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Chapter 12 — Thermodynamics

Class 11 · Physics

Overview

Chapter 12 — Thermodynamics Master Diagram

Introduction: This chapter develops the basic concepts of thermodynamics needed to describe heat, work and internal energy and to analyse simple thermodynamic processes. It starts with thermal equilibrium and the zeroth law, clarifies the difference between heat and temperature, and introduces heat capacity and calorimetry. The chapter then formulates the first law of thermodynamics (energy conservation for thermodynamic systems), applies it to ideal gases, and treats common processes (isothermal, adiabatic, isobaric, isochoric) with their PV-diagrams and formulae. Importance: Thermodynamics provides the foundation for understanding energy transfer and conversion in physical systems — essential for engines, refrigerators, heat engines, material behaviour, and many practical and theoretical problems in physics and engineering. Mastery of this chapter builds mathematical reasoning about state and path variables and prepares students for later study of the second law and statistical thermodynamics. Key themes: thermal equilibrium and temperature scales; distinction between heat, work and internal energy; heat capacity and calorimetry; first law of thermodynamics; state functions vs…

Learning Objectives

  • Define the zeroth law of thermodynamics and use it to explain thermal equilibrium and temperature measurement
  • Explain temperature scales and convert temperatures between Celsius, Kelvin and Fahrenheit scales
  • Define heat, work and internal energy and distinguish between heat and temperature in physical terms
  • State the first law of thermodynamics and apply it to calculate heat, work and internal energy changes in simple processes
  • Derive expressions for work done by an ideal gas in isothermal, isobaric, isochoric and adiabatic processes and solve related numerical problems
  • Calculate change in internal energy for monoatomic ideal gases using ΔU = nCvΔT and apply it to thermodynamic problems
  • Derive the relation Cp − Cv = R for ideal gases and compute molar and specific heat capacities from given data
  • Sketch and interpret PV diagrams for common thermodynamic processes and evaluate the work done from the area under the curve

Topics in this chapter

10 topics · tap a topic title to jump straight to it.

🔬1

Basic concepts and definitions

Fig 1 — Educational Diagram: Basic concepts and definitions

Fig 1 — Educational Diagram: Basic concepts and definitions

⚡ PHYSICAL LAW / FORMULA

Basic concepts and definitions

Key Point: Equation of state (ideal gas): PV = nRT

Introduction: Thermodynamics is the study of energy, its transformations, and the macroscopic properties of matter that describe those transformations. The subject begins with basic concepts that define what is being studied and how we measure changes.

System, Surroundings and Boundary: A system is the specific portion of the universe chosen for study (e.g. gas in a cylinder). Everything else is the surroundings. The boundary separates system and surroundings and may be fixed or movable, real or imaginary.

Types of Systems:

  • Isolated system: No exchange of mass or energy (heat/work) with surroundings (ideal thermos bottle).
  • Closed system: No mass exchange but energy (heat/work) can cross boundary (piston-cylinder with fixed mass).
  • Open system: Both mass and energy can cross boundary (flowing fluid in a pipe).

Macroscopic State and State Variables: The macroscopic state of a system is described by measurable state variables (pressure P, volume V, temperature T, internal energy U, number of moles n, etc.). A set of state variables that uniquely specifies the state is called the state. An equation of state relates state variables (e.g. ideal gas law PV = nRT).

Extensive and Intensive Variables:

  • Extensive: depend on system size (mass, volume, internal energy, entropy).
  • Intensive: independent of size (temperature, pressure, density).

Thermodynamic Equilibrium: A system is in thermodynamic equilibrium if it has mechanical, thermal and chemical equilibrium simultaneously (no net macroscopic change). In equilibrium, state variables are uniform and unchanging in time. The Zeroth Law of Thermodynamics formalizes thermal equilibrium: if A is in thermal equilibrium with B, and B with C, then A is with C; this allows the definition of temperature.

Process and Path: A process is any change of state. A process may be described by the path traced in state-space (e.g. P-V plane). If the process proceeds infinitely slowly through a succession of equilibrium states it is called quasi-static. Reversible processes are ideal quasi-static processes that can be reversed without leaving net changes in system and surroundings; most real processes are irreversible.

State Functions and Path Functions:

  • State functions (U, P, V, T, entropy S) depend only on the state, not on how the state was reached.
  • Path functions (heat Q and work W) depend on the process path; their values between two states depend on the route taken.

Heat and Work: Heat is energy transfer due to temperature difference. Work is energy transfer when a force acts through a distance (mechanical work, electrical work, etc.). For a simple compressible system, mechanical work done by the system in a quasi-static process is W = ∫ P dV (area under P–V curve).

Internal Energy: Internal energy U is the total microscopic energy (kinetic + potential) of the particles in the system. For an ideal gas, U depends only on temperature (U = nCvT for ideal gas, where Cv is molar specific heat at constant volume).

Heat Capacity and Specific Heat: Heat capacity C is the amount of heat required to change the temperature by 1 K. Specific heat c (per unit mass) and molar heat capacities (Cv, Cp) are commonly used. For small temperature change dQ = C dT (for systems held at constant parameter corresponding to that C).

Common Elementary Processes (defined by which variable is held fixed): isothermal (T constant), isochoric/isochoric (V constant), isobaric (P constant), adiabatic (no heat exchange Q = 0). Each has characteristic relations (e.g., isothermal ideal gas: PV = constant; reversible adiabatic ideal gas: PV^γ = constant, where γ = Cp/Cv).

Why these definitions matter: These basic definitions let us classify problems, write appropriate relations (equations of state, energy balances), draw process paths on P–V or T–S diagrams, and apply the laws of thermodynamics (e.g. first and second laws) correctly.

📌 Examples
  • Gas in a piston-cylinder: system = gas, boundary = piston walls; compressing the piston does work on the gas (W), heat may flow through the walls (Q).
  • Thermos flask: approximates an isolated system (no heat exchange with surroundings).
  • Boiling water in an open pot: open system if steam leaves (mass exchange); if covered, behaves more like a closed system.
  • Refrigerator: operates by cyclic processes (working fluid undergoes compression, condensation, expansion, evaporation) with heat and work exchanges between system and surroundings.
🧮 Formulas
  1. \[Equation of state (ideal gas): PV = nRT\]
  2. \[First law (energy balance): ΔU = Q − W (sign convention: Q into system positive\]
    \[W done by system positive)\]
  3. \[Work (quasi-static\]
    \[mechanical): W = ∫ P dV\]
  4. \[Isothermal reversible work for ideal gas: W = nRT ln(V_f / V_i)\]
  5. \[Adiabatic relation (ideal gas\]
    \[reversible): PV^γ = constant and TV^{γ−1} = constant\]
  6. \[Internal energy for ideal gas: U = n C_v T (ΔU = n C_v ΔT)\]
🌡️2

Zeroth law of thermodynamics and temperature

Fig 2.1 — Educational Diagram: Newton

Fig 2.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

⚡ PHYSICAL LAW / FORMULA

Zeroth law of thermodynamics and temperature

Key Point: Zeroth law (qualitative): If A ~ C and B ~ C then A ~ B (thermal equilibrium is transitive).

Zeroth law (statement): If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. In symbols: if A ~ C and B ~ C (thermal equilibrium), then A ~ B.

Thermal equilibrium means no net exchange of heat between systems when they are brought into thermal contact. Thermal equilibrium is a transitive relation — the content of the Zeroth law — and this transitivity allows the concept of temperature to be a well-defined physical property of a system.

Temperature as a physical property: The Zeroth law implies there exists a scalar quantity (temperature) that is the same for all systems in mutual thermal equilibrium. Temperature characterizes the hotness/coldness of a body and predicts the direction of spontaneous heat flow (from higher temperature to lower temperature).

Thermometers and temperature scales: A thermometer is a device that reaches thermal equilibrium with the system whose temperature is to be measured; the thermometer’s reading (after calibration) gives the temperature of the system. Empirical scales (Celsius, Fahrenheit) are based on fixed reference points; the thermodynamic (absolute) scale is the Kelvin scale, where 0 K is absolute zero. Relation: T(K) = t(°C) + 273.15.

Ideal gas thermometer and absolute temperature: For an ideal gas at fixed volume, pressure is proportional to absolute temperature (p ∝ T). Using a reference point (e.g., triple point of water), one defines the Kelvin scale so that the gas law gives an absolute measure of temperature independent of substance used in the thermometer.

Microscopic meaning: In kinetic theory for an ideal monoatomic gas, the average translational kinetic energy per particle is proportional to absolute temperature: <KE> = (3/2) k_B T, where k_B is Boltzmann’s constant. This provides a microscopic interpretation of temperature as a measure of average molecular kinetic energy.

Importance: The Zeroth law underpins all temperature measurement and thermometry. Without it, a single consistent scale of temperature that applies universally would not be possible.

📌 Examples
  • Thermometer in a glass of water: the mercury/alcohol column stops changing when the thermometer and water reach the same temperature (thermal equilibrium).
  • Room thermostat: the thermostat sensor reaches thermal equilibrium with room air; when temperature falls below set value the heater turns on.
  • Two cups of water separated by a metal plate: after some time both attain the same temperature — no further net heat flow — demonstrating thermal equilibrium.
  • Mixing hot and cold water: heat flows from hot to cold until both reach a common temperature (final equilibrium), predictable by conservation of energy and initial temperatures.
🧮 Formulas
  1. \[Zeroth law (qualitative): If A ~ C and B ~ C then A ~ B (thermal equilibrium is transitive).\]
  2. \[Ideal gas law: PV = nRT\]
  3. \[At constant V (ideal gas thermometer): p ∝ T → T = (p / p_ref) · T_ref\]
  4. \[Celsius–Kelvin conversion: T(K) = t(°C) + 273.15\]
  5. \[Mean kinetic energy (monoatomic ideal gas): ⟨KE⟩ = (3/2) k_B T\]
    \[where k_B = 1.380649×10⁻²³ J·K⁻¹\]
  6. \[Linear thermal expansion (thermometer column): ΔL = α L₀ ΔT (used in liquid-in-glass thermometers)\]
3

Heat and internal energy

Fig 3 — Educational Diagram: Heat and internal energy

Fig 3 — Educational Diagram: Heat and internal energy

⚡ PHYSICAL LAW / FORMULA

Heat and internal energy

Key Point: ΔU = Q − W (First law; W is work done by the system)

Definitions

Heat (Q) is energy in transfer between a system and its surroundings due to a temperature difference. Heat is a process quantity (path function) — it is not a property of the system.

Internal energy (U) is the total microscopic energy of a system: the sum of kinetic and potential energies of all its molecules (translational, rotational, vibrational kinetic energies and intermolecular potential energies). Internal energy is a state function: its value depends only on the state (e.g., temperature, pressure, volume, phase) of the system.

Microscopic picture

For a collection of N particles, internal energy U = sum of molecular kinetic energies + sum of potential energies due to interactions. For an ideal monatomic gas (no intermolecular potential), U is entirely kinetic and depends only on temperature: U = (3/2) nRT (or (3/2) NkT).

Key conceptual differences

  • Heat is energy in transit caused by ΔT; internal energy is energy contained in the system.
  • Heat (Q) and work (W) are process quantities; internal energy (U) is a state quantity.
  • Adding heat to a system can change its internal energy and/or do work on the surroundings.

First law of thermodynamics (energy conservation)

ΔU = Q − W

Here ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system on the surroundings (physics sign convention). This summarizes that heat supplied may increase internal energy and/or be converted to work.

Temperature vs internal energy

For ideal gases: internal energy depends only on temperature. For real substances (esp. during phase changes or when intermolecular forces are important) U depends on temperature and other variables (e.g., volume, phase).

Heat capacity and latent heat

  • Specific heat capacity (c): heat required per unit mass to raise temperature by 1 K. Q = m c ΔT
  • Molar heat capacity (Cv, Cp): heat per mole per K at constant volume or pressure. For ideal gas at constant volume: ΔU = n Cv ΔT.
  • Latent heat (L): heat required per unit mass for a phase change at constant temperature. Q = m L (during phase change temperature remains constant).

Important remarks and sign conventions

  • Heat supplied to the system: Q > 0. Heat removed: Q < 0.
  • Work done by the system: W > 0 (so ΔU = Q − W). If using the engineers’/chemists’ convention (W = work done on the system) sign changes accordingly — be consistent.
  • Because U is a state function, ΔU between two states is independent of path; Q and W individually depend on the path.

Examples of processes

  • Isothermal (ΔT = 0 for ideal gas): ΔU = 0, so Q = W (heat supplied equals work done by gas).
  • Adiabatic: Q = 0, so ΔU = −W (work done by gas reduces its internal energy and temperature).
  • At constant volume: W = 0, so ΔU = Q (all heat changes internal energy).

Units

SI unit of heat and internal energy: joule (J). Specific heat: J kg−1 K−1. Molar heat capacity: J mol−1 K−1. Latent heat: J kg−1.

📌 Examples
  • Boiling water on a stove: heat supplied raises water temperature until it reaches 100°C, then heat goes into latent heat (phase change) without temperature increase — internal energy increases through breaking of intermolecular bonds.
  • Gas in an insulated piston (adiabatic compression): no heat exchange (Q = 0), work done on gas increases its internal energy and temperature.
  • Cooling coffee in a room: heat flows from coffee to surroundings (Q < 0), internal energy of coffee decreases until thermal equilibrium.
  • Calorimetry: mixing hot water with cold water in a calorimeter and using m1 c (T1−Tf) = m2 c (Tf−T2) to find final temperature or specific heat of a substance.
  • Hand warmers (chemical packets): exothermic reaction releases heat that increases internal energy of surroundings, felt as warmth.
🧮 Formulas
  1. \[ΔU = Q − W (First law\]
    \[W is work done by the system)\]
  2. \[Q = m c ΔT (heat to change temperature\]
    \[m = mass\]
    \[c = specific heat)\]
  3. \[Q = n C_v ΔT (for ideal gas at constant volume\]
    \[n = moles\]
    \[C_v = molar heat capacity at constant volume)\]
  4. \[U_ideal_gas = (f/2) nRT (f = degrees of freedom\]
    \[for monatomic f = 3 → U = (3/2) nRT)\]
  5. \[Q_phase = m L (latent heat during phase change\]
    \[L = latent heat per unit mass)\]
  6. \[Cp − Cv = R (for ideal gases\]
    \[molar heat capacities at constant pressure and volume)\]
⚙️4

Work in thermodynamics

Fig 4 — Educational Diagram: Work in thermodynamics

Fig 4 — Educational Diagram: Work in thermodynamics

⚡ PHYSICAL LAW / FORMULA

Work in thermodynamics

Key Point: Infinitesimal work (quasi‑static): dW = P dV (work done by system).

What is work in thermodynamics?
In thermodynamics, work refers to the energy transferred when a macroscopic force acts through a distance due to a change in the system's macroscopic variables (most commonly volume). For a gas in a piston, the mechanical (PV) work done by the gas during a small quasi‑static change of volume dV against the pressure P is dW = P dV. The total work in a process from initial volume Vi to final volume Vf is W = ∫(Vi->Vf) P dV.

Sign convention
Common physics sign convention: W = ∫ P dV is the work done BY the system on the surroundings. If the system expands (dV > 0) W > 0. Work done ON the system is negative of this. Always state the sign convention before using formulas.

Reversible versus irreversible processes
For a reversible (quasi‑static) process the internal gas pressure equals the external pressure so P in the integral is the gas pressure. For an irreversible expansion against a constant external pressure Pext, the work is W = ∫ Pext dV = Pext (Vf - Vi). Reversible work is the maximum work obtainable for given end states.

Path dependence
Work is a path function (not a state function): the value of W depends on the process path connecting the same end states. This contrasts with internal energy and state functions that depend only on state variables.

Special cases for ideal gases

  • Isobaric (constant pressure): W = P (Vf - Vi).
  • Isothermal (T constant) reversible for ideal gas: W = nRT ln(Vf/Vi). Here internal energy change ΔU = 0, so heat Q = W.
  • Adiabatic (Q = 0), reversible: for ideal gas with γ = CP/CV and PV^γ = constant, W = (Pi Vi - Pf Vf)/(γ - 1). This follows from integrating P = constant × V^{-γ}.
  • Free (Joule) expansion into vacuum: Pext ≈ 0 so W ≈ 0 (no work done).

Graphical meaning
On a P–V diagram the work done by the gas during a process is the area under the process curve between Vi and Vf (with pressure on the vertical axis and volume on the horizontal axis). For a cyclic process the net work equals the area enclosed by the cycle (clockwise area positive for expansion-first cycles).

Units
SI unit of work: joule (J). Note 1 Pa·m^3 = 1 J.

Connections to first law
First law: ΔU = Q - W (with W the work done by the system). Thus W appears as energy leaving the system by mechanical work; sign conventions must be consistent.

📌 Examples
  • Piston-cylinder device: gas expansion pushes a frictionless piston upward doing work W = ∫P dV. Engines convert this work into useful mechanical work.
  • Isothermal compression in a refrigerator's compressor: for a reversible isothermal compression of n moles, W = nRT ln(Vf/Vi) (work done on the gas is negative of this if using W by system).
  • Free expansion (Joule experiment): gas expanding into vacuum in an insulated container performs essentially zero work (W ≈ 0).
  • Bicycle pump: compressing air (reducing volume) requires work done on the gas; when released the expanding air does work on surroundings.
  • Steam engine / internal combustion engine: the PV loop of one engine cycle shows the net work output as the area enclosed by the loop on the P–V diagram.
🧮 Formulas
  1. \[Infinitesimal work (quasi‑static): dW = P dV (work done by system).\]
  2. \[Total work: W = ∫_{Vi}^{Vf} P dV (path dependent).\]
  3. \[Reversible against external pressure: W_rev = ∫ P_gas dV.\]
  4. \[Irreversible constant external pressure: W = P_ext (Vf - Vi).\]
  5. \[Isobaric: W = P (Vf - Vi).\]
  6. \[Isothermal ideal gas (reversible): W = nRT ln(Vf/Vi) = RT ln(Vf/Vi) per mole.\]
💨5

Ideal gas and kinetic interpretation

Fig 5 — Educational Diagram: Ideal gas and kinetic interpretation

Fig 5 — Educational Diagram: Ideal gas and kinetic interpretation

⚡ PHYSICAL LAW / FORMULA

Ideal gas and kinetic interpretation

Key Point: Ideal gas law (molar): PV = n R T

What is an ideal gas?
An ideal gas is a model in which a large number of identical point-like particles move randomly and obey Newton's laws. Intermolecular forces are neglected except during perfectly elastic collisions. The ideal gas law relates macroscopic variables: pressure (P), volume (V) and temperature (T) by PV = nRT.

Basic assumptions of kinetic theory (ideal gas):

  • The gas consists of a very large number of identical molecules moving in random directions.
  • Molecules are point particles (their size is negligible compared with the distances between them).
  • There are no intermolecular attractive or repulsive forces except during collisions.
  • Collisions between molecules and with the walls are perfectly elastic.
  • The time spent in collisions is negligible compared with the time between collisions.
  • The average kinetic energy of molecules is proportional to the absolute temperature.

Derivation linking microscopic motion to pressure (sketch):
Consider N molecules of mass m in a cubical container of side L (volume V = L^3). For one molecule with x-component of velocity v_x, the momentum change on hitting a wall is 2m v_x and the time between successive hits on the same wall is 2L / v_x. So the average force contributed by that molecule on the wall is m v_x^2 / L. Summing over all molecules and dividing by wall area L^2 gives

p = (m N / V) average(v_x^2). By isotropy average(v_x^2) = (1/3) average(v^2), hence

p = (1/3) (N m / V) v_rms^2, where v_rms = sqrt(average(v^2)).

Using the ideal gas law in microscopic form pV = N k T (k = Boltzmann constant) we get

(1/2) m average(v^2) = (3/2) k T. This shows that the average translational kinetic energy per molecule is directly proportional to absolute temperature.

Consequences and physical meaning

  • Average translational kinetic energy per molecule: (1/2) m v_rms^2 = (3/2) k T.
  • Internal energy of a monatomic ideal gas (only translational KE) U = N (3/2 k T) = (3/2) n R T.
  • Temperature is a measure of the average kinetic energy of molecules, not of total energy of the gas (for ideal monatomic gas all internal energy is kinetic).

Limits of the model: Real gases deviate from ideal behavior at high pressure and low temperature because molecular volume and intermolecular forces become important. Ideal-gas kinetic interpretation is most accurate for dilute, high-temperature gases.

📌 Examples
  • Hot-air balloon: heating air increases molecular kinetic energy (T up) so pressure/volume relationships allow the balloon to lift.
  • Car tyre pressure: temperature change changes average molecular kinetic energy, altering pressure according to pV = nRT (visible seasonal tyre pressure variations).
  • Weather balloon: as it rises pressure decreases and the balloon expands; molecular kinetic theory explains temperature and pressure dependence.
  • Gas in a piston: compressing a gas increases collision frequency and average momentum transfer to walls, increasing pressure and temperature (work converted to molecular kinetic energy).
  • Helium in a balloon: low molar mass gives higher molecular speeds at a given temperature, explaining faster diffusion and effusion (Graham's law links to kinetic speeds).
🧮 Formulas
  1. \[Ideal gas law (molar): PV = n R T\]
  2. \[Ideal gas law (molecular): PV = N k T where N = number of molecules\]
    \[k = Boltzmann constant\]
  3. \[Pressure from molecular motion: p = (1/3) (N m / V) v_rms^2\]
  4. \[Root-mean-square speed: v_rms = sqrt(average(v^2)) = sqrt(3 k T / m) = sqrt(3 R T / M)\]
  5. \[Average translational kinetic energy per molecule: KE_avg = (1/2) m v_rms^2 = (3/2) k T\]
  6. \[Internal energy of a monatomic ideal gas: U = (3/2) n R T = (3/2) N k T\]
🌡️6

Thermodynamic processes — types and equations

Fig 6 — Educational Diagram: Thermodynamic processes — types and equations

Fig 6 — Educational Diagram: Thermodynamic processes — types and equations

⚡ PHYSICAL LAW / FORMULA

Thermodynamic processes — types and equations

Key Point: Ideal gas law: P V = n R T

Overview: A thermodynamic process is a change of state of a system (e.g. a gas) that involves heat (Q) transfer, work (W) done, and change in internal energy (ΔU). The First Law of Thermodynamics relates them: ΔU = Q - W (using the convention W = work done by the system).

Key concepts:

  • Internal energy for an ideal monoatomic gas: U = (3/2)nRT; in general ΔU = n C_v ΔT.
  • Specific heats: C_p - C_v = R and γ = C_p / C_v (ratio of specific heats).

Main types of processes (for an ideal gas):

  • Isothermal (T = constant): Temperature stays constant so ΔU = 0. Heat added equals work done by the gas: Q = W. Equation: PV = constant (hyperbola). Work: W = nRT ln(V_f / V_i).
  • Adiabatic (Q = 0): No heat exchange. ΔU = -W (work comes from internal energy). For a reversible adiabatic (ideal gas): PV^γ = constant, TV^(γ-1) = constant, or T^{γ}P^{1-γ}=const. Work between states 1 and 2: W = (P_1V_1 - P_2V_2)/(γ - 1).
  • Isobaric (P = constant): Pressure fixed. Work: W = P ΔV. Heat: Q = n C_p ΔT. Change in internal energy: ΔU = n C_v ΔT.
  • Isochoric / isovolumetric (V = constant): Volume fixed so W = 0. Heat goes into internal energy: Q = ΔU = n C_v ΔT.
  • Polytropic (PV^n = constant): A general family that includes isothermal (n=1) and adiabatic (n=γ). Work: W = (P_2V_2 - P_1V_1)/(1 - n) for n ≠ 1.
  • Cyclic process: System returns to initial state, so ΔU = 0. Net work done in a cycle equals net heat absorbed: W_net = Q_net. Heat engines operate in cyclic processes (Carnot, Otto, Diesel).
  • Free (Joule) expansion: Irreversible expansion into vacuum. For an ideal gas in an insulated container, Q = 0 and W = 0 so ΔU = 0 (temperature unchanged).

Reversible vs Irreversible / Quasi-static: A quasi-static (infinitesimally slow) process is nearly reversible and is represented by a smooth path on PV diagrams. Real processes often are irreversible.

Energy bookkeeping (First Law) — summary:

  • ΔU = Q - W
  • For ideal gas: ΔU = n C_v ΔT
  • Relate Q and W depending on the constraint (T const, P const, V const, Q=0, etc.).
📌 Examples
  • Isothermal: Slowly compressing gas in a piston while keeping it in thermal contact with a large heat reservoir (temperature constant).
  • Adiabatic: Rapid compression in a diesel engine cylinder (temperature rises because no time for heat exchange).
  • Isobaric: Heating water in an open pan — pressure remains atmospheric while volume (vapour) changes.
  • Isochoric: Heating gas sealed in a rigid, fixed-volume container — pressure increases while volume stays constant.
  • Cyclic: Carnot or Otto engine cycles — working gas undergoes a closed loop on the PV diagram and produces net work.
  • Free expansion: Gas expanding into an evacuated chamber after a partition is removed (Joule expansion).
🧮 Formulas
  1. \[Ideal gas law: P V = n R T\]
  2. \[First law: ΔU = Q - W (W = work done by the system)\]
  3. \[Internal energy (ideal gas): ΔU = n C_v ΔT\]
  4. \[Relation of heats: C_p - C_v = R, γ = C_p / C_v\]
  5. \[Isothermal (T const): ΔU = 0\]
    \[Q = W = n R T ln(V_f / V_i)\]
  6. \[Adiabatic (Q = 0\]
    \[reversible): P V^γ = constant\]
    \[T V^(γ-1) = constant\]
    \[W = (P_1V_1 - P_2V_2)/(γ - 1)\]
🌡️7

Second law of thermodynamics

Fig 7.1 — Educational Diagram: Newton

Fig 7.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

⚡ PHYSICAL LAW / FORMULA

Second law of thermodynamics

Key Point: First law for a cycle: W = Q1 - Q2 (work done = heat absorbed - heat rejected)

Statement (Kelvin–Planck): It is impossible to build a heat engine which, operating in a cycle, extracts heat from a single heat reservoir and converts all of it into work. Some heat must be rejected to a colder reservoir.

Statement (Clausius): Heat cannot spontaneously flow from a colder body to a hotter body without external work being done on the system.

Meaning and consequence: The first law (energy conservation) allows many processes that are not observed; the second law introduces directionality (time-asymmetry) in natural processes. It defines limits on the efficiency of heat engines and explains why some processes are irreversible.

Heat engines and refrigerators: A heat engine absorbs heat Q1 from a hot reservoir at temperature T1, does work W, and rejects heat Q2 to a cold reservoir at temperature T2. A refrigerator or heat pump uses work to transfer heat from cold to hot.

Reversible and irreversible processes: A reversible process is an idealization that can be reversed without net change to the system and surroundings. All real (natural) processes are irreversible and increase the total entropy of the universe.

Entropy (S): Entropy is a state function that measures energy dispersal or the unavailability of a system's energy to do work. For a reversible process, the infinitesimal change of entropy is defined as dS = dQ_rev / T. For a finite reversible change between states A and B, ΔS = ∫(A→B) dQ_rev / T.

Clausius inequality: For any cyclic process, ∮ dQ / T ≤ 0, with equality only for a reversible cycle. This leads to the conclusion that for an isolated system, entropy never decreases: ΔS_total ≥ 0.

Carnot theorem and limit on efficiency: Among all heat engines operating between the same two temperatures T1 (hot) and T2 (cold), a reversible (Carnot) engine has the maximum possible efficiency. No engine can be more efficient than a Carnot engine operating between the same reservoirs.

Intuitive examples of irreversibility: Heat flowing from hot to cold, free expansion of a gas, mixing of two different gases, friction converting organized kinetic energy to thermal energy — all increase entropy.

Applications: Establishes the theoretical maximum efficiency of engines, explains why perpetual motion machines of the second kind are impossible, and provides the basis for refrigerators, heat pumps, and understanding spontaneous processes.

Notes on temperature in Carnot relations: Temperatures T1 and T2 must be in an absolute scale (Kelvin) when used in efficiency or COP formulas.

📌 Examples
  • Internal combustion engine (car): Fuel combustion provides heat; piston-cylinder acts as a heat engine — cannot convert all heat into work, so exhaust gases carry away waste heat.
  • Refrigerator: Work is used to transfer heat from the cold interior to the warmer room; COP limited by temperatures of interior and room.
  • Heat pump for home heating: Moves heat from outside (cold) to inside (warm) using work; more efficient for heating than resistive heating because it moves existing heat.
  • Spontaneous heat flow: A hot metal rod placed in cold water cools down — heat flows from hot to cold and entropy of universe increases.
  • Mixing of gases or liquids: Two gases mixing spontaneously increases entropy and is irreversible without external work.
  • Friction: Mechanical energy lost to heat by friction increases entropy and cannot be completely recovered as useful work.
🧮 Formulas
  1. \[First law for a cycle: W = Q1 - Q2 (work done = heat absorbed - heat rejected)\]
  2. \[Thermal efficiency of an engine: η = W / Q1 = 1 - Q2 / Q1\]
  3. \[Carnot (maximum) efficiency: η_max = 1 - T2 / T1 (T in K)\]
  4. \[Coefficient of performance (COP) of refrigerator: COP_ref = Q2 / W = Q2 / (Q1 - Q2)\]
  5. \[COP of ideal (reversible) refrigerator: COP_ref,max = T2 / (T1 - T2)\]
  6. \[COP of heat pump: COP_hp = Q1 / W = Q1 / (Q1 - Q2)\]
    \[reversible COP_hp = T1 / (T1 - T2)\]
🔥8

Heat engines, refrigerators and heat pumps

Fig 8 — Educational Diagram: Heat engines, refrigerators and heat pumps

Fig 8 — Educational Diagram: Heat engines, refrigerators and heat pumps

⚡ PHYSICAL LAW / FORMULA

Heat engines, refrigerators and heat pumps

Key Point: First law for a cyclic engine: W_net = Q_h − Q_c

Overview

Heat engines, refrigerators and heat pumps are devices that transfer heat and perform work governed by the laws of thermodynamics. A heat engine converts part of heat from a hot reservoir into work while rejecting remaining heat to a cold reservoir. A refrigerator absorbs heat from a cold space and dumps it to a hotter space by doing work. A heat pump is like a refrigerator but is used to deliver heat to the hot space.

Heat engines

Working substance (gas or vapour) undergoes a cyclic process. From the first law for a full cycle, internal energy change ΔU = 0, so net work done by the system W = Q_h − Q_c, where Q_h is heat absorbed from the hot reservoir and Q_c is heat rejected to the cold reservoir.

Second law (Kelvin–Planck statement): It is impossible to construct an engine that, operating in a cycle, produces no other effect than the absorption of heat from a reservoir and the performance of an equivalent amount of work. Hence some heat must be rejected.

Maximum efficiency — Carnot engine

A reversible engine operating between two reservoirs at absolute temperatures T_h and T_c (Kelvin) is the most efficient. Its efficiency is

η_Carnot = 1 − T_c/T_h

For any engine operating between same T_h and T_c, η ≤ η_Carnot. For a general engine, efficiency η = W/Q_h = (Q_h − Q_c)/Q_h = 1 − Q_c/Q_h.

Refrigerators and heat pumps

Refrigerator: takes heat Q_c from cold reservoir and rejects Q_h to hot reservoir by input work W. From energy conservation W = Q_h − Q_c.

Heat pump: same device but goal is to deliver heat Q_h to the hot space; often used for heating homes.

Performance is measured by coefficient of performance (COP), not efficiency:

  • COP (refrigerator) = Q_c / W = Q_c / (Q_h − Q_c).
  • COP (heat pump) = Q_h / W = Q_h / (Q_h − Q_c) = 1 + COP_refrigerator.

For a reversible (Carnot) refrigerator or heat pump operating between T_h and T_c (Kelvin):

  • COP_refrigerator,Carnot = T_c / (T_h − T_c)
  • COP_heatpump,Carnot = T_h / (T_h − T_c)

Reversible vs irreversible

Reversible processes are idealizations giving maximum COP or efficiency. Real devices are irreversible due to friction, non-quasi-static processes, heat leaks, and finite-rate heat transfer; hence real performance is lower.

Key practical points

  • Engines: internal combustion engines (cars), steam turbines (power plants) — operate as heat engines but not reversible and follow specific cycles (Otto, Diesel, Rankine).
  • Refrigerators and ACs use vapour-compression cycles; heat pumps use similar cycles but aim to heat spaces.
  • Temperatures must be in absolute scale (Kelvin) when using Carnot relations.
📌 Examples
  • Car engine (internal combustion) — converts chemical energy (fuel) into work; rejects heat to atmosphere.
  • Steam turbine in thermal power plant — heat from boiler (high T) partially converted to electrical work; condenser rejects Q_c.
  • Household refrigerator — vapour-compression cycle that removes heat from inside (cold reservoir) and rejects to room (hot reservoir).
  • Air-conditioner — a refrigerator that cools an interior space; COP depends on indoor (T_c) and outdoor (T_h) temperatures.
  • Domestic heat pump — provides heating by extracting heat from outside air or ground and delivering to interior; more efficient than direct electrical heating for moderate temperature differences.
🧮 Formulas
  1. \[First law for a cyclic engine: W_net = Q_h − Q_c\]
  2. \[Efficiency (general): η = W / Q_h = 1 − Q_c / Q_h\]
  3. \[Carnot efficiency (reversible between T_h and T_c): η_Carnot = 1 − T_c / T_h (T in Kelvin)\]
  4. \[Work (net) in terms of Q_h and η: W = η Q_h\]
  5. \[COP of refrigerator: COP_R = Q_c / W = Q_c / (Q_h − Q_c)\]
  6. \[COP of heat pump: COP_HP = Q_h / W = Q_h / (Q_h − Q_c) = 1 + COP_R\]
🔬9

Entropy and its consequences

Fig 9 — Educational Diagram: Entropy and its consequences

Fig 9 — Educational Diagram: Entropy and its consequences

⚡ PHYSICAL LAW / FORMULA

Entropy and its consequences

Key Point: dS = δQ_rev / T

What is Entropy?

Entropy (S) is a thermodynamic state function that measures the degree of disorder or the number of accessible microscopic arrangements (microstates) of a system. It is a measure of energy dispersal at a given temperature. In macroscopic thermodynamics, entropy quantifies how much heat transfer is irreversible and gives a direction to spontaneous processes (the arrow of time).

Clausius definition (macroscopic)

For a reversible process the infinitesimal change of entropy is defined as dS = δQ_rev / T. For a finite reversible change between two states:

ΔS = ∫_(1)^(2) (δQ_rev / T).

Boltzmann definition (microscopic)

Entropy is related to the number of microstates Ω compatible with the macroscopic state by Boltzmann's formula: S = k_B ln Ω (where k_B is Boltzmann's constant). This links entropy to probability and disorder.

Key properties

  • State function: S depends only on the initial and final states, not on the path.
  • For an isolated system, entropy never decreases: S_total (isolated) either increases for irreversible processes or remains constant for reversible ones.
  • Clausius inequality for any cyclic process: ∮ (δQ / T) ≤ 0, equality for a reversible cycle.
  • Entropy of the universe (system + surroundings) increases for any spontaneous (irreversible) process: ΔS_univ = ΔS_sys + ΔS_surr ≥ 0.

Entropy changes in common processes

- Isothermal reversible expansion of an ideal gas: ΔS = Q_rev / T = nR ln(V_2 / V_1).
- General change for an ideal gas between states (T,V): ΔS = nC_v ln(T_2/T_1) + nR ln(V_2/V_1).
- Phase change at constant temperature (melting/boiling): ΔS = L / T (L = latent heat).
- Mixing of ideal gases (ideal, no energy change): entropy increases because number of accessible microstates increases; for ideal mixture: ΔS_mix = -nR Σ x_i ln x_i (x_i are mole fractions).

Consequences of entropy

  • Direction of spontaneous processes: Processes with ΔS_univ > 0 occur spontaneously. If ΔS_univ = 0 the process is reversible and at equilibrium.
  • Irreversibility and lost work: Entropy production corresponds to energy that cannot be converted into useful work. Reversible processes give maximum work extraction.
  • Limits on heat engines: Using entropy balance and the second law leads to maximum (Carnot) efficiency η_max = 1 - T_c/T_h. For a reversible engine, Q_h/T_h = Q_c/T_c and net entropy change is zero.
  • Arrow of time: Macroscopic irreversibility (entropy increase) gives a thermodynamic direction to time—natural processes tend toward equilibrium (maximum entropy consistent with constraints).
  • Local entropy decrease possible: A subsystem can decrease entropy (e.g., refrigerator, living organisms) only if surroundings gain at least as much entropy so that ΔS_univ ≥ 0.

Units

SI unit of entropy: joule per kelvin (J K^−1). Often molar entropy is quoted in J K^−1 mol^−1.

📌 Examples
  • Melting of ice: At 0°C, the entropy change for melting per kg (or per mole) is L/T where L is latent heat of fusion; entropy increases because liquid has more accessible microstates than solid.
  • Heat flow between two bodies: If a hot body at T_h gives heat Q to a cold body at T_c (T_h > T_c), the universe entropy change is Q(1/T_c - 1/T_h) > 0, so the process is spontaneous and irreversible.
  • Free (Joule) expansion of an ideal gas into vacuum: No work and no heat exchange with surroundings, but gas entropy increases because accessible volume increases (irreversible process).
  • Mixing of gases: Two ideal gases allowed to mix spontaneously increase total entropy even if temperature stays constant—mixing is irreversible and increases disorder.
  • Refrigerator: Local entropy of the refrigerated compartment decreases, but the compressor dumps more heat to surroundings so total entropy increases, satisfying the second law.
  • Carnot engine: An ideal reversible heat engine operating between two reservoirs has maximum possible efficiency and zero net entropy production for the cycle.
🧮 Formulas
  1. \[dS = δQ_rev / T\]
  2. \[ΔS = ∫ (δQ_rev / T) (for a reversible path)\]
  3. \[Clausius inequality: ∮ (δQ / T) ≤ 0 (equality for reversible cycle)\]
  4. \[Entropy change for ideal gas: ΔS = n C_v ln(T2/T1) + n R ln(V2/V1)\]
  5. \[Isothermal ideal gas: ΔS = n R ln(V2/V1)\]
  6. \[Phase change at constant T: ΔS = L / T (L = latent heat)\]
🔬10

Applications, problem solving and common derivations

Fig 10 — Educational Diagram: Applications, problem solving and common derivations

Fig 10 — Educational Diagram: Applications, problem solving and common derivations

⚡ PHYSICAL LAW / FORMULA

Applications, problem solving and common derivations

Key Point: First law: ΔU = Q - W

This topic explains how to apply the basic laws of thermodynamics (primarily the first law) to common processes, how to set up and solve typical problems, and gives the standard derivations used in Class 11. Key ideas: treat the system (usually an ideal gas) clearly, identify the process (isothermal, adiabatic, isobaric, isochoric, cyclic), use the ideal gas law where applicable, apply the first law (DeltaU = Q - W) with a consistent sign convention, and use specific heat relations (Cp - Cv = R).

Problem-solving strategy:

  • Step 1: Specify the system and whether the process is quasi-static (so that P at each step is defined).
  • Step 2: Identify the process type. Choose the correct relation for P, V, T or for work W = ∫ P dV.
  • Step 3: Use the ideal gas law P V = n R T to eliminate variables where needed.
  • Step 4: Apply the first law: ΔU = Q - W. For an ideal monoatomic or diatomic gas ΔU = n Cv ΔT.
  • Step 5: Compute W (area under P-V curve) and Q (from first law or Q = n Cp ΔT for processes at constant pressure).
  • Step 6: Check units and limiting cases (e.g., for isothermal ΔU = 0 so Q = W).

Common derivations and important relations (brief derivation steps):

  • Work in a quasi-static process: W = ∫_{V1}^{V2} P dV. For an ideal gas, substitute P from PV = nRT when needed.
  • Isothermal (T constant): P = nRT/V ⇒ W = nRT ∫_{V1}^{V2} dV/V = nRT ln(V2/V1). For ideal gas ΔU = 0 ⇒ Q = W.
  • Isobaric (P constant): W = P (V2 - V1). Heat Q = n Cp ΔT and ΔU = n Cv ΔT.
  • Isochoric (V constant): W = 0. So Q = ΔU = n Cv ΔT.
  • Adiabatic (Q = 0, reversible): use first law ΔU = -W. For ideal gas: PV^γ = constant (where γ = Cp/Cv). Other useful forms: TV^{γ-1} = constant and T^{γ} P^{1-γ} = constant. Work done: W = (P1 V1 - P2 V2)/(γ - 1).
  • Mayer's relation: Cp - Cv = R (per mole). For n moles: n(Cp - Cv) = nR.
  • Change in internal energy (ideal gas): ΔU = n Cv ΔT (depends only on temperature for ideal gas).
  • Carnot efficiency (reversible engine between Th and Tc): η = 1 - Tc/Th (temperatures in Kelvin). This is the maximum possible efficiency.
  • Refrigerator and heat pump COPs: COP_{refrigerator} = Qc/W = Tc/(Th - Tc); COP_{heat pump} = Qh/W = Th/(Th - Tc).

Common pitfalls:

  • Sign convention: Q positive into system, W positive when done by the system (so ΔU = Q - W).
  • Adiabatic does not always mean fast — it means no heat exchange. Reversible adiabatic = isentropic.
  • Work equals area under P-V curve; for non-quasi-static processes this area may not be well-defined with a single P(V).
  • Always use absolute (Kelvin) temperature for Carnot and other thermodynamic temperature relations.

When to use which formula:

  • Isothermal expansion/compression: use W = nRT ln(V2/V1) and ΔU = 0.
  • Adiabatic reversible: use PV^γ = constant and W = (P1V1 - P2V2)/(γ - 1).
  • Constant pressure: W = PΔV, Q = nCpΔT.
  • Constant volume: W = 0, Q = nCvΔT.

These relations and the problem strategy cover the majority of Class 11 thermodynamics problems and derivations.

📌 Examples
  • Isothermal expansion of 1 mol ideal gas from 10 L to 20 L at 300 K. Compute work: W = nRT ln(V2/V1) = (1)(8.314)(300) ln(2) ≈ 1729 J. Since ΔU = 0, Q = W = 1729 J.
  • Adiabatic compression of air (γ = 1.4): initial P1, V1 and final V2 known. Use PV^γ = constant to find P2, then W = (P1V1 - P2V2)/(γ - 1) and ΔU = -W.
  • Heating at constant pressure: 2 moles of ideal gas heated from 300 K to 400 K at constant pressure. Q = n Cp ΔT. For diatomic gas Cp ≈ (7/2)R, so Q = 2*(7/2)*8.314*(100) ≈ 5819 J.
  • Heat engine between 500 K and 300 K: maximum (Carnot) efficiency η = 1 - 300/500 = 0.4 or 40%.
🧮 Formulas
  1. \[First law: ΔU = Q - W\]
  2. \[Work (quasi-static): W = ∫_{V1}^{V2} P dV\]
  3. \[Ideal gas law: PV = nRT\]
  4. \[Isothermal work: W = nRT ln(V2/V1)\]
  5. \[Isobaric work: W = P(V2 - V1)\]
  6. \[Isochoric: W = 0\]
    \[Q = nCvΔT\]

Key Concepts

System
The part of the universe chosen for study in a thermodynamic analysis; it can be closed, open or isolated.
Surroundings
Everything outside the system that can exchange energy or matter with the system.
Boundary
The real or imaginary surface that separates the system from its surroundings.
Thermodynamic equilibrium
A state in which a system has no net macroscopic flows of matter or energy and temperature, pressure, and chemical potential are uniform and unchanging with time.
Zeroth law of thermodynamics
If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other; this defines temperature.
State function
A property whose value depends only on the current state of the system, not on how that state was reached (e.g., internal energy, pressure, temperature).
Path function
A quantity whose value depends on the specific process or path taken between two states (e.g., heat and work).
Process
Any change that takes a system from one equilibrium state to another (e.g., heating, compression).
Quasi-static process
An idealized process that proceeds infinitely slowly so the system remains approximately in equilibrium at all intermediate stages.
Reversible process
An ideal process that can be reversed by an infinitesimal change without leaving net changes in the system and surroundings; requires quasi-static and no dissipative effects.
Irreversible process
A real process that cannot be exactly reversed; it produces entropy due to dissipative effects like friction, turbulence or finite temperature differences.
Cyclic process
A process in which the system returns to its initial state at the end, so all state functions have zero net change over the cycle.
Heat
Energy transferred between system and surroundings due to a temperature difference; a path function (denoted Q).
Work
Energy transfer associated with a force acting through a distance or generalized coordinates; for quasi-static pressure–volume work, W = ∫P dV.
Internal energy
The total microscopic energy of a system (kinetic + potential) of its particles; a state function usually denoted U.
First law of thermodynamics
Statement of energy conservation for thermodynamic systems: the change in internal energy equals heat supplied minus work done by the system: ΔU = Q − W (or Q = ΔU + W with W the work done by the system).
Isothermal process
A process that occurs at constant temperature (ΔT = 0); for an ideal gas internal energy change is zero in an isothermal process.
Adiabatic process
A process in which no heat is exchanged between the system and surroundings (Q = 0); for reversible adiabatic of ideal gas, PV^γ = constant.
Isobaric process
A process that occurs at constant pressure (ΔP = 0).
Isochoric (isovolumetric) process
A process that occurs at constant volume (ΔV = 0); no PV work is done by the system.

Practice Questions

  1. Differentiate between a state function and a path function, giving one example of each. / अवस्था फलन और पथ फलन में अंतर कीजिए, प्रत्येक का एक उदाहरण दीजिए।
    Show answer

    A state function depends only on the current state of the system, not the route taken (e.g., internal energy U), whereas a path function depends on the specific process between two states (e.g., heat Q or work W). / अवस्था फलन केवल निकाय की वर्तमान अवस्था पर निर्भर करता है, अपनाए गए मार्ग पर नहीं (जैसे आंतरिक ऊर्जा U), जबकि पथ फलन दो अवस्थाओं के बीच के विशिष्ट प्रक्रम पर निर्भर करता है (जैसे ऊष्मा Q या कार्य W)।

  2. State the first law of thermodynamics and give its sign convention. / ऊष्मागतिकी का प्रथम नियम बताइए और इसकी चिह्न परिपाटी दीजिए।
    Show answer

    The first law states ΔU = Q − W, where ΔU is the change in internal energy, Q is heat added to the system (positive when into the system), and W is the work done by the system (positive when done by the system). / प्रथम नियम के अनुसार ΔU = Q − W, जहाँ ΔU आंतरिक ऊर्जा में परिवर्तन है, Q निकाय को दी गई ऊष्मा है (निकाय में जाने पर धनात्मक), और W निकाय द्वारा किया गया कार्य है (निकाय द्वारा किए जाने पर धनात्मक)।

  3. Derive the work done by an ideal gas during a reversible isothermal expansion from V_i to V_f. / आदर्श गैस द्वारा V_i से V_f तक उत्क्रमणीय समतापी प्रसार में किए गए कार्य को व्युत्पन्न कीजिए।
    Show answer

    W = ∫P dV with P = nRT/V, so W = nRT ∫(dV/V) from V_i to V_f = nRT ln(V_f/V_i). Since temperature is constant, ΔU = 0 and Q = W. / W = ∫P dV जहाँ P = nRT/V, अतः W = nRT ∫(dV/V), V_i से V_f तक = nRT ln(V_f/V_i)। चूँकि ताप स्थिर है, ΔU = 0 और Q = W।

  4. Calculate the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K (R = 8.314 J/mol·K). / 300 K पर 1 मोल आदर्श गैस के 10 L से 20 L तक समतापी प्रसार में किया गया कार्य परिकलित कीजिए (R = 8.314 J/mol·K)।
    Show answer

    W = nRT ln(V_f/V_i) = 1 × 8.314 × 300 × ln(2) = 2494.2 × 0.693 ≈ 1729 J. Since ΔU = 0, Q = W ≈ 1729 J. / W = nRT ln(V_f/V_i) = 1 × 8.314 × 300 × ln(2) = 2494.2 × 0.693 ≈ 1729 J। चूँकि ΔU = 0, Q = W ≈ 1729 J।

  5. Using the first law, show that in an adiabatic process the gas does work at the expense of its internal energy. / प्रथम नियम का उपयोग करते हुए दर्शाइए कि रुद्धोष्म प्रक्रम में गैस अपनी आंतरिक ऊर्जा की कीमत पर कार्य करती है।
    Show answer

    In an adiabatic process Q = 0, so the first law ΔU = Q − W gives ΔU = −W. Thus when the gas does positive work (expands), its internal energy decreases and temperature falls. / रुद्धोष्म प्रक्रम में Q = 0, अतः प्रथम नियम ΔU = Q − W से ΔU = −W। इस प्रकार जब गैस धनात्मक कार्य करती है (प्रसार), इसकी आंतरिक ऊर्जा घटती है और ताप गिरता है।

  6. Derive Mayer's relation Cp − Cv = R for an ideal gas and explain why Cp > Cv. / आदर्श गैस के लिए मेयर संबंध Cp − Cv = R व्युत्पन्न कीजिए और समझाइए कि Cp > Cv क्यों है।
    Show answer

    At constant volume Q = nCvΔT = ΔU; at constant pressure Q = nCpΔT = ΔU + PΔV = nCvΔT + nRΔT, giving Cp − Cv = R. Cp > Cv because at constant pressure extra heat is needed to do expansion work in addition to raising internal energy. / स्थिर आयतन पर Q = nCvΔT = ΔU; स्थिर दाब पर Q = nCpΔT = ΔU + PΔV = nCvΔT + nRΔT, जिससे Cp − Cv = R। Cp > Cv इसलिए क्योंकि स्थिर दाब पर आंतरिक ऊर्जा बढ़ाने के अतिरिक्त प्रसार कार्य करने हेतु अतिरिक्त ऊष्मा की आवश्यकता होती है।

  7. State the kinetic interpretation of temperature and write the relation for the average translational kinetic energy of a gas molecule. / ताप की गतिज व्याख्या बताइए और गैस अणु की औसत स्थानांतरीय गतिज ऊर्जा का संबंध लिखिए।
    Show answer

    Temperature is a measure of the average translational kinetic energy of gas molecules; for an ideal gas the average translational KE per molecule is (1/2)m·v_rms² = (3/2)k_B T, so it is directly proportional to absolute temperature. / ताप गैस अणुओं की औसत स्थानांतरीय गतिज ऊर्जा का माप है; आदर्श गैस के लिए प्रति अणु औसत स्थानांतरीय गतिज ऊर्जा (1/2)m·v_rms² = (3/2)k_B T है, अतः यह परम ताप के सीधे समानुपाती है।

  8. A Carnot engine operates between 500 K and 300 K. Find its maximum efficiency and explain why a real engine has lower efficiency. / एक कार्नो इंजन 500 K और 300 K के बीच कार्य करता है। इसकी अधिकतम दक्षता ज्ञात कीजिए और बताइए कि वास्तविक इंजन की दक्षता कम क्यों होती है।
    Show answer

    η_max = 1 − T_c/T_h = 1 − 300/500 = 0.4, i.e. 40%. A real engine is less efficient because of irreversibilities such as friction, finite-rate heat transfer and heat leaks, which produce entropy and reduce useful work. / η_max = 1 − T_c/T_h = 1 − 300/500 = 0.4, अर्थात 40%। वास्तविक इंजन कम दक्ष होता है क्योंकि घर्षण, परिमित दर ऊष्मा स्थानांतरण और ऊष्मा रिसाव जैसी अनुत्क्रमणीयताएँ एन्ट्रॉपी उत्पन्न करती हैं और उपयोगी कार्य घटाती हैं।

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