If Thermodynamic Processes are not clear and you are looking for a Physics Tutor, contact Kumar Sir.Phone: +91-9958461445Email: kumarsirphysics@gmail.com
Call Kumar SirWhatsApp Kumar SirVisit WebsiteEmail Kumar Sir

Thermodynamic Processes

Class 11 Physics notes covering Isothermal, Adiabatic, Isochoric and Isobaric Processes, PV Diagrams, Work Done, Process Comparison, Numericals and PYQs.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

Thermodynamic processes describe how a gas moves from one equilibrium state to another. The four standard processes are isothermal, adiabatic, isochoric and isobaric. Each process has a fixed condition, a characteristic P-V graph, a particular work formula and a different heat-internal-energy relation.

Introduction

A thermodynamic process is the path followed by a system when it changes from one state to another.

For a fixed amount of gas, the state is commonly described by pressure, volume and temperature. A process tells us how these quantities change. The same initial and final states may be connected by different paths, and the work done and heat exchanged may be different for different paths. This is why P-V diagrams are central in thermodynamics.

In exams, do not start with a formula blindly. First identify the constant quantity: temperature, heat, volume or pressure. Then choose the correct process formula. Isothermal means T constant. Adiabatic means Q = 0. Isochoric means V constant. Isobaric means P constant.

Path Matters

Work and heat depend on the path. P-V graph shape is therefore part of the physics, not decoration.

State Function

Internal energy change for an ideal gas depends only on temperature change, not on the chosen path.

Exam Habit

Mark the process condition before substitution: T constant, Q zero, V constant or P constant.

Isothermal Process

Condition T = constant   Formula PV = constant

For an ideal gas in an isothermal process, temperature remains constant. Since internal energy of an ideal gas depends only on temperature, ΔU = 0. Heat supplied during isothermal expansion is completely converted into work done by the gas. The process must be slow enough for heat exchange with surroundings to maintain temperature.

Physical and Molecular Meaning

At molecular level, the process condition controls how molecular speed, collision rate and wall pressure change. In isothermal process, the chosen constraint decides whether heat exchange, work transfer or internal energy change dominates the process.

Real-Life Examples

Examples include slow expansion of gas in a cylinder kept in contact with a heat reservoir and slow compression where heat is removed continuously.

Exam Perspective

  • Always write the defining condition first.
  • Draw the P-V graph shape before solving graph-based questions.
  • Check whether work is zero, positive or negative from volume change.
  • For ideal gas, connect internal energy change with temperature change.

Common Mistakes and Memory Tricks

Do not confuse process names. Isothermal has constant temperature, while adiabatic has zero heat exchange. Isochoric has zero work, while isobaric uses W = PΔV. The memory trick is: thermal relates to temperature, adiabatic blocks heat, choric relates to volume space, and baric relates to pressure.

Adiabatic Process

Condition Q = 0   Formula PVγ = constant

In an adiabatic process, no heat enters or leaves the system. Work is done at the cost of internal energy during expansion, so temperature falls. During adiabatic compression, work is done on the gas and temperature rises. The adiabatic P-V curve is steeper than the isothermal curve because pressure changes more rapidly with volume.

Physical and Molecular Meaning

At molecular level, the process condition controls how molecular speed, collision rate and wall pressure change. In adiabatic process, the chosen constraint decides whether heat exchange, work transfer or internal energy change dominates the process.

Real-Life Examples

Examples include rapid compression in a bicycle pump, diesel engine compression and fast expansion of gas from a nozzle.

Exam Perspective

  • Always write the defining condition first.
  • Draw the P-V graph shape before solving graph-based questions.
  • Check whether work is zero, positive or negative from volume change.
  • For ideal gas, connect internal energy change with temperature change.

Common Mistakes and Memory Tricks

Do not confuse process names. Isothermal has constant temperature, while adiabatic has zero heat exchange. Isochoric has zero work, while isobaric uses W = PΔV. The memory trick is: thermal relates to temperature, adiabatic blocks heat, choric relates to volume space, and baric relates to pressure.

Isochoric Process

Condition V = constant   Formula W = 0

In an isochoric process, the volume remains constant, so the boundary does not move and no P-V work is done. Heat supplied changes internal energy directly. Pressure and temperature change together for an ideal gas at constant volume.

Physical and Molecular Meaning

At molecular level, the process condition controls how molecular speed, collision rate and wall pressure change. In isochoric process, the chosen constraint decides whether heat exchange, work transfer or internal energy change dominates the process.

Real-Life Examples

Examples include heating gas in a rigid sealed container or pressure rise in a closed pressure cooker before steam release.

Exam Perspective

  • Always write the defining condition first.
  • Draw the P-V graph shape before solving graph-based questions.
  • Check whether work is zero, positive or negative from volume change.
  • For ideal gas, connect internal energy change with temperature change.

Common Mistakes and Memory Tricks

Do not confuse process names. Isothermal has constant temperature, while adiabatic has zero heat exchange. Isochoric has zero work, while isobaric uses W = PΔV. The memory trick is: thermal relates to temperature, adiabatic blocks heat, choric relates to volume space, and baric relates to pressure.

Isobaric Process

Condition P = constant   Formula W = P(V2 - V1)

In an isobaric process, pressure remains constant while volume and temperature may change. Heat supplied is used partly to increase internal energy and partly to do expansion work. On a P-V graph, the isobaric process is a horizontal line.

Physical and Molecular Meaning

At molecular level, the process condition controls how molecular speed, collision rate and wall pressure change. In isobaric process, the chosen constraint decides whether heat exchange, work transfer or internal energy change dominates the process.

Real-Life Examples

Examples include gas expanding under a freely moving piston carrying constant load.

Exam Perspective

  • Always write the defining condition first.
  • Draw the P-V graph shape before solving graph-based questions.
  • Check whether work is zero, positive or negative from volume change.
  • For ideal gas, connect internal energy change with temperature change.

Common Mistakes and Memory Tricks

Do not confuse process names. Isothermal has constant temperature, while adiabatic has zero heat exchange. Isochoric has zero work, while isobaric uses W = PΔV. The memory trick is: thermal relates to temperature, adiabatic blocks heat, choric relates to volume space, and baric relates to pressure.

P-V Diagrams

P-V diagrams show pressure on the vertical axis and volume on the horizontal axis. Work done is the area under the process curve.

The isothermal and adiabatic curves both slope downward for expansion, but the adiabatic curve is steeper. This is because during adiabatic expansion the gas cools, so pressure falls more rapidly than in isothermal expansion.

Isothermal Expansion in Piston

GasSlow expansionT constant, heat enters

Adiabatic Expansion

GasQ = 0Insulated boundary, T falls in expansion

Isochoric Heating

GasHeat inRigid walls: V constant, W = 0

Isobaric Expansion

GasP constantW = PΔV

Combined P-V Curves

VPIsothermalAdiabatic steeper

Molecular Interpretation

Molecular speed changes with temperature

Isothermal P-V Graph

VPPV = constant

Adiabatic P-V Graph

VPPVγ = constant

Isochoric and Isobaric P-V Graphs

VPIsochoricIsobaric

Work Done in Thermodynamic Processes

Work done by gas is the area under the P-V curve. The formula changes with the process path.

PV = constant

Isothermal ideal gas condition. Pressure decreases inversely with volume.

PVγ = constant

Adiabatic ideal gas condition. Curve is steeper than isothermal.

W = nRT ln(V2/V1)

Work for reversible isothermal ideal gas process.

W = P(V2 - V1)

Work for constant pressure process.

Q = ΔU

For isochoric process because W = 0.

ΔU = 0

For ideal gas isothermal process because temperature is constant.

Q = 0

For adiabatic process because there is no heat exchange.

W = 0

For isochoric process because volume is constant.

Process Comparison

ProcessConstant quantityHeat exchangeWork doneInternal energyFormulaP-V graphExam note
IsothermalTemperatureAllowedW = nRT ln(V2/V1)ΔU = 0 for ideal gasPV = constantRectangular hyperbolaHeat supplied equals work in ideal gas expansion.
AdiabaticHeat exchange zeroQ = 0Done at cost of internal energyChanges with temperaturePVγ = constantSteeper than isothermalRapid or insulated process.
IsochoricVolumeAllowedW = 0ΔU = QV constantVertical lineRigid container.
IsobaricPressureAllowedW = PΔVDepends on temperature changeP constantHorizontal lineMovable piston under constant load.

Isothermal vs Adiabatic

Both may appear as downward curves during expansion, but isothermal keeps temperature constant by heat exchange, while adiabatic blocks heat exchange and therefore temperature changes. For the same initial point, the adiabatic expansion curve drops more steeply.

Isochoric vs Isobaric

Isochoric has no work because volume does not change. Isobaric has easy work calculation because pressure is constant. On P-V diagrams, isochoric is vertical and isobaric is horizontal.

If Thermodynamic Processes are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

Call Kumar SirWhatsApp Kumar SirVisit WebsiteEmail Kumar Sir

50 Solved Numericals

Attempt each question first, then open the solution.

Numerical 1 Easy. An ideal gas expands isothermally at 300 K from 2 L to 4 L. If n = 1 mol, find work done. Take R = 8.314 J mol-1 K-1 and ln2 = 0.693.
Show Solution

Given: n = 1, T = 300 K, V2/V1 = 2.

Formula: W = nRT ln(V2/V1).

Solution: W = 1 × 8.314 × 300 × 0.693 = 1728 J approximately.

Final Answer: Work done = 1.73 × 103 J.

Numerical 2 Easy. A gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work done.
Show Solution

Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.

Final Answer: Work done = 1000 J.

Numerical 3 Easy. A gas is heated at constant volume. If heat supplied is 600 J, find work done and change in internal energy.
Show Solution

Given: Isochoric process, Q = 600 J.

Formula: W = 0, ΔU = Q.

Solution: At constant volume, no boundary displacement occurs. Therefore W = 0 and ΔU = 600 J.

Final Answer: W = 0, ΔU = 600 J.

Numerical 4 Easy. In an adiabatic process, a gas does 450 J work. Find heat exchanged and change in internal energy.
Show Solution

Given: Adiabatic process, W = 450 J.

Formula: Q = 0, ΔU = Q - W.

Solution: ΔU = 0 - 450 = -450 J.

Final Answer: Q = 0, ΔU = -450 J.

Numerical 5 Easy. For an isothermal process of ideal gas, heat supplied is 900 J. Find ΔU and W.
Show Solution

Given: Ideal gas isothermal process, Q = 900 J.

Formula: ΔU = 0, Q = W.

Solution: Temperature is constant, so internal energy does not change. Therefore W = 900 J.

Final Answer: ΔU = 0, W = 900 J.

Numerical 6 CBSE. A gas is compressed isobarically from 10 L to 6 L under 1 × 105 Pa. Find work by gas.
Show Solution

Given: P = 1 × 105 Pa, ΔV = -4 L = -4 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.

Final Answer: Work by gas = -400 J.

Numerical 7 CBSE. Pressure of a gas doubles at constant volume. If initial temperature is 300 K, find final temperature.
Show Solution

Given: Isochoric process, P proportional to T.

Formula: P1/T1 = P2/T2.

Solution: Since pressure doubles, temperature doubles. T2 = 600 K.

Final Answer: Final temperature = 600 K.

Numerical 8 CBSE. Volume of a gas doubles isothermally. If initial pressure is 4 atm, find final pressure.
Show Solution

Given: Isothermal process, V2 = 2V1, P1 = 4 atm.

Formula: P1V1 = P2V2.

Solution: P2 = P1V1/2V1 = 2 atm.

Final Answer: Final pressure = 2 atm.

Numerical 9 CBSE. A gas expands adiabatically. It does 300 J work. What happens to temperature for an ideal gas?
Show Solution

Given: Adiabatic expansion, W = 300 J.

Formula: Q = 0, ΔU = -W.

Solution: ΔU = -300 J. For ideal gas, internal energy depends on temperature, so temperature decreases.

Final Answer: Temperature decreases.

Numerical 10 CBSE. At constant pressure, a gas absorbs 1200 J heat and does 400 J work. Find change in internal energy.
Show Solution

Given: Q = 1200 J, W = 400 J.

Formula: ΔU = Q - W.

Solution: ΔU = 1200 - 400 = 800 J.

Final Answer: Internal energy increases by 800 J.

Numerical 11 Easy. In process set 2, an ideal gas expands isothermally at 300 K from 2 L to 4 L. If n = 1 mol, find work done. Take R = 8.314 J mol-1 K-1 and ln2 = 0.693.
Show Solution

Given: n = 1, T = 300 K, V2/V1 = 2.

Formula: W = nRT ln(V2/V1).

Solution: W = 1 × 8.314 × 300 × 0.693 = 1728 J approximately.

Final Answer: Work done = 1.73 × 103 J.

Numerical 12 Easy. In process set 2, a gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work done.
Show Solution

Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.

Final Answer: Work done = 1000 J.

Numerical 13 Easy. In process set 2, a gas is heated at constant volume. If heat supplied is 600 J, find work done and change in internal energy.
Show Solution

Given: Isochoric process, Q = 600 J.

Formula: W = 0, ΔU = Q.

Solution: At constant volume, no boundary displacement occurs. Therefore W = 0 and ΔU = 600 J.

Final Answer: W = 0, ΔU = 600 J.

Numerical 14 Easy. In an adiabatic process, a gas does 450 J work. Find heat exchanged and change in internal energy.
Show Solution

Given: Adiabatic process, W = 450 J.

Formula: Q = 0, ΔU = Q - W.

Solution: ΔU = 0 - 450 = -450 J.

Final Answer: Q = 0, ΔU = -450 J.

Numerical 15 Easy. For an isothermal process of ideal gas, heat supplied is 900 J. Find ΔU and W.
Show Solution

Given: Ideal gas isothermal process, Q = 900 J.

Formula: ΔU = 0, Q = W.

Solution: Temperature is constant, so internal energy does not change. Therefore W = 900 J.

Final Answer: ΔU = 0, W = 900 J.

Numerical 16 CBSE. In process set 2, a gas is compressed isobarically from 10 L to 6 L under 1 × 105 Pa. Find work by gas.
Show Solution

Given: P = 1 × 105 Pa, ΔV = -4 L = -4 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.

Final Answer: Work by gas = -400 J.

Numerical 17 CBSE. Pressure of a gas doubles at constant volume. If initial temperature is 300 K, find final temperature.
Show Solution

Given: Isochoric process, P proportional to T.

Formula: P1/T1 = P2/T2.

Solution: Since pressure doubles, temperature doubles. T2 = 600 K.

Final Answer: Final temperature = 600 K.

Numerical 18 CBSE. Volume of a gas doubles isothermally. If initial pressure is 4 atm, find final pressure.
Show Solution

Given: Isothermal process, V2 = 2V1, P1 = 4 atm.

Formula: P1V1 = P2V2.

Solution: P2 = P1V1/2V1 = 2 atm.

Final Answer: Final pressure = 2 atm.

Numerical 19 CBSE. In process set 2, a gas expands adiabatically. It does 300 J work. What happens to temperature for an ideal gas?
Show Solution

Given: Adiabatic expansion, W = 300 J.

Formula: Q = 0, ΔU = -W.

Solution: ΔU = -300 J. For ideal gas, internal energy depends on temperature, so temperature decreases.

Final Answer: Temperature decreases.

Numerical 20 CBSE. At constant pressure, a gas absorbs 1200 J heat and does 400 J work. Find change in internal energy.
Show Solution

Given: Q = 1200 J, W = 400 J.

Formula: ΔU = Q - W.

Solution: ΔU = 1200 - 400 = 800 J.

Final Answer: Internal energy increases by 800 J.

Numerical 21 NEET. In process set 3, an ideal gas expands isothermally at 300 K from 2 L to 4 L. If n = 1 mol, find work done. Take R = 8.314 J mol-1 K-1 and ln2 = 0.693.
Show Solution

Given: n = 1, T = 300 K, V2/V1 = 2.

Formula: W = nRT ln(V2/V1).

Solution: W = 1 × 8.314 × 300 × 0.693 = 1728 J approximately.

Final Answer: Work done = 1.73 × 103 J.

Numerical 22 NEET. In process set 3, a gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work done.
Show Solution

Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.

Final Answer: Work done = 1000 J.

Numerical 23 NEET. In process set 3, a gas is heated at constant volume. If heat supplied is 600 J, find work done and change in internal energy.
Show Solution

Given: Isochoric process, Q = 600 J.

Formula: W = 0, ΔU = Q.

Solution: At constant volume, no boundary displacement occurs. Therefore W = 0 and ΔU = 600 J.

Final Answer: W = 0, ΔU = 600 J.

Numerical 24 NEET. In an adiabatic process, a gas does 450 J work. Find heat exchanged and change in internal energy.
Show Solution

Given: Adiabatic process, W = 450 J.

Formula: Q = 0, ΔU = Q - W.

Solution: ΔU = 0 - 450 = -450 J.

Final Answer: Q = 0, ΔU = -450 J.

Numerical 25 NEET. For an isothermal process of ideal gas, heat supplied is 900 J. Find ΔU and W.
Show Solution

Given: Ideal gas isothermal process, Q = 900 J.

Formula: ΔU = 0, Q = W.

Solution: Temperature is constant, so internal energy does not change. Therefore W = 900 J.

Final Answer: ΔU = 0, W = 900 J.

Numerical 26 NEET. In process set 3, a gas is compressed isobarically from 10 L to 6 L under 1 × 105 Pa. Find work by gas.
Show Solution

Given: P = 1 × 105 Pa, ΔV = -4 L = -4 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.

Final Answer: Work by gas = -400 J.

Numerical 27 NEET. Pressure of a gas doubles at constant volume. If initial temperature is 300 K, find final temperature.
Show Solution

Given: Isochoric process, P proportional to T.

Formula: P1/T1 = P2/T2.

Solution: Since pressure doubles, temperature doubles. T2 = 600 K.

Final Answer: Final temperature = 600 K.

Numerical 28 NEET. Volume of a gas doubles isothermally. If initial pressure is 4 atm, find final pressure.
Show Solution

Given: Isothermal process, V2 = 2V1, P1 = 4 atm.

Formula: P1V1 = P2V2.

Solution: P2 = P1V1/2V1 = 2 atm.

Final Answer: Final pressure = 2 atm.

Numerical 29 NEET. In process set 3, a gas expands adiabatically. It does 300 J work. What happens to temperature for an ideal gas?
Show Solution

Given: Adiabatic expansion, W = 300 J.

Formula: Q = 0, ΔU = -W.

Solution: ΔU = -300 J. For ideal gas, internal energy depends on temperature, so temperature decreases.

Final Answer: Temperature decreases.

Numerical 30 NEET. At constant pressure, a gas absorbs 1200 J heat and does 400 J work. Find change in internal energy.
Show Solution

Given: Q = 1200 J, W = 400 J.

Formula: ΔU = Q - W.

Solution: ΔU = 1200 - 400 = 800 J.

Final Answer: Internal energy increases by 800 J.

Numerical 31 JEE Main. In process set 4, an ideal gas expands isothermally at 300 K from 2 L to 4 L. If n = 1 mol, find work done. Take R = 8.314 J mol-1 K-1 and ln2 = 0.693.
Show Solution

Given: n = 1, T = 300 K, V2/V1 = 2.

Formula: W = nRT ln(V2/V1).

Solution: W = 1 × 8.314 × 300 × 0.693 = 1728 J approximately.

Final Answer: Work done = 1.73 × 103 J.

Numerical 32 JEE Main. In process set 4, a gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work done.
Show Solution

Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.

Final Answer: Work done = 1000 J.

Numerical 33 JEE Main. In process set 4, a gas is heated at constant volume. If heat supplied is 600 J, find work done and change in internal energy.
Show Solution

Given: Isochoric process, Q = 600 J.

Formula: W = 0, ΔU = Q.

Solution: At constant volume, no boundary displacement occurs. Therefore W = 0 and ΔU = 600 J.

Final Answer: W = 0, ΔU = 600 J.

Numerical 34 JEE Main. In an adiabatic process, a gas does 450 J work. Find heat exchanged and change in internal energy.
Show Solution

Given: Adiabatic process, W = 450 J.

Formula: Q = 0, ΔU = Q - W.

Solution: ΔU = 0 - 450 = -450 J.

Final Answer: Q = 0, ΔU = -450 J.

Numerical 35 JEE Main. For an isothermal process of ideal gas, heat supplied is 900 J. Find ΔU and W.
Show Solution

Given: Ideal gas isothermal process, Q = 900 J.

Formula: ΔU = 0, Q = W.

Solution: Temperature is constant, so internal energy does not change. Therefore W = 900 J.

Final Answer: ΔU = 0, W = 900 J.

Numerical 36 JEE Main. In process set 4, a gas is compressed isobarically from 10 L to 6 L under 1 × 105 Pa. Find work by gas.
Show Solution

Given: P = 1 × 105 Pa, ΔV = -4 L = -4 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.

Final Answer: Work by gas = -400 J.

Numerical 37 JEE Main. Pressure of a gas doubles at constant volume. If initial temperature is 300 K, find final temperature.
Show Solution

Given: Isochoric process, P proportional to T.

Formula: P1/T1 = P2/T2.

Solution: Since pressure doubles, temperature doubles. T2 = 600 K.

Final Answer: Final temperature = 600 K.

Numerical 38 JEE Main. Volume of a gas doubles isothermally. If initial pressure is 4 atm, find final pressure.
Show Solution

Given: Isothermal process, V2 = 2V1, P1 = 4 atm.

Formula: P1V1 = P2V2.

Solution: P2 = P1V1/2V1 = 2 atm.

Final Answer: Final pressure = 2 atm.

Numerical 39 JEE Main. In process set 4, a gas expands adiabatically. It does 300 J work. What happens to temperature for an ideal gas?
Show Solution

Given: Adiabatic expansion, W = 300 J.

Formula: Q = 0, ΔU = -W.

Solution: ΔU = -300 J. For ideal gas, internal energy depends on temperature, so temperature decreases.

Final Answer: Temperature decreases.

Numerical 40 JEE Main. At constant pressure, a gas absorbs 1200 J heat and does 400 J work. Find change in internal energy.
Show Solution

Given: Q = 1200 J, W = 400 J.

Formula: ΔU = Q - W.

Solution: ΔU = 1200 - 400 = 800 J.

Final Answer: Internal energy increases by 800 J.

Numerical 41 JEE Advanced. In process set 5, an ideal gas expands isothermally at 300 K from 2 L to 4 L. If n = 1 mol, find work done. Take R = 8.314 J mol-1 K-1 and ln2 = 0.693.
Show Solution

Given: n = 1, T = 300 K, V2/V1 = 2.

Formula: W = nRT ln(V2/V1).

Solution: W = 1 × 8.314 × 300 × 0.693 = 1728 J approximately.

Final Answer: Work done = 1.73 × 103 J.

Numerical 42 JEE Advanced. In process set 5, a gas expands at constant pressure 2 × 105 Pa from 3 L to 8 L. Find work done.
Show Solution

Given: P = 2 × 105 Pa, ΔV = 5 L = 5 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 2 × 105 × 5 × 10-3 = 1000 J.

Final Answer: Work done = 1000 J.

Numerical 43 JEE Advanced. In process set 5, a gas is heated at constant volume. If heat supplied is 600 J, find work done and change in internal energy.
Show Solution

Given: Isochoric process, Q = 600 J.

Formula: W = 0, ΔU = Q.

Solution: At constant volume, no boundary displacement occurs. Therefore W = 0 and ΔU = 600 J.

Final Answer: W = 0, ΔU = 600 J.

Numerical 44 JEE Advanced. In an adiabatic process, a gas does 450 J work. Find heat exchanged and change in internal energy.
Show Solution

Given: Adiabatic process, W = 450 J.

Formula: Q = 0, ΔU = Q - W.

Solution: ΔU = 0 - 450 = -450 J.

Final Answer: Q = 0, ΔU = -450 J.

Numerical 45 JEE Advanced. For an isothermal process of ideal gas, heat supplied is 900 J. Find ΔU and W.
Show Solution

Given: Ideal gas isothermal process, Q = 900 J.

Formula: ΔU = 0, Q = W.

Solution: Temperature is constant, so internal energy does not change. Therefore W = 900 J.

Final Answer: ΔU = 0, W = 900 J.

Numerical 46 JEE Advanced. In process set 5, a gas is compressed isobarically from 10 L to 6 L under 1 × 105 Pa. Find work by gas.
Show Solution

Given: P = 1 × 105 Pa, ΔV = -4 L = -4 × 10-3 m3.

Formula: W = PΔV.

Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.

Final Answer: Work by gas = -400 J.

Numerical 47 JEE Advanced. Pressure of a gas doubles at constant volume. If initial temperature is 300 K, find final temperature.
Show Solution

Given: Isochoric process, P proportional to T.

Formula: P1/T1 = P2/T2.

Solution: Since pressure doubles, temperature doubles. T2 = 600 K.

Final Answer: Final temperature = 600 K.

Numerical 48 JEE Advanced. Volume of a gas doubles isothermally. If initial pressure is 4 atm, find final pressure.
Show Solution

Given: Isothermal process, V2 = 2V1, P1 = 4 atm.

Formula: P1V1 = P2V2.

Solution: P2 = P1V1/2V1 = 2 atm.

Final Answer: Final pressure = 2 atm.

Numerical 49 JEE Advanced. In process set 5, a gas expands adiabatically. It does 300 J work. What happens to temperature for an ideal gas?
Show Solution

Given: Adiabatic expansion, W = 300 J.

Formula: Q = 0, ΔU = -W.

Solution: ΔU = -300 J. For ideal gas, internal energy depends on temperature, so temperature decreases.

Final Answer: Temperature decreases.

Numerical 50 JEE Advanced. At constant pressure, a gas absorbs 1200 J heat and does 400 J work. Find change in internal energy.
Show Solution

Given: Q = 1200 J, W = 400 J.

Formula: ΔU = Q - W.

Solution: ΔU = 1200 - 400 = 800 J.

Final Answer: Internal energy increases by 800 J.

50 PYQs and Exam Questions

Answers are hidden so this works like a real question bank.

Q1 CBSE. Define an isothermal process.
Show Answer

Explanation: An isothermal process occurs at constant temperature. For an ideal gas, internal energy remains constant.

Answer: T = constant.

Q2 CBSE. Define an adiabatic process.
Show Answer

Explanation: An adiabatic process has no heat exchange between system and surroundings.

Answer: Q = 0.

Q3 CBSE. What is work done in an isochoric process?
Show Answer

Explanation: Volume remains constant, so boundary displacement is zero.

Answer: W = 0.

Q4 CBSE. Write the work formula for an isobaric process.
Show Answer

Explanation: Pressure remains constant, so work is pressure times volume change.

Answer: W = P(V2 - V1).

Q5 NEET. Which process has ΔU = 0 for an ideal gas? (A) Isothermal (B) Adiabatic (C) Isochoric (D) Isobaric
Show Answer

Explanation: For ideal gas, internal energy depends only on temperature. Isothermal means temperature constant.

Answer: (A) Isothermal.

Q6 NEET. Which process has Q = 0? (A) Isothermal (B) Adiabatic (C) Isobaric (D) Isochoric
Show Answer

Explanation: No heat exchange is the defining condition of adiabatic process.

Answer: (B) Adiabatic.

Q7 NEET. The adiabatic curve is steeper than isothermal curve because:
Show Answer

Explanation: For the same volume increase, pressure falls faster in adiabatic expansion because the gas cools while doing work.

Answer: Temperature changes in adiabatic expansion.

Q8 JEE Main. On a P-V graph, a vertical line represents which process?
Show Answer

Explanation: Vertical line means volume is constant.

Answer: Isochoric process.

Q9 JEE Main. On a P-V graph, a horizontal line represents which process?
Show Answer

Explanation: Horizontal line means pressure is constant.

Answer: Isobaric process.

Q10 JEE Main. For isothermal expansion of an ideal gas, what is the relation between Q and W?
Show Answer

Explanation: Internal energy change is zero, so heat supplied equals work done.

Answer: Q = W.

Q11 JEE Advanced. Why does temperature fall during adiabatic expansion?
Show Answer

Explanation: No heat enters, so work done by gas comes from internal energy.

Answer: Internal energy decreases.

Q12 JEE Advanced. Compare work in reversible isothermal and adiabatic expansion between same initial volume and final volume.
Show Answer

Explanation: The isothermal curve lies above the adiabatic curve during expansion, so area under it is larger.

Answer: Isothermal work is greater.

Q13 IB. Explain molecular meaning of isothermal expansion.
Show Answer

Explanation: Average molecular kinetic energy remains constant while heat enters to replace energy leaving as work.

Answer: Molecular average kinetic energy remains constant.

Q14 ICSE. State Boyle's law connection with isothermal process.
Show Answer

Explanation: For a fixed amount of gas at constant temperature, pressure varies inversely with volume.

Answer: PV = constant.

Q15 IGCSE. Which process involves no volume change?
Show Answer

Explanation: The constant-volume process is called isochoric.

Answer: Isochoric.

Q16 A-Level. Write the integral expression for work done in any quasi-static process.
Show Answer

Explanation: Work is the area under P-V graph.

Answer: W = ∫PdV.

Q17 Assertion-Reason. Assertion: Work done in an isochoric process is zero. Reason: Volume remains constant.
Show Answer

Explanation: Both are true and the reason correctly explains the assertion.

Answer: Both true; reason is correct.

Q18 Assertion-Reason. Assertion: In isothermal ideal gas expansion, ΔU = 0. Reason: Internal energy of ideal gas depends only on temperature.
Show Answer

Explanation: Both are true and the reason correctly explains the assertion.

Answer: Both true; reason is correct.

Q19 True/False. Adiabatic process always means temperature remains constant.
Show Answer

Explanation: False. Adiabatic means no heat exchange; temperature can change due to work.

Answer: False.

Q20 True/False. Isochoric process has zero boundary work.
Show Answer

Explanation: True because volume does not change.

Answer: True.

Q21 Conceptual. Why must isothermal expansion be slow?
Show Answer

Explanation: Slow expansion allows heat exchange with surroundings to maintain constant temperature.

Answer: To maintain thermal equilibrium and constant T.

Q22 Conceptual. Why is rapid compression approximately adiabatic?
Show Answer

Explanation: There is very little time for heat exchange with surroundings.

Answer: Heat transfer is negligible.

Q23 Difficult Numerical. A gas expands isothermally from V to 2V at temperature T. What is proportionality of work?
Show Answer

Explanation: Using W = nRT ln(V2/V1), work is nRT ln2.

Answer: W = nRT ln2.

Q24 Case Study. Bicycle pump becomes warm during rapid compression. Which process is approximated?
Show Answer

Explanation: Rapid compression gives little time for heat loss, so it is nearly adiabatic.

Answer: Adiabatic compression.

Q25 Case Study. Diesel engine ignition occurs due to rapid compression. Explain.
Show Answer

Explanation: Rapid adiabatic compression raises air temperature enough to ignite fuel.

Answer: Temperature rises in adiabatic compression.

Q26 CBSE. Clearly define an isothermal process.
Show Answer

Explanation: An isothermal process occurs at constant temperature. For an ideal gas, internal energy remains constant.

Answer: T = constant.

Q27 CBSE. Clearly define an adiabatic process.
Show Answer

Explanation: An adiabatic process has no heat exchange between system and surroundings.

Answer: Q = 0.

Q28 CBSE. What is work done in an isochoric process?
Show Answer

Explanation: Volume remains constant, so boundary displacement is zero.

Answer: W = 0.

Q29 CBSE. Write the work formula for an isobaric process.
Show Answer

Explanation: Pressure remains constant, so work is pressure times volume change.

Answer: W = P(V2 - V1).

Q30 NEET. In a process question, which process has ΔU = 0 for an ideal gas? (A) Isothermal (B) Adiabatic (C) Isochoric (D) Isobaric
Show Answer

Explanation: For ideal gas, internal energy depends only on temperature. Isothermal means temperature constant.

Answer: (A) Isothermal.

Q31 NEET. In a process question, which process has Q = 0? (A) Isothermal (B) Adiabatic (C) Isobaric (D) Isochoric
Show Answer

Explanation: No heat exchange is the defining condition of adiabatic process.

Answer: (B) Adiabatic.

Q32 NEET. The adiabatic curve is steeper than isothermal curve because:
Show Answer

Explanation: For the same volume increase, pressure falls faster in adiabatic expansion because the gas cools while doing work.

Answer: Temperature changes in adiabatic expansion.

Q33 JEE Main. On a P-V graph, a vertical line represents which process?
Show Answer

Explanation: Vertical line means volume is constant.

Answer: Isochoric process.

Q34 JEE Main. On a P-V graph, a horizontal line represents which process?
Show Answer

Explanation: Horizontal line means pressure is constant.

Answer: Isobaric process.

Q35 JEE Main. For isothermal expansion of an ideal gas, what is the relation between Q and W?
Show Answer

Explanation: Internal energy change is zero, so heat supplied equals work done.

Answer: Q = W.

Q36 JEE Advanced. Explain why does temperature fall during adiabatic expansion?
Show Answer

Explanation: No heat enters, so work done by gas comes from internal energy.

Answer: Internal energy decreases.

Q37 JEE Advanced. Compare work in reversible isothermal and adiabatic expansion between same initial volume and final volume.
Show Answer

Explanation: The isothermal curve lies above the adiabatic curve during expansion, so area under it is larger.

Answer: Isothermal work is greater.

Q38 IB. Explain molecular meaning of isothermal expansion.
Show Answer

Explanation: Average molecular kinetic energy remains constant while heat enters to replace energy leaving as work.

Answer: Molecular average kinetic energy remains constant.

Q39 ICSE. State Boyle's law connection with isothermal process.
Show Answer

Explanation: For a fixed amount of gas at constant temperature, pressure varies inversely with volume.

Answer: PV = constant.

Q40 IGCSE. In a process question, which process involves no volume change?
Show Answer

Explanation: The constant-volume process is called isochoric.

Answer: Isochoric.

Q41 A-Level. Write the integral expression for work done in any quasi-static process.
Show Answer

Explanation: Work is the area under P-V graph.

Answer: W = ∫PdV.

Q42 Assertion-Reason. Assertion: Work done in an isochoric process is zero. Reason: Volume remains constant.
Show Answer

Explanation: Both are true and the reason correctly explains the assertion.

Answer: Both true; reason is correct.

Q43 Assertion-Reason. Assertion: In isothermal ideal gas expansion, ΔU = 0. Reason: Internal energy of ideal gas depends only on temperature.
Show Answer

Explanation: Both are true and the reason correctly explains the assertion.

Answer: Both true; reason is correct.

Q44 True/False. Adiabatic process always means temperature remains constant.
Show Answer

Explanation: False. Adiabatic means no heat exchange; temperature can change due to work.

Answer: False.

Q45 True/False. Isochoric process has zero boundary work.
Show Answer

Explanation: True because volume does not change.

Answer: True.

Q46 Conceptual. Explain why must isothermal expansion be slow?
Show Answer

Explanation: Slow expansion allows heat exchange with surroundings to maintain constant temperature.

Answer: To maintain thermal equilibrium and constant T.

Q47 Conceptual. Explain why is rapid compression approximately adiabatic?
Show Answer

Explanation: There is very little time for heat exchange with surroundings.

Answer: Heat transfer is negligible.

Q48 Difficult Numerical. A gas expands isothermally from V to 2V at temperature T. What is proportionality of work?
Show Answer

Explanation: Using W = nRT ln(V2/V1), work is nRT ln2.

Answer: W = nRT ln2.

Q49 Case Study. Bicycle pump becomes warm during rapid compression. In a process question, which process is approximated?
Show Answer

Explanation: Rapid compression gives little time for heat loss, so it is nearly adiabatic.

Answer: Adiabatic compression.

Q50 Case Study. Diesel engine ignition occurs due to rapid compression. Explain.
Show Answer

Explanation: Rapid adiabatic compression raises air temperature enough to ignite fuel.

Answer: Temperature rises in adiabatic compression.

Case Study Practice

Bicycle Pump

Rapid compression traps air and raises its temperature. This is an approximate adiabatic compression.

Diesel Engine

Air is compressed rapidly, temperature rises sharply, and injected fuel ignites without spark plug.

Refrigerator

The working fluid undergoes compression, cooling, expansion and evaporation in a cycle.

Pressure Cooker

Before steam release, volume is nearly fixed and pressure rises with temperature.

Hot Air Balloon

Heating air at nearly atmospheric pressure increases volume and lowers density.

Gas Cylinder

A rigid gas cylinder is close to constant-volume behavior when warmed.

Quick Revision Notes

Important Formula Sheet

  • Isothermal: T constant
  • Isothermal ideal gas: PV = constant
  • Isothermal work: W = nRT ln(V2/V1)
  • Adiabatic: Q = 0
  • Adiabatic ideal gas: PVγ = constant
  • Isochoric: V constant, W = 0
  • Isobaric: P constant, W = PΔV

Most Common Mistakes

  • Confusing isothermal with adiabatic.
  • Forgetting that adiabatic curve is steeper.
  • Using PΔV when pressure is not constant.
  • Writing W nonzero in isochoric process.
  • Thinking Q = 0 means temperature cannot change.
  • Forgetting litre to metre cube conversion.

Exam Tips

  • Identify the constant quantity first.
  • Draw the P-V graph before calculating work.
  • For ideal gas isothermal process, write ΔU = 0 immediately.
  • For adiabatic process, write Q = 0 immediately.
  • For isochoric process, write W = 0 immediately.
  • For isobaric process, use rectangular area PΔV.

If Thermodynamic Processes are not clear and you are looking for a Physics Tutor, contact Kumar Sir.

Phone: +91-9958461445   Email: kumarsirphysics@gmail.com

Call Kumar SirWhatsApp Kumar SirVisit WebsiteEmail Kumar Sir
Scroll to Top