PV = constant (Boyle's Law), W = μRT ln(V₂/V₁).

Q5

In an adiabatic process, which condition holds? Moderate

(A)ΔQ = 0, ΔU = −ΔW

(B)ΔW = 0, ΔU = ΔQ

(C)ΔU = 0, ΔQ = ΔW

(D)P = constant

Solution

✓ Correct: (A)

ΔQ = 0 (no heat exchange) ΔQ = ΔU + ΔW → 0 = ΔU + ΔW → ΔU = −ΔW

Adiabatic: thermally insulated system. Work done by gas decreases internal energy → temperature falls (expansion) or rises (compression). PVᵞ = constant, where γ = Cp/Cv.

Q6

The adiabatic relation for an ideal gas is: Moderate

(A)PV = constant

(B)PVᵞ = constant

(C)TV = constant

(D)P/V = constant

Solution

✓ Correct: (B)

PVᵞ = constant, where γ = Cp/Cv

PV = constant is Boyle's Law (isothermal). The adiabatic exponent γ > 1 always, so the adiabatic curve is steeper than the isothermal curve on a P-V diagram. For monatomic gas γ = 5/3; diatomic γ = 7/5.

Q7

In an isochoric (isovolumetric) process: Easy

(A)ΔW = 0, ΔQ = ΔU

(B)ΔQ = 0, ΔU = −ΔW

(C)ΔU = 0, ΔQ = ΔW

(D)ΔP = 0

Solution

✓ Correct: (A)

Isochoric = V = constant → no work done (ΔW = PΔV = 0). From First Law: ΔQ = ΔU. All heat absorbed goes entirely to increase internal energy. Heat supplied = m·Cv·ΔT.

Q8

For an ideal gas, the relation between Cp and Cv (molar specific heats) is: Moderate

(A)Cp = Cv

(B)Cp − Cv = R

(C)Cp + Cv = R

(D)Cp × Cv = R

Solution

✓ Correct: (B)

Cp − Cv = R (Mayer's relation)

At constant pressure, extra heat is needed to do work against expansion (PΔV = μRΔT for 1 mole). So Cp > Cv always. R = 8.31 J mol⁻¹K⁻¹. This is known as Mayer's relation. γ = Cp/Cv > 1.

Q9

In a cyclic thermodynamic process, the net change in internal energy ΔU is: Easy

(A)Positive

(B)Negative

(C)Zero

(D)Depends on the path

Solution

✓ Correct: (C)

In a cyclic process the system returns to its initial state. Since U is a state variable (path-independent), ΔU = U_final − U_initial = 0. Therefore from First Law: ΔQ = ΔW — the total heat absorbed equals the total work done by the system.

Q10

Which of the following is a state variable (state function) in thermodynamics? Easy

(A)Heat (Q)

(B)Work (W)

(C)Internal energy (U)

(D)Both (A) and (B)

Solution

✓ Correct: (C)

Internal energy U is a state variable — its value depends only on the current state, not on the path taken. Heat Q and Work W are path-dependent quantities (not state variables). P, V, T, and m are also state variables.

Q11

The Kelvin-Planck statement of the Second Law of Thermodynamics says: Moderate

(A)Heat cannot flow from cold to hot spontaneously

(B)No process can convert heat entirely into work

(C)Entropy always decreases

(D)Internal energy is conserved

Solution

✓ Correct: (B)

Kelvin-Planck: "No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion to work." → No perfect heat engine (η = 100%) exists. Clausius: Heat cannot spontaneously flow from cold to hot (option A is Clausius statement).

Q12

The efficiency of a Carnot engine working between hot reservoir T₁ and cold reservoir T₂ is: Moderate

(A)η = 1 − T₁/T₂

(B)η = T₂/T₁

(C)η = 1 − T₂/T₁

(D)η = (T₁ − T₂)/T₂

Solution

✓ Correct: (C)

η = 1 − T₂/T₁ (temperatures in Kelvin!)

T₁ = source (hot), T₂ = sink (cold). Since T₂ < T₁, η < 1 always. For η = 1 (100%), T₂ = 0 K (absolute zero) — physically impossible. Carnot efficiency is the maximum possible for any engine between two temperatures.

Q13

The Carnot cycle consists of: Moderate

(A)Two isothermal + two adiabatic processes

(B)Two isobaric + two isochoric processes

(C)Four isothermal processes

(D)Two adiabatic + two isobaric processes

Solution

✓ Correct: (A)

Step 1→2: Isothermal expansion at T₁ (absorbs Q₁) Step 2→3: Adiabatic expansion (T₁→T₂) Step 3→4: Isothermal compression at T₂ (releases Q₂) Step 4→1: Adiabatic compression (T₂→T₁)

All four steps must be reversible (quasi-static, no friction) making the Carnot engine the most efficient possible engine.

Q14

A reversible process must be: Moderate

(A)Quasi-static and dissipation-free

(B)Very fast

(C)Adiabatic only

(D)Isothermal only

Solution

✓ Correct: (A)

A reversible process requires: (1) Quasi-static — infinitely slow, system in equilibrium at every stage; (2) No dissipative effects — no friction, viscosity, etc. In practice, all real processes are irreversible to some degree.

Q15

Which of the following is an irreversible process? Easy

(A)Quasi-static isothermal expansion in frictionless cylinder

(B)Free expansion of gas into vacuum

(C)Quasi-static adiabatic compression

(D)Slow compression with heat reservoir contact

Solution

✓ Correct: (B)

Free expansion of gas into vacuum is irreversible — it is a rapid, non-quasi-static process where the gas passes through non-equilibrium states. It cannot be reversed spontaneously. Other irreversible examples: diffusion, combustion, heat flow across finite temperature differences.

Q16

Work done by an ideal gas in an isothermal expansion from V₁ to V₂ at temperature T is: Moderate

(A)W = μRT(V₂ − V₁)

(B)W = μRT ln(V₂/V₁)

(C)W = μR(T₁ − T₂)/(γ − 1)

(D)W = P(V₂ − V₁)

Solution

✓ Correct: (B)

W = μRT ln(V₂/V₁) [isothermal ideal gas]

Since PV = μRT → P = μRT/V. W = ∫P dV = μRT∫dV/V = μRT ln(V₂/V₁). For expansion V₂ > V₁ → W > 0 (gas does work). Option (C) is for adiabatic process. Option (D) is for isobaric.

Q17

An adiabatic wall between two systems: Easy

(A)Allows heat flow but not mass

(B)Allows both heat and mass flow

(C)Does not allow heat flow

(D)Allows heat flow in one direction only

Solution

✓ Correct: (C)

An adiabatic wall is a perfect insulator — it does not allow heat (energy) flow. A diathermic wall is the opposite — it allows heat to flow until thermal equilibrium is reached. The Zeroth Law uses diathermic walls conceptually.

Q18

For an ideal gas compressed adiabatically, the temperature: Moderate

(A)Decreases

(B)Increases

(C)Remains constant

(D)First increases then decreases

Solution

✓ Correct: (B)

In adiabatic compression: ΔQ = 0. Work is done on the gas (ΔW < 0) → ΔU = −ΔW > 0 → internal energy increases → temperature increases. This is why a bicycle pump heats up when inflating a tyre rapidly.

Q19

The ratio γ = Cp/Cv for a monatomic ideal gas is: Moderate

(A)1.0

(B)1.4 (7/5)

(C)1.67 (5/3)

(D)1.3

Solution

✓ Correct: (C)

Monatomic (He, Ar): Cv = 3R/2, Cp = 5R/2, γ = 5/3 ≈ 1.67 Diatomic (N₂, O₂): Cv = 5R/2, Cp = 7R/2, γ = 7/5 = 1.4

Monatomic gases have only 3 translational degrees of freedom. Diatomic gases have 2 additional rotational degrees of freedom (5 total). More degrees of freedom → higher Cv → lower γ.

Q20

Which of the following processes on a P-V diagram is represented by a steeper curve? Moderate

(A)Isothermal process

(B)Isobaric process

(C)Adiabatic process

(D)Isochoric process

Solution

✓ Correct: (C)

Adiabatic curve is steeper than isothermal. Isothermal: PV = const → slope = −P/V. Adiabatic: PVᵞ = const → slope = −γP/V. Since γ > 1, the adiabatic slope is steeper. This is because during adiabatic expansion, temperature also falls, reducing pressure more.

Q21

In a free expansion of an ideal gas into vacuum (no heat exchange), which is true? Moderate

(A)ΔU > 0, temperature rises

(B)ΔU = 0, temperature unchanged

(C)ΔU < 0, temperature falls

(D)ΔW > 0, temperature rises

Solution

✓ Correct: (B)

Free expansion into vacuum: ΔQ = 0 (insulated), ΔW = 0 (no external pressure to work against). By First Law: ΔU = ΔQ − ΔW = 0. For ideal gas, U depends only on T → temperature remains unchanged. (This is NOT a reversible process.)

Q22

According to Carnot's theorem, no engine operating between temperatures T₁ and T₂ can have efficiency: Moderate

(A)Greater than that of the Carnot engine

(B)Equal to that of the Carnot engine

(C)Less than that of the Carnot engine

(D)None of the above

Solution

✓ Correct: (A)

Carnot's theorem: No engine working between two given temperatures can have efficiency greater than that of the Carnot engine. The Carnot efficiency η = 1 − T₂/T₁ is the maximum possible efficiency for any heat engine operating between T₁ and T₂. All real engines are less efficient due to irreversibility.

Q23

Heat and work are NOT state variables because they: Moderate

(A)Are forms of energy

(B)Depend on the path taken

(C)Depend only on initial state

(D)Are conserved quantities

Solution

✓ Correct: (B)

Q and W are path-dependent — different paths from the same initial to final state give different Q and W values. Only their difference (ΔQ − ΔW = ΔU) is path-independent. Saying "a gas has a certain amount of heat" is meaningless — heat is energy in transit.

Q24

In the Carnot cycle, Q₁/Q₂ = T₁/T₂. This means: Moderate

(A)Heat absorbed equals heat released

(B)The ratio of heats equals the ratio of temperatures

(C)The process is irreversible

(D)Efficiency is always 100%

Solution

✓ Correct: (B)

Q₁/Q₂ = T₁/T₂ (universal relation for Carnot cycle)

This is a universal relation independent of the nature of the working substance. It is used to define a truly universal thermodynamic temperature scale. Efficiency η = 1 − Q₂/Q₁ = 1 − T₂/T₁.

Q25

The molar specific heat capacity at constant pressure for an ideal monatomic gas is: Moderate

(A)3R/2

(B)5R/2

(C)7R/2

(D)R

Solution

✓ Correct: (B)

Monatomic: Cv = 3R/2, Cp = Cv + R = 3R/2 + R = 5R/2 Diatomic: Cv = 5R/2, Cp = 7R/2

Using Mayer's relation Cp = Cv + R. For monatomic gas (3 translational DOF): Cv = (3/2)R → Cp = 5R/2 ≈ 20.8 J mol⁻¹K⁻¹. This matches experimental value for He (20.8).

N1

A geyser heats water flowing at 3.0 litres/minute from 27°C to 77°C. If the heat of combustion of fuel is 4.0 × 10⁴ J/g, find the rate of fuel consumption. (s_water = 4.2 × 10³ J/kg·K) Moderate

Solution

Flow rate = 3 L/min = 3 × 10⁻³ m³/min = 3 kg/min (water density = 1000 kg/m³) ΔT = 77 − 27 = 50°C Heat required per minute: Q = m × s × ΔT = 3 × 4200 × 50 = 630,000 J/min Rate of fuel consumption = Q / Heat of combustion = 630,000 / 40,000 = 15.75 g/min ≈ 15.75 g/min

The geyser burns approximately 15.75 g of fuel per minute to heat the water.

N2

What amount of heat must be supplied to 2.0 × 10⁻² kg of nitrogen at room temperature to raise its temperature by 45°C at constant pressure? (M = 28, R = 8.3 J mol⁻¹K⁻¹) Moderate

Solution

N₂ is diatomic: Cv = 5R/2, Cp = 7R/2 = 7 × 8.3/2 = 29.05 J/mol·K Number of moles μ = mass/M = (2.0 × 10⁻²)/(28 × 10⁻³) = 0.714 mol At constant pressure: ΔQ = μ × Cp × ΔT = 0.714 × 29.05 × 45 ΔQ = 0.714 × 1307.25 ΔQ ≈ 933 J ≈ 933.4 J

N3

A Carnot engine works between temperatures 800 K (source) and 300 K (sink). (a) Find efficiency. (b) If the engine absorbs 1000 J from source per cycle, how much work is done and how much heat is rejected to sink? Moderate

Solution

(a) η = 1 − T₂/T₁ = 1 − 300/800 = 1 − 0.375 = 0.625 Efficiency = 62.5% (b) Work done per cycle: W = η × Q₁ = 0.625 × 1000 = 625 J Heat rejected to sink: Q₂ = Q₁ − W = 1000 − 625 = 375 J Verification: Q₁/Q₂ = 1000/375 = 2.67 = T₁/T₂ = 800/300 = 2.67 ✓

N4

A cylinder with a movable piston contains 3 moles of hydrogen (diatomic, γ = 1.4) at STP. The walls are insulated. By what factor does pressure increase if gas is compressed to half its original volume? Moderate

Solution

Insulated walls + piston with sand = adiabatic process. γ for H₂ = 1.4.

Adiabatic: P₁V₁ᵞ = P₂V₂ᵞ V₂ = V₁/2 P₂/P₁ = (V₁/V₂)ᵞ = (V₁/(V₁/2))^1.4 = (2)^1.4 (2)^1.4 = 2^1 × 2^0.4 = 2 × 1.319 = 2.638 Pressure increases by a factor of ≈ 2.64

N5

An electric heater supplies heat at 100 W. If the system performs work at 75 J/s, at what rate is the internal energy increasing? Easy

Solution

First Law: ΔQ/dt = ΔU/dt + ΔW/dt ΔQ/dt = 100 W (heat supplied per second) ΔW/dt = 75 J/s (work done per second) ΔU/dt = ΔQ/dt − ΔW/dt = 100 − 75 = 25 W Internal energy increases at 25 J/s = 25 W

N6

A gas changes state adiabatically from A to B with 22.3 J of work done ON the system. If the gas goes from A to B via a process where net heat absorbed = 9.35 cal, find the net work done by the system. (1 cal = 4.19 J) Hard

Solution

Adiabatic process A→B: ΔQ = 0, Work done ON system = 22.3 J → ΔW = −22.3 J ΔU = ΔQ − ΔW = 0 − (−22.3) = +22.3 J Since U is a state function, ΔU = +22.3 J for any path A→B. Second path A→B: ΔQ = 9.35 cal = 9.35 × 4.19 = 39.18 J ΔU = 22.3 J (same, state function) Net work done BY system: ΔW = ΔQ − ΔU = 39.18 − 22.3 = 16.88 J ≈ 16.9 J

N7

Find the efficiency of a Carnot engine operating between 127°C and 27°C. Also find the work done per cycle if heat absorbed from source = 500 J. Moderate

Solution

T₁ = 127 + 273 = 400 K (hot source) T₂ = 27 + 273 = 300 K (cold sink) η = 1 − T₂/T₁ = 1 − 300/400 = 1 − 0.75 = 0.25 = 25% Work done per cycle: W = η × Q₁ = 0.25 × 500 = 125 J Heat rejected to sink: Q₂ = Q₁ − W = 500 − 125 = 375 J

Always convert temperatures to Kelvin before using the Carnot efficiency formula.

N8

1 g of water (liquid) is converted to steam at 100°C. The latent heat of vaporisation = 2256 J/g. At atmospheric pressure (1.013 × 10⁵ Pa), volume changes from 1 cm³ to 1671 cm³. Find ΔU, ΔW, and ΔQ. Moderate

Solution

ΔQ = 2256 J (latent heat for 1 g water→steam) ΔW = PΔV = 1.013 × 10⁵ × (1671 − 1) × 10⁻⁶ = 1.013 × 10⁵ × 1670 × 10⁻⁶ = 1.013 × 10⁵ × 1.67 × 10⁻³ = 169.2 J ΔU = ΔQ − ΔW = 2256 − 169.2 = 2086.8 J

Most of the heat (93%) goes to increase internal energy (breaking intermolecular bonds). Only 7% does work against atmospheric pressure.

N9

A Carnot engine has efficiency of 40%. If the sink temperature is 27°C, find the source temperature. If the efficiency is to be increased to 60%, what should be the new source temperature (keeping sink at 27°C)? Moderate

Solution

T₂ = 27 + 273 = 300 K, η₁ = 0.40 Case 1: η = 1 − T₂/T₁ 0.40 = 1 − 300/T₁ 300/T₁ = 0.60 T₁ = 300/0.60 = 500 K = 227°C Case 2: η = 0.60 0.60 = 1 − 300/T₁' 300/T₁' = 0.40 T₁' = 300/0.40 = 750 K = 477°C

To increase efficiency from 40% to 60%, the source temperature must be raised from 500 K to 750 K (an increase of 250 K).

N10

A P-V diagram shows a process D→E→F. D→E is a linear process with pressure changing from 600 N/m² to 300 N/m² while volume goes from 2 m³ to 5 m³. E→F is isobaric compression from 5 m³ back to 2 m³ at 300 N/m². Find total work done. Moderate

Solution

D→E (linear process, trapezoid under P-V curve): P goes 600→300 N/m², V goes 2→5 m³ W_DE = Area of trapezoid = ½(P_D + P_E)(V_E − V_D) = ½(600 + 300)(5 − 2) = ½ × 900 × 3 = 1350 J E→F (isobaric compression, P = 300 N/m²): W_EF = P × (V_F − V_E) = 300 × (2 − 5) = 300 × (−3) = −900 J (work done ON gas) Total work = W_DE + W_EF = 1350 − 900 = 450 J

N11

2 moles of an ideal monatomic gas (γ = 5/3) undergo adiabatic expansion from (P₁ = 10⁵ Pa, V₁ = 10 L, T₁ = 300 K) to final volume V₂ = 80 L. Find T₂ and work done by the gas. (R = 8.3 J/mol·K) Hard

Solution

For adiabatic: T₁V₁^(γ-1) = T₂V₂^(γ-1) γ = 5/3, γ−1 = 2/3 T₂ = T₁ × (V₁/V₂)^(γ-1) = 300 × (10/80)^(2/3) = 300 × (1/8)^(2/3) = 300 × (1/8)^(2/3) (1/8)^(1/3) = 1/2, so (1/8)^(2/3) = 1/4 T₂ = 300 × 1/4 = 75 K Work done by gas (adiabatic): W = μR(T₁ − T₂)/(γ − 1) = 2 × 8.3 × (300 − 75) / (5/3 − 1) = 2 × 8.3 × 225 / (2/3) = 2 × 8.3 × 225 × 3/2 = 5602.5 J ≈ 5600 J

N12

A Carnot refrigerator works between 0°C (cold) and 27°C (hot). What is the coefficient of performance (COP)? If 1000 J of heat is removed from the cold reservoir, how much work is done and how much heat is delivered to the hot reservoir? Moderate

Solution

T₂ = 0 + 273 = 273 K (cold reservoir) T₁ = 27 + 273 = 300 K (hot reservoir) COP of refrigerator = T₂/(T₁ − T₂) = 273/(300 − 273) = 273/27 ≈ 10.11 Q₂ = 1000 J (heat removed from cold reservoir) Work done: W = Q₂/COP = 1000/10.11 ≈ 98.9 J Heat delivered to hot reservoir: Q₁ = Q₂ + W = 1000 + 98.9 = 1098.9 J ≈ 1099 J

High COP (≈10) means very efficient refrigerator — just ~99 J of work removes 1000 J of heat from cold space.

N13

1 mole of an ideal diatomic gas (γ = 1.4) at 300 K is compressed isothermally to half its volume. Find (a) work done ON the gas, (b) heat released by gas, (c) change in internal energy. (R = 8.3 J/mol·K) Moderate

Solution

μ = 1 mol, T = 300 K, V₂ = V₁/2 Work done BY gas (isothermal): W = μRT ln(V₂/V₁) = 1 × 8.3 × 300 × ln(1/2) = 2490 × (−0.693) = −1724.6 J Work done ON gas = +1724.6 J ≈ 1725 J (c) Isothermal ideal gas: ΔU = 0 (b) From First Law: ΔQ = ΔU + W = 0 + (−1724.6) Heat released = 1724.6 J ≈ 1725 J

In isothermal compression, all work done on the gas is released as heat to surroundings. Temperature remains constant.

N14

Two cylinders A and B of equal volume are connected via a stopcock. A contains gas at 1 atm, 27°C. B is evacuated. The system is thermally insulated and the stopcock is opened. Find (a) final pressure, (b) change in internal energy, (c) change in temperature. Moderate

Solution

(a) Final pressure: Gas expands to double volume at constant T (for ideal gas) P₁V₁ = P₂(2V₁) [isothermal, since ΔU = 0] P₂ = P₁/2 = 1/2 atm = 0.5 atm (b) ΔU = 0: ΔQ = 0 (insulated) and ΔW = 0 (expansion into vacuum, no external pressure) → ΔU = ΔQ − ΔW = 0 (c) ΔT = 0: Since ΔU = 0 for ideal gas → temperature unchanged at 27°C Note: This is FREE EXPANSION — irreversible process. Intermediate states are NOT equilibrium states (not on P-V-T surface)

N15

A heat engine absorbs 1000 J from a hot reservoir and rejects 600 J to cold reservoir in each cycle. (a) Find efficiency. (b) Compare with Carnot efficiency if T_hot = 500 K and T_cold = 300 K. (c) Is this engine reversible? Moderate

Solution

(a) Actual efficiency: W = Q₁ − Q₂ = 1000 − 600 = 400 J η_actual = W/Q₁ = 400/1000 = 0.40 = 40% (b) Carnot efficiency: η_Carnot = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.40 = 40% (c) Since η_actual = η_Carnot = 40%, this engine achieves the maximum possible efficiency → it IS a reversible engine (Carnot engine). Any engine with η < 40% would be irreversible.

This is a remarkable result — only a Carnot (reversible) engine can achieve the theoretical maximum efficiency. All real engines have lower efficiency due to irreversible losses.