Follow the energy, identify the path, then ask whether the second law permits it.
JEE Main · JEE Advanced · NEET UG80 min study guide9 indexed sections
BY THE END, YOU CAN
Apply a declared first-law sign convention and distinguish state functions from path transfers.
Derive work, heat and energy changes for ideal-gas processes.
Analyse cycles, engine efficiency and refrigerator performance.
Use entropy and Carnot bounds to test physical possibility.
BEFORE YOU START
Ideal-gas equation, absolute temperature and elementary logarithms.
Areas under graphs and simple integration.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2014. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
An externally insulated vessel contains 2 mol monatomic ideal gas at 700 K below a fixed conducting partition, and 2 mol diatomic ideal gas at 400 K above it. A frictionless top piston maintains the upper gas at constant external pressure. Use C_V,mono=3R/2 and C_P,dia=7R/2. Find the common equilibrium temperature.
Reveal answer & worked solution
Answer: 490 K (independently derived; archived option D).
The lower chamber has fixed volume, so its heat loss is 2(3R/2)(700−T). The upper gas expands at constant pressure, so its heat gain is 2(7R/2)(T−400).
The outer insulation means these exchanged heats balance: 3R(700−T)=7R(T−400).
Hence 2100−3T=7T−2800, giving 10T=4900 and T=490 K. Using C_V for both gases would omit the piston work.
Source: JEE (Advanced) 2014, Paper 2, Code 1, Q11; independently derived solution; archived-key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. A state is a snapshot; a process is a route
Thermodynamics describes bulk energy exchanges without following every molecule. Define the system boundary first: the gas inside a cylinder, a complete engine, or two bodies exchanging heat. A closed system exchanges energy but not matter; an isolated system exchanges neither. Many apparent energy paradoxes disappear when the boundary and surroundings are explicitly named.
Pressure, volume, temperature and internal energy describe an equilibrium state. A process links states; a cycle returns to the initial state. Internal energy U is a state function, so its total change around a cycle is zero. Heat Q and work W describe energy crossing a boundary during a process. They are not stored substances or properties attached to a state.
Use a consistent sign convention throughout this lesson: Q is positive into the system and W is positive when the system does work on its surroundings. Then ΔU = Q − W. Some chemistry conventions define work positive on the system, giving a plus sign instead. Either convention works if every sign and interpretation follows it.
Temperature and heat are not synonyms. Temperature characterises thermal state; heat is energy transfer driven by a temperature difference. A gas can gain internal energy through compression without heat transfer. A gas can absorb heat while its temperature stays constant if the energy leaves through work. Always ask which transfer mechanism occurred.
ΔU = Q − W
Q > 0 for heat entering; W > 0 for work done by the system. Changes in bulk kinetic/potential energy are neglected unless explicitly included.
ORIGINAL WORKED EXAMPLE
Read the signs before calculating
A gas absorbs 700 J of heat while surroundings do 200 J of work on it. Find its internal-energy change.
Heat entering gives Q = +700 J.
Work done on the gas corresponds to W = −200 J in the by-system convention.
Thus ΔU = 700 − (−200) = 900 J. Both transfers raise the gas’s energy.
Answer ΔU = +900 J.
02 / UNDERSTAND & APPLY
2. Pressure–volume diagrams make path dependence visible
For quasistatic expansion against a well-defined opposing pressure, a small volume change does boundary work δW = pext dV. In the reversible limit the gas pressure differs only infinitesimally from external pressure, so W = ∫p dV along the equilibrium curve. Expansion gives positive work and compression negative work under our convention.
On a pressure–volume diagram, the signed area under a quasistatic path is work by the gas. A vertical path has zero boundary work because volume does not change. A horizontal path at constant pressure gives pΔV. A straight sloping segment gives average endpoint pressure times volume change. Curved paths require the actual pressure function or a valid integral.
Different paths between the same endpoints can enclose different areas, so their work and heat differ even though ΔU is the same. A clockwise cycle has positive net work by the gas; an anticlockwise cycle requires net work input. The enclosed area is net work, not the total heat absorbed on the heating portions.
An irreversible process need not have a unique uniform gas pressure at every intermediate instant. Do not draw an invented equilibrium curve and integrate the gas pressure along it. Free expansion into vacuum is a crucial example: external pressure is zero, so boundary work is zero even though the volume increases. Adiabatic free expansion of an ideal gas has unchanged internal energy and temperature, yet is not a reversible adiabatic path.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative scaled coordinates. Both paths have the same state-function changes. Vertical segments do no pV work.
ORIGINAL WORKED EXAMPLE
Two paths, same endpoints
An ideal monatomic gas goes from (p₀,V₀) to (2p₀,2V₀). Path A first doubles pressure at fixed volume, then doubles volume at fixed pressure. Path B first doubles volume at p₀, then doubles pressure. Find W and Q for each in units of p₀V₀.
For a monatomic ideal gas U = 3pV/2, so ΔU = (3/2)(4 − 1)p₀V₀ = 4.5p₀V₀ for either path.
Path A has work only in its second leg: Wₐ = 2p₀(V₀) = 2p₀V₀. Thus Qₐ = 6.5p₀V₀.
Path B has work only in its first leg: Wᵦ = p₀V₀. Thus Qᵦ = 5.5p₀V₀.
Answer A: W = 2p₀V₀, Q = 6.5p₀V₀. B: W = p₀V₀, Q = 5.5p₀V₀.
03 / UNDERSTAND & APPLY
3. Internal energy and heat capacities
For a fixed amount n of ideal gas, pV = nRT with absolute temperature T in kelvin. The ideal-gas internal energy depends only on temperature. Over a range where the molar constant-volume heat capacity Cᵥ is effectively constant, ΔU = nCᵥΔT. This is a state-function relation even when the actual path is not at constant volume.
Heat capacity depends on the process used to add energy. At constant volume no pV boundary work occurs, so Qᵥ = nCᵥΔT. At constant pressure an expanding gas also does work nRΔT, making Qp = nCpΔT with Cp − Cᵥ = R for an ideal gas. This relation uses molar capacities; for mass-specific capacities replace R by the specific gas constant.
An ideal monatomic gas with translational degrees of freedom has Cᵥ = 3R/2, Cp = 5R/2 and γ = Cp/Cᵥ = 5/3. A rigid diatomic model with active rotation but negligible vibration gives 5R/2, 7R/2 and 7/5. These values depend on which molecular modes are thermally active; they are not universal at all temperatures.
A common mistake is to write Q = nCᵥΔT for every process. That expression is the internal-energy change for the stated ideal-gas model; it equals heat only when the net work term vanishes. Likewise ΔU = 0 does not mean Q = 0. The first law says heat and work may be equal and nonzero.
ΔU = nCᵥΔT; Cp − Cᵥ = R; γ = Cp/Cᵥ
Ideal gas with fixed amount; use constant heat capacities only over a suitable temperature interval.
04 / UNDERSTAND & APPLY
4. Four named processes, four different constraints
Isochoric means constant volume. Then W = 0 for pV work and Q = ΔU. Pressure changes in proportion to absolute temperature for fixed n. Isobaric means constant pressure. Then W = p(V₂ − V₁) = nR(T₂ − T₁), and heat must cover both internal-energy change and expansion work.
Isothermal means constant temperature. For an ideal gas, ΔU = 0. In reversible isothermal expansion, p = nRT/V and integrating gives W = nRT ln(V₂/V₁). The first law then gives Q = W. The logarithm’s argument is a dimensionless volume ratio. For compression the ratio is less than one, making both W and Q negative.
Adiabatic means no heat transfer, Q = 0. It does not mean constant temperature. If an ideal gas expands adiabatically while doing positive work, ΔU = −W is negative and it cools. A sufficiently insulated process or a process too rapid for significant thermal exchange may be approximately adiabatic, but rapidity alone does not establish reversibility.
Names are constraints, not visual labels. A straight line on a p–V diagram is not automatically isothermal or adiabatic. Read the equation of the path. If p is proportional to V, then T is proportional to V² for a fixed ideal gas, and heat/work must be calculated from that path and the first law.
ORIGINAL WORKED EXAMPLE
Isobaric heating
Two moles of ideal monatomic gas are heated at constant pressure from 300 K to 400 K. Express W, ΔU and Q using R, then take R = 8.31 J mol⁻¹ K⁻¹.
Temperature change is 100 K. Work is nRΔT = 200R J.
Internal-energy increase is n(3R/2)ΔT = 300R J.
Heat is ΔU + W = 500R J. Numerically these are 1662, 2493 and 4155 J.
Answer W = 1662 J; ΔU = 2493 J; Q = 4155 J.
ORIGINAL WORKED EXAMPLE
Reversible isothermal doubling
One mole of ideal gas expands reversibly at 300 K from V to 2V. Find work and heat with R = 8.31 and ln2 = 0.693.
Temperature is unchanged, so ideal-gas ΔU = 0.
Integrate nRT/V: W = nRT ln2 = 8.31 × 300 × 0.693 ≈ 1728 J.
Heat must enter by the same amount to maintain temperature while work leaves.
Answer W = Q ≈ 1.73 kJ; ΔU = 0.
05 / UNDERSTAND & APPLY
5. Derive the reversible adiabatic curve
For a reversible adiabatic change of an ideal gas with constant Cᵥ, the first law gives nCᵥ dT = −p dV. Substitute p = nRT/V and divide by T: Cᵥ dT/T = −R dV/V. Integration yields TV^(γ−1) = constant, using R/Cᵥ = γ − 1. Combining with pV = nRT gives pV^γ = constant.
The derivation contains important conditions: fixed gas amount, ideal-gas behaviour, reversible quasistatic pV work, no heat exchange and effectively constant heat capacities. Adiabatic free expansion does not obey this curve. A process with friction can be adiabatic while generating entropy and violating pV^γ = constant.
Between two states on the reversible adiabat, work by the gas is (p₁V₁ − p₂V₂)/(γ − 1), equivalently nCᵥ(T₁ − T₂). An expansion cools; a compression heats. At the same point in a p–V diagram, an adiabat is steeper than an isotherm because its slope magnitude is γp/V rather than p/V.
A polytropic path pV^k = constant is a broader mathematical model. For k ≠ 1, integrate to get W = (p₂V₂ − p₁V₁)/(1 − k). The k = 1 limit is logarithmic and corresponds to an ideal-gas isotherm. A polytropic exponent equals γ only for the reversible adiabatic model under the stated assumptions; the notation itself does not establish the heat exchange.
ORIGINAL WORKED EXAMPLE
An adiabatic volume change
An ideal monatomic gas expands reversibly and adiabatically from V₁ to 8V₁ starting at 480 K. Find final temperature.
For a monatomic ideal gas γ = 5/3, so γ − 1 = 2/3.
Use T₂/T₁ = (V₁/V₂)^(γ−1) = (1/8)^(2/3) = 1/4.
Thus T₂ = 120 K. This assumes the ideal monatomic constant-capacity model over the interval.
Answer 120 K.
06 / UNDERSTAND & APPLY
6. Cycles, engines and refrigerators
A heat engine completes a cycle while receiving heat Qh from a hot reservoir, rejecting heat Qc to a cold reservoir and delivering net work Wout. Here Qh and Qc are positive magnitudes. Since the working substance returns to its initial state, ΔUcycle = 0 and Wout = Qh − Qc. Thermal efficiency is η = Wout/Qh = 1 − Qc/Qh.
The denominator is heat absorbed, not algebraic net heat. On a multi-leg cycle, identify and add every positive heat input when defining efficiency. Do not assume that all expansion legs absorb heat or all compression legs reject heat without calculation; the first law determines each leg.
A refrigerator uses work input to move heat from a colder region to a hotter one. Its useful output is removed heat Qc, so coefficient of performance is COPR = Qc/Win. A heat pump’s useful output is delivered heat Qh, giving COPH = Qh/Win = COPR + 1. COP can exceed one without creating energy because the device moves heat as well as consuming work.
For a reversible Carnot engine between reservoirs at Th and Tc, ηC = 1 − Tc/Th. Temperatures must be in kelvin. No engine between those two reservoirs can exceed this efficiency. A real engine is lower because irreversibility generates entropy. The Carnot value depends on reservoir temperatures, not on the working substance or engineering cleverness alone.
η = 1 − Qc/Qh ≤ 1 − Tc/Th
Qh and Qc are positive heat magnitudes; the temperature bound applies to an engine operating between those two thermal reservoirs.
COPR,Carnot = Tc/(Th − Tc)
Ideal reversible refrigerator, with absolute temperatures and Th > Tc.
ORIGINAL WORKED EXAMPLE
Check a claimed engine
An engine takes 1200 J from a 600 K reservoir and rejects 500 J to a 300 K reservoir per cycle. Is the claim possible?
Energy conservation alone gives Wout = 700 J and claimed efficiency 700/1200 ≈ 58.3%.
The Carnot limit between the reservoirs is 1 − 300/600 = 50%.
The claim exceeds that limit, so it is impossible as stated despite satisfying the first law. Equivalently reservoir entropy change is 500/300 − 1200/600 = −1/3 J K⁻¹.
Answer Impossible: claimed 58.3% exceeds the 50% upper bound.
07 / UNDERSTAND & APPLY
7. The second law adds direction to energy accounting
Energy conservation allows many processes that never occur spontaneously. A hot object warming a cold one conserves energy, but the reverse isolated transfer would also conserve energy. The second law distinguishes the direction: the total entropy of an isolated system cannot decrease. Reversible ideal processes leave total entropy unchanged; irreversible processes increase it.
Entropy S is a state function. Its change is calculated along a reversible path connecting the same endpoints: ΔS = ∫δQrev/T. The actual process need not be reversible, but using its irreversible heat directly in this formula is generally invalid. For an ideal gas with constant Cᵥ, ΔS = nCᵥ ln(T₂/T₁) + nR ln(V₂/V₁).
An individual system’s entropy can decrease if entropy is transferred to its surroundings. A freezing liquid can lose entropy while the released heat raises the surroundings’ entropy by at least as much. Thus “entropy always increases” is an incomplete statement unless the system boundary is isolated or system and surroundings are considered together.
For heat Q transferred from a reservoir at Th to another at Tc < Th, total entropy change is Q/Tc − Q/Th > 0. The colder reservoir gains more entropy than the hotter one loses. For adiabatic free expansion of an ideal gas into a larger volume, Q = W = ΔU = 0 but ΔS = nR ln(V₂/V₁) > 0. This sharply separates adiabatic from isentropic.
ORIGINAL WORKED EXAMPLE
Entropy production during heat flow
600 J flows directly from a 400 K reservoir to a 300 K reservoir. Find the entropy change of each reservoir and the total.
The hot reservoir loses heat: ΔSh = −600/400 = −1.5 J K⁻¹.
Total change is +0.5 J K⁻¹. Energy is conserved but the transfer is irreversible.
Answer Hot −1.5 J K⁻¹; cold +2.0 J K⁻¹; total +0.5 J K⁻¹.
08 / UNDERSTAND & APPLY
8. Original practice across the three thermodynamic questions
For every problem ask separately: what is conserved, what is the path constraint, and is the proposed direction or efficiency allowed?
ORIGINAL PRACTICE · 1
A gas loses 80 J of heat and does 30 J of work. Find ΔU.
Reveal answer & reasoning
−110 J.
Q = −80 and W = +30, so ΔU = Q − W = −110 J.
ORIGINAL PRACTICE · 2
One mole of monatomic ideal gas warms by 20 K at fixed volume. Find W and Q in terms of R.
Reveal answer & reasoning
W = 0; Q = 30R J.
No volume change means no pV work. ΔU = (3R/2)(20) = 30R J and Q = ΔU.
ORIGINAL PRACTICE · 3
In a reversible isothermal ideal-gas compression, volume halves. What are the signs of W, Q and ΔU?
Reveal answer & reasoning
W negative; Q negative; ΔU zero.
Work nRT ln(1/2) is negative. Energy supplied as compression work must leave as heat to keep temperature fixed.
ORIGINAL PRACTICE · 4
Does Q = 0 imply pV^γ is constant?
Reveal answer & reasoning
No.
That relation also needs reversible quasistatic ideal-gas behaviour with suitable constant heat capacities. Adiabatic free expansion is a counterexample.
ORIGINAL PRACTICE · 5
A clockwise rectangular p–V cycle spans pressures 1 and 3 kPa and volumes 2 and 5 litres. Find net work.
Reveal answer & reasoning
6 J.
Enclosed area is (2 kPa)(3 L) = 6 J, because 1 kPa L = 1 J.
ORIGINAL PRACTICE · 6
A cyclic engine absorbs 900 J and rejects 600 J. Find work and efficiency.
Reveal answer & reasoning
300 J and 1/3 ≈ 33.3%.
ΔUcycle = 0 gives work = 900 − 600. Divide by absorbed heat 900.
ORIGINAL PRACTICE · 7
What is the Carnot efficiency between 27°C and 327°C?
Reveal answer & reasoning
50%.
Convert temperatures to approximately 300 and 600 K; η = 1 − 300/600.
ORIGINAL PRACTICE · 8
A refrigerator removes 800 J from its cold compartment with 200 J work input. Find COP and heat delivered to the room.
Reveal answer & reasoning
COP = 4; room receives 1000 J.
COPR = Qc/W = 800/200. Energy balance gives Qh = Qc + W.
ORIGINAL PRACTICE · 9
One mole of ideal gas freely expands adiabatically into twice its original volume. Find ΔU, ΔT and ΔS.
Reveal answer & reasoning
ΔU = 0; ΔT = 0; ΔS = R ln2.
Vacuum provides no opposing pressure, and insulation makes Q zero. Ideal-gas U depends only on T, but the increased volume raises entropy.
ORIGINAL PRACTICE · 10
Can a system’s entropy decrease during an irreversible process?
Reveal answer & reasoning
Yes, if sufficient entropy leaves to the surroundings.
The nondecrease rule applies to total entropy of an isolated system, not every subsystem individually.
09 / UNDERSTAND & APPLY
9. A compact audit of a thermodynamics solution
Write the sign convention at the top, label states with absolute temperature, and identify the process equation before integrating work. Compute ΔU from endpoints for the ideal-gas model, then obtain heat from the first law. For a cycle, verify that all internal-energy changes sum to zero and that net work matches the signed enclosed area.
Finally apply a second-law check. An engine’s efficiency must respect the reservoir-temperature limit. A reversible adiabat has no entropy change, but an arbitrary adiabatic process may generate entropy. A result satisfying energy conservation is not automatically physically possible. The best solution explains both the energy balance and the direction of the process.
Heat and work are transfers; internal energy and entropy are state functions.
Use external pressure for irreversible boundary work.
Isothermal and adiabatic are distinct constraints.
State the assumptions behind pV^γ = constant.
Use positive heat magnitudes consistently in efficiency and COP.
Entropy can decrease locally while increasing overall.
KEEP THE BIG IDEAS
Quick revision
The first law accounts for energy; the second law restricts direction.
The path sets work and heat, while endpoints set changes in state functions.
Reversible adiabatic formulas have stronger conditions than Q = 0.
Efficiency compares work with total absorbed heat, and reservoir temperatures must be absolute.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.