📌 Snapshot
- First law, adiabatic and isothermal work, and Carnot efficiency explained with solved numericals.
- Why a bicycle pump heats up, why heat and work are not stored in a gas, and why no engine can turn heat completely into work.
- Zeroth law, internal energy, Mayer's relation, the five standard processes and the Carnot cycle compared side by side.
- Five worked examples with the common wrong step shown, a formula sheet and twelve NTA style MCQs.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 11.
Why does a bicycle pump get hot?
Squeeze the air in a bicycle pump and the barrel warms up, though you never held a flame to it. Let air rush out of a spray can and the can turns cold. In 1798 Count Rumford noticed that boring a brass cannon produced enough heat to boil water, and the heat depended on how long the drill was turned, not on how sharp it was. Heat was not a fluid called caloric flowing between bodies. It was energy, and work could turn into it. Thermodynamics is the study of that exchange: how heat, work and internal energy convert into one another, and where nature puts limits on the conversion.
Systems, equilibrium and the Zeroth Law
Thermodynamics deals with big, measurable variables such as pressure P, volume V, temperature T and internal energy U, and ignores what each molecule is doing. The object under study is the system, and everything else is the surroundings. A system is in thermodynamic equilibrium when its variables stop changing with time.
Two systems separated by an adiabatic wall (an insulating one) cannot exchange heat, so they do not come to a common state. Across a diathermic wall (a conducting one) heat flows until they reach thermal equilibrium.
The Zeroth Law: if A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. The quantity that is equal for all systems in mutual thermal equilibrium is called temperature. It is what lets a thermometer (C) tell you whether two bodies (A and B) would exchange heat without ever bringing them together. R. H. Fowler stated it in 1931, long after the first and second laws were numbered, but logically it comes first.
State variables describe the condition of a system. Extensive variables such as volume, internal energy and mass scale with its size. Intensive variables such as pressure, temperature and density do not. The relation joining them is the equation of state, which for an ideal gas is $PV = \mu RT$.
Internal energy, heat and work
Internal energy U is the sum of the kinetic and potential energies of the molecules, measured in the frame where the centre of mass is at rest. It depends only on the state of the system, and not on how the system got there. For an ideal gas, U depends only on temperature.
Heat is energy transferred because of a temperature difference. Work is energy transferred by other means, such as a moving piston. Both are energy in transit, and neither is a state variable. A gas has a definite internal energy, but it does not "have" a definite amount of heat or work.
| State variable | Path variable | |
|---|---|---|
| Examples | P, V, T, U | Q (heat), W (work) |
| Depends on | The present state only | The route taken between states |
| Over a full cycle | Net change is zero | Can be non-zero |
For a gas pushing a piston against a constant pressure, the work done by the gas is $W = P\Delta V$. In general it is the area under the P–V curve. It is positive when the gas expands and negative when it is compressed.
The First Law
Energy is conserved, and heat is one of its forms. The First Law says
$$\Delta Q = \Delta U + \Delta W$$
where $\Delta Q$ is the heat supplied to the system, $\Delta W$ is the work done by the system and $\Delta U$ is the change in internal energy. Heat added either raises the internal energy or lets the system do work. Some books write $\Delta U = Q + W$ with W as work done on the system. The physics is identical, and only the sign of W changes, so always check which sign your question uses.
Worked example: boiling one gram of water. When 1 g of water boils at atmospheric pressure, it becomes 1671 cm³ of steam and absorbs 2256 J. Find the work done by the system and the change in internal energy. Take $P_{a}$ = $1.013 \times 10^{5}$ Pa.
| Step | What we do | Result |
|---|---|---|
| 1 | Volume change, 1671 cm³ − 1 cm³ | $\Delta V = 1670 \times 10^{-6}$ m³ |
| 2 | Work against the atmosphere, P$\Delta V$ | $1.013 \times 10^{5} \times 1.670 \times 10^{-3} = 169.2$ J |
| 3 | First law, $\Delta U = \Delta Q - \Delta W$ | 2256 − 169.2 = 2086.8 J |
Only about 7.5% of the heat goes into pushing the atmosphere back. The rest, 2086.8 J, goes into overcoming the attractions between molecules. The common wrong step is to say $\Delta U$ equals the 2256 J supplied, forgetting that the expanding steam did work.
Specific heats of a gas and Mayer's relation
A gas can be heated in many ways, so it has many specific heats. Two matter: $C_{V}$, the molar specific heat at constant volume, and $C_{P}$, at constant pressure. At constant volume no work is done, so every joule goes into U, and for an ideal gas $\Delta U = \mu C_{V}\Delta T$ in every process. At constant pressure the gas must also do work $P\Delta V = \mu R\Delta T$, so extra heat is needed:
$\mu C_{P}\Delta T = \mu C_{V}\Delta T + \mu R\Delta T$, giving $C_{P} - C_{V} = R$ (Mayer's relation)
The ratio $\gamma = C_{P}/C_{V}$ appears in all adiabatic formulae. For a solid, $\Delta V$ is negligible, and equipartition gives $U = 3RT$ per mole, so $C = 3R$. A calorie is the heat that raises 1 g of water from 14.5 °C to 15.5 °C, and 1 cal = 4.186 J.
The five standard processes
A quasi-static process is so slow that the gas is in equilibrium at every stage, which is what lets us draw it as a curve on a P–V diagram. For an ideal gas the main processes are:
| Process | What is fixed | Work done by gas | Heat Q | Change in U |
|---|---|---|---|---|
| Isothermal | T, so $PV = \text{constant}$ | $\mu RT \ln(V_{2}/V_{1})$ | Q = W | 0 |
| Adiabatic | $Q = 0$, so $PV^{\gamma} = \text{constant}$ | $\mu R(T_{1} - T_{2})/(\gamma - 1) = (P_{1}V_{1} - P_{2}V_{2})/(\gamma - 1)$ | 0 | −W |
| Isochoric | V | 0 | $\mu C_{V}\Delta T$ | $\mu C_{V}\Delta T$ |
| Isobaric | P | $P(V_{2} - V_{1}) = \mu R\Delta T$ | $\mu C_{P}\Delta T$ | $\mu C_{V}\Delta T$ |
| Cyclic | Returns to start | Area enclosed by loop | Q = W | 0 |
The adiabatic process also obeys $TV^{\gamma-1} = \text{constant}$. In an isothermal expansion the gas absorbs exactly as much heat as it spends in work, so its temperature stays fixed. In an adiabatic expansion the gas gets no heat, so the work comes out of its internal energy and the gas cools. In an adiabatic compression the work done on the gas raises U and the gas heats up. That is the bicycle pump.
Pause and predict: on a P–V diagram, an isothermal curve and an adiabatic curve for the same gas cross at one point. Which one is steeper?
Check your answer: the adiabatic. Along an isothermal curve $dP/dV = -P/V$. Along an adiabatic curve $dP/dV = -\gamma P/V$, which is steeper by the factor $\gamma$. Physically, when you compress a gas adiabatically it also heats up, so the pressure climbs faster than it would at fixed temperature.
Worked example: isothermal work. Two moles of an ideal gas expand isothermally at 300 K until the volume doubles. Find the work done and the heat absorbed (R = 8.31 J mol⁻¹ K⁻¹, $\ln 2 = 0.693$).
| Step | What we do | Result |
|---|---|---|
| 1 | Use $W = \mu RT \ln(V_{2}/V_{1})$ | $2 \times 8.31 \times 300 \times \ln 2$ |
| 2 | Evaluate | 4986 × 0.693 ≈ 3456 J |
| 3 | Isothermal, so $\Delta U = 0$ and Q = W | Q ≈ 3.46 kJ absorbed |
The common wrong step is $W = P\Delta V$ with the initial pressure. The pressure falls as the gas expands, so the integral gives the logarithm. Another is to think "$\Delta U = 0$ means no heat". It means heat in equals work out.
Worked example: compressing air. Air ($\gamma$ = 1.4) at 300 K is compressed adiabatically to one sixteenth of its volume, as in a diesel engine. Find the final temperature and the work done on one mole.
| Step | What we do | Result |
|---|---|---|
| 1 | Adiabatic, so $T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$ | $T_{2} = 300 \times 16^{0.4}$ |
| 2 | $16^{0.4} = 2^{1.6}$ | $\approx 3.03$ |
| 3 | Final temperature | $T_{2} \approx 909$ K |
| 4 | Work by gas, $\mu R(T_{1} - T_{2})/(\gamma - 1)$ | $8.31 \times (300 - 909)/0.4 \approx -1.27 \times 10^{4}$ J |
The negative sign means 12.7 kJ of work is done on the gas, and all of it goes into raising U. The common wrong step is to use the isothermal result and say the temperature does not change. With no time for heat to leave, the temperature must rise.
Beyond the textbook. For the exam: in an adiabatic compression the temperature rises, with $T_{2}$ found from $TV^{\gamma-1} = \text{constant}$. In reality: a diesel engine uses exactly this. It compresses air by a large factor until it is hot enough to ignite the fuel as it is sprayed in, so there is no spark plug. Real engines lose some heat to the cylinder walls and the compression is not perfectly slow, so the actual temperature is lower than the ideal estimate, though still high enough to ignite diesel. Why it matters: it shows the ideal formula predicts the right behaviour, even when the numbers are only approximate.
Worked example: work in a cycle. A gas goes round a rectangular loop in a P–V diagram. Starting at P = $1 \times 10^{5}$ Pa and V = $2 \times 10^{-3}$ m³, it is heated at constant volume to $3 \times 10^{5}$ Pa, expanded at constant pressure to $5 \times 10^{-3}$ m³, cooled at constant volume to $1 \times 10^{5}$ Pa and compressed at constant pressure back to the start. Find the net work and net heat.
| Step | What we do | Result |
|---|---|---|
| 1 | Work on the two constant-volume steps | 0 |
| 2 | Expansion at $3 \times 10^{5}$ Pa | $3 \times 10^{5} \times 3 \times 10^{-3} = +900$ J |
| 3 | Compression at $1 \times 10^{5}$ Pa | $1 \times 10^{5} \times (-3 \times 10^{-3}) = -300$ J |
| 4 | Net work (area of rectangle) | 900 − 300 = 600 J |
| 5 | Cycle, so $\Delta U = 0$ and net Q = W | 600 J of net heat absorbed |
The wrong step is to say the net heat is zero "because the gas returns to its start". The internal energy returns to its start. The heat and work do not, and the loop area is the work extracted per cycle.
Pause and predict: a gas expands into an empty, insulated container when a partition is removed (free expansion). Does an ideal gas change temperature?
Check your answer: no. The gas does no work, since there is nothing to push against, and no heat flows, so $\Delta U = 0$. An ideal gas's U depends only on T, so T stays the same. This process is fast and not quasi-static, so it cannot be drawn as a curve on a P–V diagram.
The Second Law
The First Law allows a heat engine that turns all its heat into work, or a refrigerator that works with no input. Nature does neither. The Second Law says so in two equivalent forms:
| Statement | Says | About |
|---|---|---|
| Kelvin–Planck | No process has, as its sole result, the absorption of heat from a reservoir and its complete conversion into work | Heat engines |
| Clausius | No process has, as its sole result, the transfer of heat from a colder body to a hotter body | Refrigerators |
A heat engine takes heat $Q_{1}$ from a hot source at $T_{1}$, converts part into work W and releases $Q_{2}$ to a cold sink at $T_{2}$. Its efficiency is $\eta = W/Q_{1} = 1 - Q_{2}/Q_{1}$. A refrigerator runs the process backwards, using work to pull heat out of a cold space.
A reversible process can be undone so that both the system and its surroundings return exactly to their original states. For that it must be quasi-static and free of friction and other dissipative effects. Natural processes such as heat flowing down a temperature gradient, free expansion and friction are irreversible.
The Carnot engine
The ideal engine is the Carnot engine, a reversible engine working between two temperatures. Its cycle has four steps:
| Step | Process | What happens |
|---|---|---|
| 1 → 2 | Isothermal expansion at $T_{1}$ | Gas absorbs $Q_{1}$ from the source |
| 2 → 3 | Adiabatic expansion | Gas cools from $T_{1}$ to $T_{2}$ |
| 3 → 4 | Isothermal compression at $T_{2}$ | Gas rejects $Q_{2}$ to the sink |
| 4 → 1 | Adiabatic compression | Gas warms from $T_{2}$ back to $T_{1}$ |
Carnot's theorem states that no engine working between two given temperatures can be more efficient than a Carnot engine, and that all Carnot engines between those temperatures have the same efficiency, whatever the working substance. For a Carnot engine $Q_{2}/Q_{1} = T_{2}/T_{1}$, so
$$\eta = 1 - \dfrac{T_{2}}{T_{1}}$$
with $T_{1}$ and $T_{2}$ in kelvin. Raising the source temperature or lowering the sink temperature raises the efficiency.
Pause and predict: can a Carnot engine reach 100% efficiency?
Check your answer: only if $T_{2}$ = 0 K, since then $\eta = 1 - 0$. Absolute zero cannot be reached, and an engine with $Q_{2}$ = 0 would violate the Kelvin–Planck statement. So the efficiency is always below 100%.
Worked example: a Carnot engine. A Carnot engine works between a source at 500 K and a sink at 300 K, absorbing 2000 J per cycle. Find its efficiency, the work per cycle and the heat rejected. Then find the source temperature that would give 50% with the sink unchanged.
| Step | What we do | Result |
|---|---|---|
| 1 | Efficiency, $\eta = 1 - T_{2}/T_{1}$ | $1 - 300/500 = 0.40$ |
| 2 | Work, $W = \eta Q_{1}$ | 0.40 × 2000 = 800 J |
| 3 | Heat rejected, $Q_{2} = Q_{1} - W$ | 1200 J |
| 4 | For $\eta = 0.50$, $T_{1} = T_{2}/(1 - \eta)$ | $300/0.5 = 600$ K |
The wrong step is to put the temperatures in Celsius. With 227 °C and 27 °C, the formula would give $1 - 27/227$ = 88%, wildly wrong. Because $Q_{2}/Q_{1} = T_{2}/T_{1}$ comes from the gas laws, absolute temperatures are essential.
Beyond the textbook. For the exam: $\eta = 1 - T_{2}/T_{1}$ is the maximum efficiency between two temperatures. In reality: a Carnot engine would need infinitely slow steps to stay reversible, so it would deliver no useful power. Real power stations, working between a steam temperature of a few hundred degrees and a cooling temperature near room temperature, typically convert only about a third to two fifths of the fuel's energy into electricity. Why it matters: the Carnot limit tells engineers how much room for improvement exists, and the gap between it and real engines is the price of friction, heat loss and speed.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "A gas contains heat." | We say "hot gas" | It contains internal energy; heat is energy in transit |
| "In $\Delta Q = \Delta U + \Delta W$, $\Delta W$ is work done on the gas." | Many books use work on the system | Here $\Delta W$ is work done by the system |
| "Isothermal means no heat is exchanged." | Both isothermal and adiabatic sound like 'no change' | Isothermal: $\Delta U = 0$ and Q = W. Adiabatic: Q = 0 |
| "Use $PV = \text{constant}$ in an adiabatic process." | It is the more familiar law | Adiabatic: $PV^{\gamma} = \text{constant}$ |
| "Use Celsius in Carnot efficiency." | Temperatures are often given in Celsius | Always use kelvin |
| "Efficiency depends on the working gas." | Real engines differ by fuel | Carnot efficiency depends only on $T_{1}$ and $T_{2}$ |
| "Cyclic means no work." | The system returns to its start | $\Delta U = 0$, but net Q = net W, which equals the loop area |
| "The Zeroth Law was discovered first." | It is numbered zero | It was formulated last, in 1931, but logically precedes the others |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| First law | $\Delta Q = \Delta U + \Delta W$ | J | Any process |
| Work at constant pressure | $W = P\Delta V = \mu R\Delta T$ | J | Isobaric |
| Internal energy of ideal gas | $\Delta U = \mu C_{V}\Delta T$ | J | Any process |
| Mayer's relation | $C_{P} - C_{V}$ = R | J mol⁻¹ K⁻¹ | Any ideal gas |
| Ratio of specific heats | $\gamma = C_{P}/C_{V}$ | none | Adiabatic problems |
| Isothermal work | $W = \mu RT \ln(V_{2}/V_{1})$ | J | Constant T |
| Adiabatic relations | $PV^{\gamma} = \text{const}$, $TV^{\gamma-1} = \text{const}$ | none | Q = 0 |
| Adiabatic work | $W = \mu R(T_{1} - T_{2})/(\gamma - 1)$ | J | Q = 0 |
| Efficiency | $\eta = W/Q_{1} = 1 - Q_{2}/Q_{1}$ | none | Any heat engine |
| Carnot efficiency | $\eta = 1 - T_{2}/T_{1}$ | none | Reversible engine, T in kelvin |
60-second revision
| Topic | Lock this in |
|---|---|
| Zeroth law | Defines temperature, basis of thermometers |
| State vs path | U, P, V, T are state variables; Q and W depend on path |
| First law | $\Delta Q = \Delta U + \Delta W$ with W done by the system |
| Mayer | $C_{P} - C_{V}$ = R, $\gamma = C_{P}/C_{V}$ |
| Isothermal | $\Delta U = 0$, Q = $W = \mu RT \ln(V_{2}/V_{1})$ |
| Adiabatic | Q = 0, PV^$\gamma$ = const, expansion cools, steeper than isothermal |
| Isochoric, isobaric | $W = 0$ and $Q = \Delta U$; $W = P\Delta V$ and $Q = \mu C_{P}\Delta T$ |
| Cyclic | $\Delta U = 0$, work = enclosed area |
| Second law | Kelvin–Planck for engines, Clausius for refrigerators |
| Carnot | Two isothermal and two adiabatic steps, $\eta = 1 - T_{2}/T_{1}$, kelvin only |
Where this chapter connects
- Previous: Thermal Properties of Matter introduces heat, specific heat and temperature scales that this chapter turns into laws.
- Next: Kinetic Theory explains internal energy and the values of $C_{V}$, $C_{P}$ and $\gamma$ from molecular motion.
- Same ideas elsewhere: Work, Energy and Power is where energy conservation, which the First Law extends to heat, comes from.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Bodies A and B are each in thermal equilibrium with a body C. A and B are now placed in contact. What happens?
▸ Show answer & explanation
Answer: C
This is the Zeroth Law. Two systems in thermal equilibrium with a third are in thermal equilibrium with each other, so they share the same temperature and no net heat flows. The law is what justifies defining temperature as the quantity that is equal in equilibrium.
Q2. An ideal gas is compressed adiabatically. In the form $\Delta Q = \Delta U + \Delta W$, where $\Delta W$ is work done by the gas, which statement is correct?
▸ Show answer & explanation
Answer: B
In an adiabatic process $\Delta Q = 0$, so $\Delta U = -\Delta W$. During compression the surroundings do work on the gas, so the work done by the gas is negative and $\Delta U$ is positive. The gas heats up. Option A describes expansion, and option D describes an isothermal process.
Q3. An electric heater supplies heat to a system at the rate of 100 W. The system does work at the rate of 75 J s⁻¹. The rate at which the internal energy of the system increases is
▸ Show answer & explanation
Answer: A
By the first law, the rate of change of U is the heat rate minus the work rate: 100 − 75 = 25 W. Option D adds the two rates, which would be the case only if the work were being done on the system.
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Q4. When 1 g of water at 100 °C changes to steam at atmospheric pressure ($1.013 \times 10^{5}$ Pa), its volume increases from 1 cm³ to 1671 cm³ and it absorbs 2256 J. The change in internal energy is about
▸ Show answer & explanation
Answer: B
The work done by the expanding steam is $P\Delta V = 1.013 \times 10^{5} \times 1670 \times 10^{-6} \approx 169$ J. Then $\Delta U = \Delta Q - \Delta W = 2256 - 169 \approx 2087$ J. Option A is the work alone, C is the heat alone, and D adds the work instead of subtracting it.
Q5. The molar specific heat at constant volume of helium is 12.5 J mol⁻¹ K⁻¹. Take R = 8.31 J mol⁻¹ K⁻¹. Its molar specific heat at constant pressure is about
▸ Show answer & explanation
Answer: C
Mayer's relation gives $C_{P} = C_{V} + R = 12.5 + 8.31 = 20.8$ J mol⁻¹ K⁻¹. At constant pressure the gas does expansion work in addition to warming up, so $C_{P}$ must be larger than $C_{V}$. Option A subtracts R by mistake.
Q6. Match List I with List II. List I: (i) Isothermal expansion (ii) Adiabatic expansion (iii) Isochoric heating (iv) Isobaric expansion. List II: (p) Work done is zero (q) Heat absorbed equals work done (r) Gas does work at the cost of its internal energy (s) Work done is $P\Delta V$
▸ Show answer & explanation
Answer: A
In an isothermal process $\Delta U = 0$, so Q = W. In an adiabatic expansion Q = 0, so the work comes out of internal energy. With the volume fixed there is no work, and at constant pressure the work is P$\Delta V$. Options B and C swap pairs, and D mixes up all four.
Q7. Assertion (A): When an ideal gas expands isothermally, the heat absorbed by the gas equals the work done by it. Reason (R): The internal energy of an ideal gas depends only on its temperature.
▸ Show answer & explanation
Answer: A
At constant temperature an ideal gas has $\Delta U = 0$ because U depends only on T. The first law then reduces to $\Delta Q = \Delta W$. So R is exactly the reason why A holds.
Q8. Two moles of an ideal gas expand isothermally at 300 K from 2 L to 6 L. Take R = 8.3 J mol⁻¹ K⁻¹ and $\ln 3 = 1.10$. The heat absorbed is closest to
▸ Show answer & explanation
Answer: C
$Q = W = \mu RT \ln(V_{2}/V_{1}) = 2 \times 8.3 \times 300 \times 1.10 \approx 5.47 \times 10^{3}$ J. Option A is the answer for one mole, B leaves out the logarithm ($\mu RT$ alone), and D doubles the correct value.
Q9. At the point where an isothermal curve and an adiabatic curve of the same ideal gas cross on a P–V diagram, the magnitude of the slope of the adiabatic curve is how many times that of the isothermal curve?
▸ Show answer & explanation
Answer: B
Along the isothermal $dP/dV = -P/V$, and along the adiabatic $dP/dV = -\gamma P/V$ at the same P and V. The ratio is $\gamma$, which is greater than 1 for every gas, so the adiabatic curve is the steeper one.
Q10. A monatomic ideal gas ($\gamma = 5/3$) at 27 °C is compressed adiabatically to one eighth of its initial volume. The final temperature is
▸ Show answer & explanation
Answer: C
$T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$ with $\gamma - 1 = 2/3$, so $T_{2} = 300 \times 8^{2/3} = 300 \times 4 = 1200$ K. Option B uses a factor of 2, which is the answer for $\gamma = 4/3$ with a volume ratio of 8, and A would be an isothermal process.
Q11. A Carnot engine with its sink at 300 K has an efficiency of 40%. To raise the efficiency to 50% with the sink unchanged, the source temperature must be raised by
▸ Show answer & explanation
Answer: B
At 40%, $0.6 = T_{2}/T_{1}$, so $T_{1} = 300/0.6 = 500$ K. At 50%, $T_{1} = 300/0.5 = 600$ K. The rise is 100 K. Option A is the tempting 10% of 500 K, and efficiency is not linear in $T_{1}$.
Q12. Which statement is the Clausius form of the Second Law of Thermodynamics?
▸ Show answer & explanation
Answer: B
The Clausius statement concerns refrigerators: heat does not flow from cold to hot by itself. Option A is the Kelvin–Planck statement about heat engines, C is the First Law and D is a property of ideal gases.
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