Is it correct to say: Mixing sign conventions, e.g. writing ΔU = Q + W while treating W as work done by the gas?

With the textbook convention (W = work done BY the system), the first law is Q = ΔU + W. In compression W is negative.

Is it correct to say: Thinking an isothermal process involves no heat exchange?

That describes an adiabatic process. In an isothermal process ΔT = 0, so ΔU = 0 and Q = W. Heat does flow.

Is it correct to say: Using temperatures in °C in η = 1 − T_C/T_H?

Use kelvin. Between 227 °C and 27 °C the efficiency is 1 − 300/500 = 40%, not 1 − 27/227 ≈ 88%.

Is it correct to say: Using nC_PΔT for the change in internal energy in an isobaric process?

ΔU = nC_VΔT always, for an ideal gas. nC_PΔT is the heat supplied at constant pressure.

Is it correct to say: Writing ln as log₁₀ without the factor 2.303?

nRT ln(V_f/V_i) = 2.303 nRT log₁₀(V_f/V_i). Dropping 2.303 makes the answer 2.3 times too small.

Is it correct to say: Believing a refrigerator's coefficient of performance cannot exceed 1, like efficiency?

K = Q_C/|W| is a ratio of heat moved to work done, not an efficiency. The textbook quotes K ≈ 5 for a household refrigerator and 2.5 to 3 for a room air conditioner.

📄 Formula sheet (PDF)

Thermodynamics

Zeroth and first laws, work done by a gas, isothermal, isobaric, isochoric and adiabatic processes, heat engines, refrigerators, second law and the Carnot cycle — Maharashtra HSC Physics Ch 4

📐 15 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Fix the sign convention before you start. Q is positive when heat is supplied to the system, and W is positive when work is done BY the system (expansion). Then Q = ΔU + W.
→For each process, write what is zero: isothermal ΔT = 0 (so ΔU = 0), isochoric W = 0, adiabatic Q = 0, cyclic ΔU = 0. This reduces the first law to a single relation.
→Learn the derivations of work done in an isothermal process and in an adiabatic process. Both start from W = ∫P dV, using PV = nRT for isothermal and PV^γ = constant for adiabatic.
→Carnot efficiency and coefficient of performance must use temperatures in kelvin. Convert first; Celsius values give wrong answers, sometimes even negative ones.
→On a P–V diagram, work is the area under the curve. For a cycle, net work is the enclosed area: positive for a clockwise cycle (engine), negative for anticlockwise (refrigerator).

📐 Formulas(15)

Work Done by a Gas

W = \int_{V_{i}}^{V_{f}} P d V, W = P (V_{f} - V_{i}) (constant P)

Symbol	Meaning
$W$	Work done by the gas (J)
$P$	Pressure of the gas (Pa)
$V_{i}, V_{f}$	Initial and final volume (m³)

First Law of Thermodynamics★ Board fav

Q = Δ U + W

Symbol	Meaning
$Q$	Heat supplied to the system (J); negative if heat is given out
$Δ U$	Change in internal energy (J)
$W$	Work done by the system (J); negative if work is done on it

Change in Internal Energy of an Ideal Gas

Δ U = n C_{V} Δ T

Symbol	Meaning
$n$	Number of moles
$C_{V}$	Molar specific heat at constant volume (J mol⁻¹ K⁻¹)
$Δ T$	Change in temperature (K)

Isothermal Process — Work Done★ Board fav

W = n R T ln \frac{V _{f}}{V _{i}} = 2.303 n R T lo g_{10} \frac{V _{f}}{V _{i}}

Symbol	Meaning
$T$	Constant absolute temperature (K)
$V_{i}, V_{f}$	Initial and final volume (m³ or L, same unit for both)

Isobaric Process

W = P Δ V = n R Δ T, Q = n C_{P} Δ T

Symbol	Meaning
$Δ V$	Change in volume (m³)
$C_{P}$	Molar specific heat at constant pressure (J mol⁻¹ K⁻¹)

Isochoric Process

W = 0, Q = Δ U = n C_{V} Δ T

Adiabatic Process — Equation of State★ Board fav

P V^{γ} = constant, T V^{γ - 1} = constant

Symbol	Meaning
$γ$	Ratio of specific heats C_P/C_V

Adiabatic Process — Work Done

W = \frac{P _{i} V _{i} - P _{f} V _{f}}{γ - 1} = \frac{n R ( T _{i} - T _{f} )}{γ - 1}

Symbol	Meaning
$P_{i}, V_{i}, T_{i}$	Initial pressure, volume and temperature
$P_{f}, V_{f}, T_{f}$	Final pressure, volume and temperature

Cyclic Process

Δ U = 0, Q = W = area enclosed on the P - V diagram

Free Expansion

Q = 0, W = 0, Δ U = 0

Efficiency of a Heat Engine★ Board fav

η = \frac{W}{Q _{H}} = 1 - \frac{∣ Q _{C} ∣}{Q _{H}}

Symbol	Meaning
$η$	Efficiency (fraction or %)
$Q_{H}$	Heat absorbed from the hot reservoir (J)
$∣ Q_{C} ∣$	Magnitude of the heat rejected to the cold reservoir (J)
$W$	Net work done per cycle (J)

Coefficient of Performance of a Refrigerator

K = \frac{Q _{C}}{∣ W ∣} = \frac{Q _{C}}{∣ Q _{H} ∣ - Q _{C}}

Symbol	Meaning
$K$	Coefficient of performance (no unit)
$Q_{C}$	Heat extracted from the cold reservoir (J)
$∣ W ∣$	Magnitude of the work done on the refrigerant (J)

Carnot Cycle — Heat and Temperature Ratio

\frac{∣ Q _{C} ∣}{Q _{H}} = \frac{T _{C}}{T _{H}}

Symbol	Meaning
$T_{H}$	Temperature of the hot reservoir (K)
$T_{C}$	Temperature of the cold reservoir (K)

Efficiency of a Carnot Engine★ Board fav

η = 1 - \frac{T _{C}}{T _{H}}

Symbol	Meaning
$T_{H}$	Source temperature (K)
$T_{C}$	Sink temperature (K)

Coefficient of Performance of a Carnot Refrigerator

K = \frac{T _{C}}{T _{H} - T _{C}}

✏️ Solved Examples

1Solved Exampleeasy2 steps

A gas absorbs 500 J of heat and does 200 J of work on the surroundings. Find the change in its internal energy.

Heat is supplied (Q = +500 J) and work is done by the gas (W = +200 J).

Q = Δ U + W \Rightarrow Δ U = Q - W

2Solved Exampleboard4 steps

Two moles of an ideal gas at 27 °C expand isothermally from 10 L to 20 L. Find the work done by the gas and the heat absorbed. (R = 8.314 J mol⁻¹ K⁻¹, log₁₀ 2 = 0.3010)

T = 27 + 273 = 300 K. Only the volume ratio matters, so litres can be used directly.

W = 2.303 n R T lo g_{10} \frac{V _{f}}{V _{i}}

3Solved Exampleboard3 steps

A Carnot engine works between a source at 227 °C and a sink at 27 °C. If it absorbs 2000 J from the source in each cycle, find its efficiency, the work done per cycle and the heat rejected to the sink.

Convert to kelvin: T_H = 500 K, T_C = 300 K.

η = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{500} = 0.4

4Solved ExampleHOTS4 steps

One mole of air (γ = 1.4) at 27 °C is compressed adiabatically to one-eighth of its volume. Find the final temperature and the work done by the gas. (R = 8.314 J mol⁻¹ K⁻¹, 8^0.4 ≈ 2.297)

Use TV^(γ−1) = constant with T_i = 300 K and V_i/V_f = 8.

T_{f} = T_{i} (\frac{V _{i}}{V _{f}})^{γ - 1} = 300 \times 8^{0.4}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Mixing sign conventions, e.g. writing ΔU = Q + W while treating W as work done by the gas
✓With the textbook convention (W = work done BY the system), the first law is Q = ΔU + W. In compression W is negative.
2
Thinking an isothermal process involves no heat exchange
✓That describes an adiabatic process. In an isothermal process ΔT = 0, so ΔU = 0 and Q = W. Heat does flow.
3
Using temperatures in °C in η = 1 − T_C/T_H
✓Use kelvin. Between 227 °C and 27 °C the efficiency is 1 − 300/500 = 40%, not 1 − 27/227 ≈ 88%.
4
Using nC_PΔT for the change in internal energy in an isobaric process
✓ΔU = nC_VΔT always, for an ideal gas. nC_PΔT is the heat supplied at constant pressure.
5
Writing ln as log₁₀ without the factor 2.303
✓nRT ln(V_f/V_i) = 2.303 nRT log₁₀(V_f/V_i). Dropping 2.303 makes the answer 2.3 times too small.
6
Believing a refrigerator's coefficient of performance cannot exceed 1, like efficiency
✓K = Q_C/|W| is a ratio of heat moved to work done, not an efficiency. The textbook quotes K ≈ 5 for a household refrigerator and 2.5 to 3 for a room air conditioner.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A Carnot refrigerator keeps its freezer at −13 °C in a room at 27 °C. Find its coefficient of performance.
Q2
An engine absorbs 800 J from the source and rejects 600 J to the sink in each cycle. Find the work done per cycle and the efficiency.
Q3
A gas expands at a constant pressure of 2 × 10⁵ Pa from 0.01 m³ to 0.03 m³. Find the work done by the gas.
Q4
300 J of heat is supplied to a gas kept in a rigid container. Find the work done and the change in internal energy.
Q5
Two moles of a monatomic ideal gas are heated at constant pressure through 10 K. Find Q, ΔU and W. (R = 8.314 J mol⁻¹ K⁻¹)
Q6
A gas goes through a cycle that forms a rectangle on the P–V diagram, with P between 1 × 10⁵ Pa and 3 × 10⁵ Pa and V between 1 × 10⁻³ m³ and 3 × 10⁻³ m³. Find the net work done per cycle.

📝 Notes

Thermodynamics — Maharashtra HSC Overview

Chapter 4 of the Maharashtra Board (Balbharati) Std XII Physics textbook builds from thermal equilibrium (zeroth law) to energy conservation (first law) and then to the limits on converting heat into work (second law and the Carnot engine).

One law, many processes

The first law $Q = \Delta U + W$ , with $W$ as work done by the system, is the key idea of the chapter. Every named process is the first law with one quantity set to zero:

Isothermal: $\Delta U = 0$ , so $Q = W = 2.303\,nRT\log_{10}(V_f/V_i)$ .
Isochoric: $W = 0$ , so $Q = \Delta U = nC_V\Delta T$ .
Adiabatic: $Q = 0$ , so $\Delta U = -W$ , with $PV^\gamma$ = constant.
Cyclic: $\Delta U = 0$ , so $Q = W$ = area enclosed on the $P$ – $V$ diagram.

Remember $\Delta U = nC_V\Delta T$ for an ideal gas in any process. It is the fastest way to find $\Delta U$ .

Heat engines and refrigerators

An engine takes $Q_H$ from a hot reservoir, converts part of it into work $W$ , and rejects heat $|Q_C|$ , so $\eta = 1 - |Q_C|/Q_H$ . A refrigerator runs the cycle backwards and is rated by its coefficient of performance $K = Q_C/|W|$ . The Kelvin–Planck and Clausius statements of the second law say $Q_C$ can never be zero, and that heat cannot flow from cold to hot without work.

Carnot limit

For a reversible Carnot cycle $|Q_C|/Q_H = T_C/T_H$ . This gives the highest efficiency possible between two temperatures, $\eta = 1 - T_C/T_H$ , and the corresponding $K = T_C/(T_H - T_C)$ . In descriptive answers, name the four strokes in order: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.