Topics
6📖 1. Chapter Overview
Overview
Thermodynamics studies heat, work, temperature and internal energy in macroscopic systems such as gases, engines and refrigerators. It begins with thermal equilibrium and the zeroth law, which gives the scientific basis of temperature measurement. The first law of thermodynamics is energy conservation applied to heat and work: heat supplied to a system changes internal energy and may do work. Different thermodynamic processes such as isothermal, adiabatic, isobaric and isochoric processes are best understood using PV diagrams. The second law explains why natural processes have a direction and why heat engines cannot convert all absorbed heat into work. Entropy measures disorder or energy dispersal. Carnot engine gives the maximum possible efficiency between two temperatures, making this chapter very important for NEET.
- 1Thermodynamics deals with macroscopic variables like pressure, volume, temperature and internal energy.
- 2A thermodynamic system may be open, closed or isolated depending on exchange of matter and energy.
- 3Work done by gas is positive during expansion and negative during compression under the common NEET convention.
- 4Isothermal ideal gas process has ΔU = 0 because temperature is constant.
- 5Adiabatic process has ΔQ = 0 because no heat is exchanged.
- 6Entropy of an isolated system never decreases in natural processes.
- 7Carnot engine is reversible and has maximum efficiency for given source and sink temperatures.
First Law
Heat given becomes internal energy plus work: Q goes to U and W.
Processes
Iso means same: isothermal same temperature, isobaric same pressure, isochoric same volume.
Daily Life Example
A car engine takes heat from burning fuel, converts part of it into mechanical work and rejects remaining heat to surroundings.
NEET Quick Example
If 500 J heat is supplied to a gas and it does 200 J work, then ΔU = 500 - 200 = 300 J.
Confusing Heat with Internal Energy
Heat is energy in transit due to temperature difference; internal energy belongs to the state of the system.
Using Celsius in Carnot Efficiency
Carnot efficiency requires absolute temperatures in kelvin.
Ignoring Sign Convention
Under NEET convention, work done by gas during expansion is positive and during compression is negative.
Heat supplied to a system is used to increase internal energy and to do work by the system.
Variables
ΔQ=Heat supplied to the system
ΔU=Change in internal energy
ΔW=Work done by the system
Work done by a gas equals the area under the PV curve.
Variables
W=Work done by gas
P=Pressure
dV=Small change in volume
🌡️ 2. Thermal Equilibrium & Zeroth Law
Overview
Thermal equilibrium is the condition in which two bodies in thermal contact do not exchange net heat. This happens when they have the same temperature. Temperature is therefore the property that decides the direction of heat flow. The zeroth law of thermodynamics states that if two systems are separately in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law makes temperature measurement possible because a thermometer acts as the third system. Thermodynamic systems may be open, closed or isolated depending on exchange of matter and energy. A state of a system is described by state variables such as pressure, volume, temperature and internal energy.
- 1Thermal equilibrium does not mean equal internal energy; it means equal temperature.
- 2Temperature is a state variable.
- 3Pressure, volume, temperature and internal energy define thermodynamic state for many systems.
- 4State variables depend only on current state, not on path.
- 5A system plus surroundings together form the universe in thermodynamics.
- 6Zeroth law logically precedes the first and second laws because it defines temperature comparison.
Zeroth Law
If A matches C and B matches C, then A matches B.
System Types
Open exchanges all, closed exchanges energy only, isolated exchanges nothing.
Solved Concept Example
If a thermometer is in equilibrium with water and the same thermometer is in equilibrium with oil, both water and oil have the same temperature.
System Example
Gas in a sealed conducting cylinder is a closed system because it can exchange heat but not matter.
Confusing Thermal Equilibrium with Equal Heat
Thermal equilibrium means equal temperature, not equal heat or equal internal energy.
Calling Heat a State Variable
Heat is a path function, not a property stored in a system.
Ignoring System Boundary
Whether a system is open, closed or isolated depends on what boundary you choose.
Two bodies A and B are in thermal equilibrium when their temperatures are equal.
Variables
TA=Temperature of body A
TB=Temperature of body B
If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other.
Variables
TA, TB, TC=Temperatures of systems A, B and C
🔥 3. Heat, Work & Internal Energy
Overview
Heat, work and internal energy are central to thermodynamics. Heat is energy transferred due to temperature difference. Work is energy transferred when a system expands or compresses against external pressure. Internal energy is the total microscopic kinetic and potential energy of molecules in the system. For an ideal gas, internal energy depends only on temperature. Heat and work are path functions because their values depend on the process path, not only initial and final states. Internal energy is a state function because its change depends only on initial and final states. In the common NEET convention, heat supplied to the system is positive and work done by the system during expansion is positive.
- 1A system does not contain heat; heat crosses the boundary during a process.
- 2PV work occurs when gas volume changes.
- 3At constant volume, work done by gas is zero.
- 4Area under PV curve gives work done by gas.
- 5Internal energy of ideal gas depends only on temperature, not pressure or volume separately.
- 6Same initial and final states can have different heat and work values along different paths.
State vs Path
State is where you are; path is how you got there. U is state, Q and W are path.
Work Sign
Gas expands and gives work out, so W is positive.
Numerical Problem
A gas expands at constant pressure 2 × 10^5 Pa from 1 L to 3 L. W = PΔV = 2×10^5×2×10^-3 = 400 J.
Path Function Example
A gas can go from the same initial state to the same final state by different paths, giving different work values, but the same ΔU.
Calling Heat a State Function
Heat depends on the path of energy transfer, so it is not a state function.
Forgetting Work Is Zero at Constant Volume
If volume does not change, dV = 0 and W = 0.
Assuming Internal Energy Depends on Volume for Ideal Gas
For an ideal gas, internal energy depends only on temperature.
Work done by gas during volume change equals area under PV curve.
Variables
W=Work done by gas
P=Gas pressure
dV=Small change in volume
Work done during an isobaric process.
Variables
P=Constant pressure
V1=Initial volume
V2=Final volume
Change in internal energy of ideal gas depends only on temperature change.
Variables
n=Number of moles
Cv=Molar heat capacity at constant volume
ΔT=Temperature change
⚙️ 4. First Law of Thermodynamics
Overview
The first law of thermodynamics is the law of conservation of energy applied to thermodynamic systems. It states that heat supplied to a system is used partly to increase its internal energy and partly to do work by the system. Mathematically, ΔQ = ΔU + ΔW under the common convention where work done by the system is positive. This law shows that heat and work are two ways of transferring energy across a system boundary. It is used in every thermodynamic process: isothermal, adiabatic, isochoric and isobaric. In numerical problems, identifying which quantity is zero or known is the main step. It is one of the most important NEET formulas.
- 1The first law does not tell the direction of natural processes; that is done by the second law.
- 2For an ideal gas, ΔU depends only on ΔT.
- 3For isothermal ideal gas process, Q = W.
- 4For adiabatic process, ΔU = -W.
- 5For isochoric process, Q = ΔU.
- 6For cyclic process, heat absorbed over cycle equals net work done.
First Law Rearrangement
If Q = U + W, then U = Q - W. Subtract work done by gas from heat supplied.
Zero Conditions
Isochoric means W zero, isothermal ideal gas means U zero, adiabatic means Q zero.
Solved Example
A gas absorbs 800 J heat and does 300 J work. From ΔQ = ΔU + ΔW, ΔU = 800 - 300 = 500 J.
Previous NEET Style
In a cyclic process, if gas absorbs 1000 J heat in total, net work done by gas is also 1000 J because ΔU = 0.
Wrong Work Sign
Expansion work by gas is positive; compression work by gas is negative.
Forgetting ΔU = 0 in a Cycle
Internal energy is a state function, so it returns to initial value after a complete cycle.
Applying ΔU = 0 to All Isothermal Processes
ΔU = 0 for isothermal ideal gas, because ideal gas internal energy depends only on temperature.
Heat supplied equals change in internal energy plus work done by the system.
Variables
ΔQ=Heat supplied to system
ΔU=Change in internal energy
ΔW=Work done by system
Internal energy change of an ideal gas depends only on temperature change.
Variables
n=Number of moles
Cv=Molar heat capacity at constant volume
ΔT=Temperature change
At constant volume, no boundary work is done.
Variables
ΔV=Change in volume
W=Work done
📈 5. Thermodynamic Processes
Overview
A thermodynamic process is a path by which a system changes from one state to another. In an isothermal process, temperature remains constant; for an ideal gas, internal energy remains constant and heat supplied equals work done. In an adiabatic process, no heat is exchanged with surroundings, so work is done at the cost of internal energy. In an isochoric process, volume remains constant and work done is zero. In an isobaric process, pressure remains constant and work is PΔV. PV diagrams visually show these processes, and the area under a PV curve gives work done by gas. NEET frequently asks process identification, work comparison and first-law applications.
- 1For ideal gas in isothermal process, ΔU = 0.
- 2For adiabatic process, PV^γ = constant.
- 3For isochoric process, heat supplied only changes internal energy.
- 4For isobaric process, heat changes internal energy and also does work.
- 5Expansion gives positive area and positive work by gas.
- 6Compression gives negative work by gas.
- 7Adiabatic expansion cools the gas; adiabatic compression heats it.
Iso Processes
Thermal = temperature, baric = pressure, choric = volume.
Adiabatic
A-diabatic means no heat can pass across the boundary.
Isobaric Work Example
A gas expands at pressure 10^5 Pa from 2 L to 5 L. W = PΔV = 10^5 × 3 × 10^-3 = 300 J.
Isochoric Example
If a gas is heated in a rigid container, volume remains constant, so W = 0 and Q = ΔU.
Confusing Isothermal and Adiabatic
Isothermal keeps temperature constant; adiabatic keeps heat exchange zero.
Forgetting Isochoric Work Is Zero
Work needs volume change, so at constant volume W = 0.
Using PV = constant for Adiabatic
PV = constant is isothermal for ideal gas; adiabatic uses PV^γ = constant.
Work done by ideal gas during reversible isothermal process.
Variables
W=Work done by gas
n=Number of moles
R=Universal gas constant
T=Constant absolute temperature
V1, V2=Initial and final volumes
Pressure-volume relation for reversible adiabatic process of an ideal gas.
Variables
P=Pressure
V=Volume
γ=Ratio of heat capacities Cp/Cv
Work done by gas in a reversible adiabatic process.
Variables
P1, V1=Initial pressure and volume
P2, V2=Final pressure and volume
γ=Adiabatic index
♻️ 6. Second Law & Entropy
Overview
The first law tells us energy is conserved, but it does not tell which processes are naturally possible. The second law of thermodynamics gives the direction of natural processes. One statement says heat cannot spontaneously flow from a colder body to a hotter body without external work. Another says no heat engine can convert all absorbed heat into work in a complete cycle. Reversible processes are ideal and can be exactly retraced, while irreversible processes occur naturally with friction, heat flow through finite temperature difference, mixing and free expansion. Entropy is a state function that measures energy dispersal or disorder. For an isolated system, entropy never decreases, which explains why natural processes are irreversible.
- 1The second law does not violate the first law; it adds direction and limitations.
- 2A heat engine must reject some heat to a sink.
- 3Friction and unrestrained expansion make processes irreversible.
- 4Entropy change of the universe is zero for reversible processes and positive for irreversible processes.
- 5Heat cannot flow from cold to hot unless external work is supplied, as in a refrigerator.
- 6Natural processes move toward greater total entropy.
Second Law
Energy is conserved by first law; direction is decided by second law.
Entropy
Entropy tracks spreading: energy and matter naturally spread out.
Concept Example
A hot cup of tea cools in a room, but a cold cup never spontaneously becomes hotter than the room without external work.
Entropy Example
When ice melts in a warm room, entropy of ice-water system may increase, and total entropy of universe increases.
Thinking First Law Allows 100% Engine
First law allows energy conservation, but second law forbids complete conversion of heat into work in a cycle.
Saying Entropy Always Increases for System Alone
Entropy of a system can decrease, but entropy of the universe cannot decrease.
Calling Fast Real Processes Reversible
Reversible processes are ideal, quasi-static and free from dissipative effects.
Entropy change when heat Qrev is reversibly transferred at absolute temperature T.
Variables
ΔS=Change in entropy
Qrev=Heat transferred reversibly
T=Absolute temperature
Efficiency of a heat engine in terms of work output and heat exchanges.
Variables
η=Efficiency of engine
W=Work output
Q1=Heat absorbed from source
Q2=Heat rejected to sink
🚗 7. Carnot Engine
Overview
A Carnot engine is an ideal reversible heat engine operating between two reservoirs at temperatures T1 and T2. Its cycle consists of two isothermal and two adiabatic processes. During isothermal expansion at the hot reservoir, the gas absorbs heat Q1. During adiabatic expansion, its temperature falls from T1 to T2. During isothermal compression at the cold reservoir, it rejects heat Q2. During adiabatic compression, its temperature rises back to T1. Carnot theorem states that no engine working between the same two temperatures can be more efficient than a Carnot engine. Its efficiency is η = 1 - T2/T1. A refrigerator is a reversed heat engine that uses work to transfer heat from cold to hot.
- 1Carnot engine gives maximum theoretical efficiency, not practical engine efficiency.
- 2A Carnot engine can never have 100% efficiency unless sink temperature is 0 K, which is unattainable.
- 3The working substance returns to its initial state after each cycle, so ΔU = 0 per cycle.
- 4Net work per cycle equals Q1 - Q2.
- 5On PV diagram, area enclosed by cycle gives net work done.
- 6On TS diagram, heat transfer during reversible process equals area under T-S curve.
- 7Coefficient of performance is used for refrigerators and heat pumps instead of efficiency.
Carnot Cycle Order
Hot iso expansion, adiabatic cooling, cold iso compression, adiabatic heating.
Efficiency
Carnot efficiency depends only on temperatures, not on working gas.
Solved Numerical
A Carnot engine works between 600 K and 300 K. Efficiency η = 1 - 300/600 = 0.5 = 50%.
Refrigerator Example
A Carnot refrigerator works between 300 K and 270 K. COP = T2/(T1 - T2) = 270/(30) = 9.
Previous NEET Style
If source temperature is doubled while sink temperature remains same, Carnot efficiency increases because T2/T1 decreases.
Using Celsius in η = 1 - T2/T1
Always convert source and sink temperatures to kelvin.
Expecting 100% Efficiency
A heat engine must reject some heat to a sink; 100% efficiency is impossible for finite sink temperature.
Confusing Refrigerator COP with Engine Efficiency
Refrigerator performance is Q removed per work input, not work output per heat input.
Maximum possible efficiency of a heat engine between hot reservoir T1 and cold reservoir T2.
Variables
η=Efficiency of Carnot engine
T1=Temperature of source in kelvin
T2=Temperature of sink in kelvin
Efficiency of any heat engine in terms of heat absorbed, heat rejected and work output.
Variables
W=Work output per cycle
Q1=Heat absorbed from hot reservoir
Q2=Heat rejected to cold reservoir
For a reversible Carnot cycle, heat exchanged is proportional to reservoir temperature.
Variables
Q1, Q2=Heat absorbed and heat rejected
T1, T2=Absolute temperatures of reservoirs
Formula Sheet
10Heat supplied to a system is used to increase internal energy and to do work by the system.
Variables
ΔQ=Heat supplied to the system
ΔU=Change in internal energy
ΔW=Work done by the system
Work done by a gas equals the area under the PV curve.
Variables
W=Work done by gas
P=Pressure
dV=Small change in volume
For an ideal gas, internal energy change depends only on temperature change.
Variables
n=Number of moles
Cv=Molar heat capacity at constant volume
ΔT=Change in temperature
Maximum possible efficiency of a heat engine operating between source temperature T1 and sink temperature T2.
Variables
η=Efficiency
T1=Absolute temperature of hot reservoir
T2=Absolute temperature of cold reservoir
Two bodies A and B are in thermal equilibrium when their temperatures are equal.
Variables
TA=Temperature of body A
TB=Temperature of body B
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NEET PYQs — Thermodynamics
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At a certain temperature T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:
Consider the following reaction: 2A(g) + B(g) → 2D(g) ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (Given: R = 8.31 J mol⁻¹ K⁻¹)
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:
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