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WBB • Class XI • Physics • Ch 8
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Thermodynamics

Thermodynamics is the branch of physics that governs the macroscopic transformations between heat, mechanical work, and internal energy in physical and chemical systems. Unlike mechanics, which tracks individual particle trajectories, thermodynamics characterizes matter in bulk using macroscopic thermodynamic coordinates: pressure, volume, temperature, and internal energy. The study begins with the Zeroth Law of Thermodynamics, which establishes the formal definition of temperature as a universal indicator of thermal equilibrium. We then formulate the First Law of Thermodynamics, the foundational principle of mechanical energy conservation applied to thermal systems, distinguishing between internal energy as a state function and heat and work as path-dependent boundary phenomena. We analyze classical thermodynamic processes, including isothermal, adiabatic, isobaric, and isochoric changes of state, deriving analytical expressions for work done and examining their trajectories on indicator P-V diagrams. We establish Mayer's relation connecting molar heat capacities at constant pressure and volume, demonstrating how atomic degrees of freedom dictate the adiabatic index gamma. Addressing the unidirectional nature of natural spontaneous processes, we formulate the Second Law of Thermodynamics through the Kelvin-Planck and Clausius statements, investigating cyclic heat engines, refrigerators, and heat pumps. Finally, we explore Sadi Carnot's idealized reversible cycle, proving Carnot's Theorem and deriving the maximum theoretical thermal efficiency attainable by any heat engine operating between two fixed thermal reservoirs. This rigorous curriculum equips West Bengal Board (WBCHSE) Class 11 students with foundational analytical mastery indispensable for higher secondary examinations and competitive entrance tests like WBJEE, JEE Main, and NEET.

Why This Chapter Matters

The industrial revolution, modern aviation, space exploration, and global power generation are all engineered upon the laws of thermodynamics. Internal combustion engines in automobiles, gas and steam turbines in power plants, and rocket propulsion systems convert thermal energy released from chemical fuels into mechanical work according to the First and Second Laws. Domestic refrigerators, commercial deep freezers, and centralized air conditioning systems function as reversed heat engines, extracting heat from lower-temperature enclosures and rejecting it to the surrounding environment via external work input. Understanding thermodynamic efficiency limits enables materials scientists and mechanical engineers to optimize combined-cycle power generation, design low-emission hybrid powertrains, and develop next-generation cryogenic cooling systems for quantum computing and magnetic resonance imaging (MRI).

Chapter Roadmap & Progression

1 1. Thermal Equilibrium, State Varia...
2 2. First Law of Thermodynamics, Int...
3 3. Thermodynamic Processes: Isother...
4 4. Molar Heat Capacities, Degrees o...
5 5. Second Law of Thermodynamics, He...
6 6. Carnot Engine, Carnot Cycle & Ca...

Complete Concept Guide (100% Curriculum Coverage)

1. Thermal Equilibrium, State Variables & Zeroth Law of Thermodynamics

Thermodynamic Systems and Boundaries

A thermodynamic system is a specified macroscopic quantity of matter bounded by an arbitrary closed surface separating it from the external surroundings. Systems are classified by the permeability of their boundaries:

  • Open System: Exchanges both matter and energy (heat/work) with surroundings (e.g., boiling water in an open beaker).
  • Closed System: Exchanges energy but NOT matter with surroundings (e.g., gas enclosed in a cylinder fitted with a sealed piston).
  • Isolated System: Exchanges neither matter nor energy with surroundings (e.g., hot liquid in an ideally insulated vacuum flask).

Diathermic Wall: An impermeable boundary that readily permits heat flow (thermal conductor like copper). Adiabatic Wall: An impermeable boundary that completely prevents heat exchange (ideal thermal insulator like asbestos or vacuum jacket).

Thermodynamic State Variables & Equation of State

The thermodynamic state of an equilibrium system is completely described by macroscopic parameters called state variables:

  • Intensive Variables: Independent of the size or mass of the system (e.g., Pressure $P$, Temperature $T$, Density $\rho$).
  • Extensive Variables: Directly proportional to the size or mass of the system (e.g., Volume $V$, Internal Energy $U$, Entropy $S$, Mass $M$).

The functional relationship connecting state variables in equilibrium is the Equation of State. For $n$ moles of an ideal gas:

$$P V = n R T$$

where $R = 8.314\,\text{J}/(\text{mol}\cdot\text{K})$ is the Universal Gas Constant.

Thermal Equilibrium & The Zeroth Law of Thermodynamics

Two thermodynamic systems are said to be in thermal equilibrium if, when placed in thermal contact across a diathermic wall, no net exchange of heat occurs between them and their macroscopic thermodynamic coordinates remain constant in time.

Statement of the Zeroth Law (R.H. Fowler, 1931):
If two thermodynamic systems $A$ and $B$ are each separately in thermal equilibrium with a third system $C$, then systems $A$ and $B$ are in thermal equilibrium with each other.

Physical Significance: The Zeroth Law implies the existence of a scalar state property called Temperature ($T$). Temperature is the physical property whose equality between two systems determines whether they are in mutual thermal equilibrium. This law provides the theoretical foundation for thermometry: system $C$ serves as a thermometer to calibrate systems $A$ and $B$.

2. First Law of Thermodynamics, Internal Energy & Work Done

Internal Energy ($U$) as a State Function

The Internal Energy ($U$) of a thermodynamic system is the total energy possessed by its constituent molecules, comprising molecular kinetic energies (translational, rotational, vibrational) and intermolecular potential energies:

$$U = E_K + E_P$$

For an ideal gas, intermolecular attractive forces are negligible ($E_P = 0$), so internal energy is purely kinetic and depends exclusively on temperature (Joule's Law):

$$U = U(T) \implies dU = n C_v dT$$

State Function: $U$ depends solely on the current equilibrium state ($P, V, T$) of the system and is independent of the past path taken to reach that state. In any cyclic process returning to the initial state, the net change in internal energy is strictly zero: $\oint dU = 0$.

Mechanical Work Done ($W$) and P-V Indicator Diagrams

When a gas enclosed in a cylinder of cross-sectional area $A$ pushes a frictionless piston outward by distance $dx$ against external pressure $P$, the incremental mechanical work done by the system is:

$$dW = F\,dx = (P A)\,dx = P\,dV$$

Total work done during a quasi-static expansion from initial volume $V_1$ to final volume $V_2$:

$$W = \int_{V_1}^{V_2} P\,dV$$

Sign Convention in Physics:

  • $W > 0$ (Positive): Work done by the gas on surroundings during expansion ($dV > 0$).
  • $W < 0$ (Negative): Work done on the gas by external agents during compression ($dV < 0$).
  • $Q > 0$ (Positive): Heat absorbed or added to the system.
  • $Q < 0$ (Negative): Heat released or rejected by the system.
P-V Indicator Diagram:
The work done during any quasi-static process is geometrically equal to the area under the process curve projected onto the volume axis of a $P\text{--}V$ diagram.
  • For a cyclic process: Net work done is the enclosed area of the cycle.
  • Clockwise Cycle: Expansion occurs at higher pressure than compression $\implies W_{\text{net}} > 0$ (Heat engine delivering net positive work).
  • Counter-Clockwise Cycle: Compression occurs at higher pressure $\implies W_{\text{net}} < 0$ (Refrigerator/Heat pump consuming external work).
Statement of the First Law of Thermodynamics

The First Law is the formulation of the Law of Conservation of Energy applied to thermodynamic systems:

$$\Delta Q = \Delta U + \Delta W$$

In differential form for an infinitesimal quasi-static process:

$$dQ = dU + dW = dU + P\,dV$$

Key Insight: Heat ($dQ$) and work ($dW$) are path-dependent functions (inexact differentials), but their difference $dU = dQ - dW$ is a path-independent exact differential representing the change in a fundamental state function ($U$).

3. Thermodynamic Processes: Isothermal, Adiabatic, Isobaric & Isochoric

Classification of Quasi-Static Processes

A quasi-static process is an idealized infinitely slow succession of equilibrium states, allowing state variables to remain well-defined throughout.

ProcessGoverning ConditionFirst Law FormulationWork Done ($W$)
Isothermal$T = \text{const}, \Delta T = 0$
$P V = \text{const}$
$\Delta U = 0 \implies Q = W$$$W = nRT\ln\left(\frac{V_2}{V_1}\right) = nRT\ln\left(\frac{P_1}{P_2}\right)$$
Adiabatic$dQ = 0$
$P V^\gamma = \text{const}$
$\Delta U = -W$$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$
Isobaric$P = \text{const}, \Delta P = 0$
$V / T = \text{const}$
$Q = \Delta U + P\Delta V = n C_p \Delta T$$$W = P(V_2 - V_1) = n R(T_2 - T_1)$$
Isochoric$V = \text{const}, \Delta V = 0$
$P / T = \text{const}$
$W = 0 \implies Q = \Delta U = n C_v \Delta T$$$W = 0$$
Comparison of Slopes on P-V Indicator Diagram

Differentiating the process equations with respect to volume $V$:

  • Isothermal Curve: $P V = \text{const} \implies P\,dV + V\,dP = 0 \implies \left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$
  • Adiabatic Curve: $P V^\gamma = \text{const} \implies V^\gamma\,dP + \gamma P V^{\gamma-1}\,dV = 0 \implies \left(\frac{dP}{dV}\right)_{\text{adi}} = -\gamma \frac{P}{V}$
Core Theorem:
$$\left(\frac{dP}{dV}\right)_{\text{adiabatic}} = \gamma \times \left(\frac{dP}{dV}\right)_{\text{isothermal}}$$Since $\gamma = C_p / C_v > 1$, the slope of an adiabatic curve is strictly $\gamma$ times steeper than that of an isothermal curve passing through the same $(P,V)$ point. During expansion, adiabatic curves drop more steeply because internal energy is depleted to perform external work.
Adiabatic State Relations

Using the ideal gas equation $P = nRT/V$, Poisson's adiabatic relations can be expressed in three equivalent formulations:

$$P V^\gamma = \text{constant}$$$$T V^{\gamma-1} = \text{constant}$$$$P^{1-\gamma} T^\gamma = \text{constant} \iff \frac{T^\gamma}{P^{\gamma-1}} = \text{constant}$$

4. Molar Heat Capacities, Degrees of Freedom & Mayer's Relation

Molar Specific Heat Capacities ($C_v$ and $C_p$)

Molar Heat Capacity ($C$): The amount of heat required to raise the temperature of 1 mole of a substance by 1 Kelvin ($1\,^\circ\text{C}$):

$$C = \frac{1}{n}\frac{dQ}{dT}$$
  • Molar Heat Capacity at Constant Volume ($C_v$): Since $dV = 0 \implies dW = 0$, all added heat increases internal energy: $C_v = \frac{1}{n}\left(\frac{dU}{dT}\right)_V$.
  • Molar Heat Capacity at Constant Pressure ($C_p$): The added heat must not only raise the temperature (increase $U$) but also perform external mechanical work ($P\,dV$) as the gas expands against constant pressure: $C_p = \frac{1}{n}\left(\frac{dQ}{dT}\right)_P$. Hence $C_p > C_v$.
Rigorous Derivation of Mayer's Relation ($C_p - C_v = R$)

Consider 1 mole ($n = 1$) of an ideal gas. By the First Law of Thermodynamics:

$$dQ = dU + P\,dV$$

At constant volume: $dV = 0 \implies dQ_v = dU = C_v\,dT$.

Since the internal energy of an ideal gas depends solely on temperature, the expression $dU = C_v\,dT$ applies to any process where temperature changes by $dT$.

At constant pressure: $dQ_p = C_p\,dT$. Substituting into the First Law:

$$C_p\,dT = C_v\,dT + P\,dV$$

From the ideal gas equation for 1 mole, $P V = R T$. Differentiating at constant pressure $P$:

$$P\,dV = R\,dT$$

Substituting $P\,dV = R\,dT$ into the First Law expression:

$$C_p\,dT = C_v\,dT + R\,dT$$

Dividing throughout by $dT$ yields Mayer's Relation:

$$C_p - C_v = R$$
Degrees of Freedom ($f$) & Equipartition Theorem

Law of Equipartition of Energy: For any dynamical system in thermal equilibrium at absolute temperature $T$, the average kinetic energy associated with each independent degree of freedom is $\frac{1}{2} k_B T$, where $k_B$ is Boltzmann's constant.

For 1 mole of gas ($N_A k_B = R$) with $f$ degrees of freedom:

$$U = \frac{f}{2} R T \implies C_v = \frac{dU}{dT} = \frac{f}{2} R, \quad C_p = C_v + R = \left(\frac{f}{2} + 1\right) R, \quad \gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$$
Gas AtomicityDegrees of Freedom ($f$)$C_v$$C_p$$\gamma = C_p/C_v$
Monatomic (He, Ne, Ar)3 (translational)$\frac{3}{2}R$$\frac{5}{2}R$$\frac{5}{3} \approx 1.67$
Diatomic (rigid: $\text{N}_2, \text{O}_2$)5 (3 trans + 2 rot)$\frac{5}{2}R$$\frac{7}{2}R$$\frac{7}{5} = 1.40$
Polyatomic Non-linear ($\text{H}_2\text{O}, \text{NH}_3$)6 (3 trans + 3 rot)$3R$$4R$$\frac{4}{3} \approx 1.33$

5. Second Law of Thermodynamics, Heat Engines & Refrigerators

Limitations of the First Law & Need for Second Law

The First Law enforces total energy conservation but imposes no directionality on physical processes. For instance, when a bullet stops inside a wooden target, mechanical kinetic energy converts entirely into heat. Yet the First Law cannot explain why heat from the block never spontaneously reassembles into kinetic energy to fire the bullet backward. The Second Law of Thermodynamics specifies the permissible direction of spontaneous thermodynamic processes.

Classical Statements of the Second Law
1. Kelvin-Planck Statement (Governs Heat Engines):
It is impossible to construct an engine operating in a complete cycle whose sole effect is the extraction of heat from a single thermal reservoir and its complete conversion into an equivalent amount of mechanical work ($W = Q_1, \eta = 100\%$ is impossible).
2. Clausius Statement (Governs Refrigerators):
It is impossible to construct a cyclic machine whose sole effect is the spontaneous transfer of heat from a body at a lower temperature to a body at a higher temperature without external work input.

Equivalence of the Two Statements: A violation of the Kelvin-Planck statement directly leads to a violation of the Clausius statement, and vice-versa. They are two manifestations of the same underlying physical law.

Thermodynamics of Heat Engines

A Heat Engine is a cyclic thermodynamic device that converts thermal energy into mechanical work:

  1. Absorbs heat $Q_1$ from a high-temperature heat source at absolute temperature $T_1$.
  2. Performs external mechanical work $W$.
  3. Rejects remaining waste heat $Q_2$ to a low-temperature heat sink at absolute temperature $T_2$.

By energy conservation over a complete cycle ($\Delta U = 0$): $W = Q_1 - Q_2$. The Thermal Efficiency ($\eta$) is:

$$\eta = \frac{\text{Net Work Output}}{\text{Heat Input}} = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$
Thermodynamics of Refrigerators and Heat Pumps

A Refrigerator operates as a reversed heat engine. Working in a cycle, it absorbs heat $Q_2$ from a cold refrigerated interior at $T_2$, consumes external mechanical/electrical work $W$, and discharges heat $Q_1 = Q_2 + W$ to the warmer surroundings at $T_1$.

The performance of a refrigerator is characterized by its Coefficient of Performance (COP, $\beta$):

$$\beta = \frac{\text{Heat extracted from cold reservoir}}{\text{Work input}} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}$$

Relationship between Heat Engine efficiency $\eta$ and Refrigerator COP $\beta$:

$$\beta = \frac{Q_2 / Q_1}{(Q_1 - Q_2) / Q_1} = \frac{1 - \eta}{\eta}$$

6. Carnot Engine, Carnot Cycle & Carnot's Theorem

The Ideal Reversible Carnot Cycle

In 1824, Nicolas Léonard Sadi Carnot conceived an idealized, perfectly reversible cyclic heat engine operating between a source at $T_1$ and a sink at $T_2$. The working substance (an ideal gas) undergoes four successive reversible operations on a $P\text{--}V$ diagram:

  1. Step 1 $\to$ 2: Reversible Isothermal Expansion (at Source Temperature $T_1$):
    The cylinder is placed on the source at $T_1$. The gas expands slowly from $(P_1, V_1)$ to $(P_2, V_2)$. Since $\Delta T = 0 \implies \Delta U = 0$, heat absorbed from source equals work done:$$Q_1 = W_{1\to 2} = n R T_1 \ln\left(\frac{V_2}{V_1}\right)$$
  2. Step 2 $\to$ 3: Reversible Adiabatic Expansion (from $T_1$ to Sink Temperature $T_2$):
    The cylinder is placed on an insulating stand ($dQ = 0$). Gas expands from $(P_2, V_2)$ to $(P_3, V_3)$, cooling from $T_1$ to $T_2$:$$W_{2\to 3} = \frac{n R (T_1 - T_2)}{\gamma - 1}, \quad \text{with } T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$$
  3. Step 3 $\to$ 4: Reversible Isothermal Compression (at Sink Temperature $T_2$):
    The cylinder is placed on the sink at $T_2$. Gas is slowly compressed from $(P_3, V_3)$ to $(P_4, V_4)$, rejecting heat $Q_2$:$$Q_2 = -W_{3\to 4} = n R T_2 \ln\left(\frac{V_3}{V_4}\right)$$
  4. Step 4 $\to$ 1: Reversible Adiabatic Compression (from $T_2$ back to $T_1$):
    The cylinder is placed back on the insulating stand ($dQ = 0$). Gas is compressed from $(P_4, V_4)$ back to original state $(P_1, V_1)$, heating up from $T_2$ to $T_1$:$$W_{4\to 1} = -\frac{n R (T_1 - T_2)}{\gamma - 1}, \quad \text{with } T_2 V_4^{\gamma-1} = T_1 V_1^{\gamma-1}$$
Derivation of Carnot Efficiency

Net work done per cycle: $W = W_{1\to 2} + W_{2\to 3} + W_{3\to 4} + W_{4\to 1}$. Since $W_{2\to 3} = -W_{4\to 1}$, the adiabatic works cancel identically:

$$W = Q_1 - Q_2$$

From the adiabatic expansion and compression equations:

$$T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1} \quad \text{and} \quad T_1 V_1^{\gamma-1} = T_2 V_4^{\gamma-1}$$

Dividing the two relations yields the volume ratio equivalence:

$$\left(\frac{V_2}{V_1}\right)^{\gamma-1} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} \implies \frac{V_2}{V_1} = \frac{V_3}{V_4}$$

Therefore, the ratio of rejected heat to absorbed heat simplifies to:

$$\frac{Q_2}{Q_1} = \frac{n R T_2 \ln(V_3/V_4)}{n R T_1 \ln(V_2/V_1)} = \frac{T_2}{T_1}$$
Carnot Efficiency Formula:
$$\eta_{\text{Carnot}} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$$For an ideal Carnot Refrigerator, the maximum COP is:$$\beta_{\text{Carnot}} = \frac{T_2}{T_1 - T_2}$$
Carnot's Theorem & Absolute Thermodynamic Temperature Scale

Carnot's Theorem:

  1. No real engine operating between two given constant thermal reservoirs can have a higher efficiency than a reversible Carnot engine operating between the same two reservoirs ($\eta_{\text{real}} \le \eta_{\text{Carnot}}$).
  2. All reversible engines operating between the same two temperatures have the identical efficiency, completely independent of the nature of the working substance.

Since $\eta = 1 - T_2/T_1$ depends solely on temperatures, Lord Kelvin utilized Carnot's relation to establish the Absolute Thermodynamic Temperature Scale, which is independent of any specific thermometric substance.

Key Formulas, Laws & Physical Constants

First Law of Thermodynamics & Mechanical Work
dQ = dU + dW = dU + P·dV, W = ∫ P dV, ΔU = n·Cv·ΔT
Isothermal & Adiabatic Process Formulations
$$W_iso = nRT·ln(V₂/V₁), W_adi = (P₁V₁ - P₂V₂) / (γ - 1) = nR(T₁ - T₂) / (γ - 1)$$
Mayer's Relation & Specific Heat Indices
Cp - Cv = R, Cv = f/2 · R, Cp = (f/2 + 1)R, γ = 1 + 2/f
Heat Engine Thermal Efficiency
η = W / Q₁ = (Q₁ - Q₂) / Q₁ = 1 - Q₂ / Q₁
Refrigerator & Heat Pump Coefficient of Performance (COP)
β = Q₂ / W = Q₂ / (Q₁ - Q₂), β = (1 - η) / η
Carnot Cycle Efficiency & Maximum COP
$$η_Carnot = 1 - T₂/T₁, β_Carnot = T₂ / (T₁ - T₂)$$

Conceptual Solved Examples & Case Studies

Example 1
One mole of an ideal gas expands isothermally at a constant temperature of 300 K until its volume is doubled. Calculate the work done by the gas during this expansion. (Take R = 8.314 J/(mol·K) and ln(2) = 0.693).
Step-by-Step Solution:
Step 1: Identify given parameters in SI units:
Number of moles $n = 1\,\text{mol}$
Temperature $T = 300\,\text{K}$ (constant)
Volume ratio $\frac{V_2}{V_1} = 2$
Gas constant $R = 8.314\,\text{J}/(\text{mol}\cdot\text{K})$

Step 2: State the formula for work done in an isothermal process:
$$W = n R T \ln\left(\frac{V_2}{V_1}\right)$$
Step 3: Substitute values and calculate:
$$W = (1) \times 8.314 \times 300 \times \ln(2)$$
$$W = 2494.2 \times 0.693 = 1728.48\,\text{J}$$
Final Answer: The work done by the expanding gas is approximately $1728.5\,\text{J}$ (or $1.73\,\text{kJ}$).
Example 2
A diatomic gas with γ = 1.4 at an initial temperature of 27°C and pressure 1 atm is compressed adiabatically to 1/8th of its original volume. Calculate: (a) the final temperature of the gas, and (b) the final pressure.
Step-by-Step Solution:
Step 1: Write down initial parameters:
Initial temperature $T_1 = 27^\circ\text{C} = 27 + 273 = 300\,\text{K}$
Initial pressure $P_1 = 1\,\text{atm}$
Volume ratio $\frac{V_1}{V_2} = 8 \implies \frac{V_2}{V_1} = \frac{1}{8}$
Adiabatic index $\gamma = 1.4 = \frac{7}{5}$

Step 2: Find final temperature using $T V^{\gamma-1} = \text{constant}$:
$$T_2 V_2^{\gamma-1} = T_1 V_1^{\gamma-1} \implies T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}$$
$$T_2 = 300 \times (8)^{1.4 - 1} = 300 \times (8)^{0.4} = 300 \times (2^3)^{2/5} = 300 \times 2^{1.2} = 300 \times 2.2974 = 689.2\,\text{K}$$
In Celsius: $T_2 = 689.2 - 273 = 416.2^\circ\text{C}$.

Step 3: Find final pressure using $P V^\gamma = \text{constant}$:
$$P_2 V_2^\gamma = P_1 V_1^\gamma \implies P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$$
$$P_2 = 1 \times (8)^{1.4} = (2^3)^{7/5} = 2^{21/5} = 2^{4.2} = 16 \times 2^{0.2} \approx 16 \times 1.1487 = 18.38\,\text{atm}$$
Final Answer: (a) The final temperature is $689.2\,\text{K}$ (approx. $416.2^\circ\text{C}$); (b) The final pressure is approximately $18.38\,\text{atm}$.
Example 3
A cyclic heat engine operates with a thermal efficiency of 40%. During each cycle, it absorbs 800 J of heat from the high-temperature reservoir. (a) What is the mechanical work performed per cycle? (b) How much heat is rejected to the cold reservoir per cycle?
Step-by-Step Solution:
Step 1: Given values:
Thermal efficiency $\eta = 40\% = 0.40$
Heat absorbed from source $Q_1 = 800\,\text{J}$

Step 2: Calculate mechanical work output:
$$\eta = \frac{W}{Q_1} \implies W = \eta \cdot Q_1$$
$$W = 0.40 \times 800\,\text{J} = 320\,\text{J}$$
Step 3: Calculate heat rejected to the sink:
By energy conservation over a complete cycle ($\Delta U = 0$):
$$W = Q_1 - Q_2 \implies Q_2 = Q_1 - W$$
$$Q_2 = 800\,\text{J} - 320\,\text{J} = 480\,\text{J}$$
Alternatively: $Q_2 = Q_1 (1 - \eta) = 800 \times (1 - 0.40) = 800 \times 0.60 = 480\,\text{J}$.

Final Answer: (a) The mechanical work delivered is $320\,\text{J}$ per cycle; (b) The heat rejected to the sink is $480\,\text{J}$ per cycle.
Example 4
A Carnot engine operates between temperatures of 500 K and 300 K. In each cycle, it absorbs 1200 J of heat from the source. (a) Calculate the efficiency of the engine. (b) Calculate the work delivered per cycle. (c) If the source temperature is increased by 100 K while keeping the sink temperature constant, calculate the new efficiency.
Step-by-Step Solution:
Part (a): Efficiency of the Carnot Engine:
Given: $T_1 = 500\,\text{K}$ (source), $T_2 = 300\,\text{K}$ (sink).
$$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.60 = 0.40 = 40\%$$
Part (b): Mechanical Work Output per Cycle:
Given: $Q_1 = 1200\,\text{J}$.
$$W = \eta \cdot Q_1 = 0.40 \times 1200\,\text{J} = 480\,\text{J}$$
Part (c): New Efficiency with Elevated Source Temperature:
New source temperature: $T_1' = 500 + 100 = 600\,\text{K}$. Sink temperature: $T_2 = 300\,\text{K}$.
$$\eta' = 1 - \frac{T_2}{T_1'} = 1 - \frac{300}{600} = 1 - 0.50 = 0.50 = 50\%$$
Final Answer: (a) Initial efficiency is $40\%$; (b) Work output is $480\,\text{J}$ per cycle; (c) The new efficiency increases to $50\%$.
Example 5
(a) State the First Law of Thermodynamics and write its mathematical formula. (b) Explain why the molar specific heat at constant pressure (Cp) is always greater than that at constant volume (Cv). (c) Derive Mayer's relation Cp - Cv = R for 1 mole of an ideal gas.
Step-by-Step Solution:
Part (a): Statement of the First Law:
The First Law of Thermodynamics states that the heat energy supplied to a closed system ($dQ$) is equal to the sum of the increase in its internal energy ($dU$) and the external mechanical work performed by the system ($dW$):
$$dQ = dU + dW = dU + P\,dV$$
Part (b): Why Cp > Cv:
When 1 mole of gas is heated at constant volume ($dV = 0$), no external work is done ($dW = 0$). All the heat supplied goes exclusively toward increasing the internal energy and raising the temperature: $dQ_v = dU = C_v\,dT$.
However, when heated at constant pressure, the gas expands as temperature rises ($dV > 0$). The supplied heat must not only increase internal energy by the same amount ($dU$) to achieve the identical temperature rise, but also perform external mechanical work ($P\,dV$) against the surrounding pressure.
Therefore, extra heat is required at constant pressure, establishing that $C_p > C_v$.

Part (c): Derivation of Mayer's Relation:
For 1 mole of an ideal gas heated by $dT$ at constant pressure:
$$dQ = C_p\,dT$$
From the First Law: $C_p\,dT = dU + P\,dV$.
Since internal energy depends only on temperature: $dU = C_v\,dT$.
$$C_p\,dT = C_v\,dT + P\,dV$$
From the ideal gas equation for 1 mole: $P V = R T$.
Differentiating at constant pressure $P$: $P\,dV = R\,dT$.
Substituting $P\,dV = R\,dT$ into the First Law expression:
$$C_p\,dT = C_v\,dT + R\,dT \implies (C_p - C_v)dT = R\,dT$$
Dividing by $dT$ gives:
$$C_p - C_v = R$$
Hence proved.
Example 6
(a) Describe the four successive operations of a Carnot cycle for an ideal gas, illustrating the complete cycle on a P-V indicator diagram. (b) Prove analytically that the thermal efficiency of a Carnot engine is given by η = 1 - T₂/T₁. (c) State Carnot's Theorem.
Step-by-Step Solution:
Part (a): Four Operations of the Carnot Cycle:
A Carnot engine operates using 1 mole of ideal gas enclosed in a frictionless cylinder through 4 reversible steps:
  1. Isothermal Expansion (1 $\to$ 2 at $T_1$): The cylinder is placed on the source at $T_1$. Gas expands from $(P_1, V_1)$ to $(P_2, V_2)$. Since $\Delta T = 0 \implies \Delta U = 0$, heat absorbed is:$$Q_1 = W_{1\to 2} = R T_1 \ln\left(\frac{V_2}{V_1}\right)$$
  2. Adiabatic Expansion (2 $\to$ 3): Placed on an insulating stand ($dQ = 0$). Gas expands from $(P_2, V_2)$ to $(P_3, V_3)$, cooling from $T_1$ to $T_2$:$$W_{2\to 3} = \frac{R(T_1 - T_2)}{\gamma - 1}, \quad \text{with } T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$$
  3. Isothermal Compression (3 $\to$ 4 at $T_2$): Placed on sink at $T_2$. Gas is compressed from $(P_3, V_3)$ to $(P_4, V_4)$, rejecting heat $Q_2$:$$Q_2 = -W_{3\to 4} = R T_2 \ln\left(\frac{V_3}{V_4}\right)$$
  4. Adiabatic Compression (4 $\to$ 1): Placed on insulating stand ($dQ = 0$). Compressed from $(P_4, V_4)$ back to $(P_1, V_1)$, warming from $T_2$ to $T_1$:$$W_{4\to 1} = -\frac{R(T_1 - T_2)}{\gamma - 1}, \quad \text{with } T_2 V_4^{\gamma-1} = T_1 V_1^{\gamma-1}$$
Part (b): Analytical Derivation of Efficiency:
The net work output per cycle is the sum of works in all 4 steps:
$$W_{\text{net}} = W_{1\to 2} + W_{2\to 3} + W_{3\to 4} + W_{4\to 1}$$
Notice that $W_{2\to 3} + W_{4\to 1} = \frac{R(T_1 - T_2)}{\gamma - 1} - \frac{R(T_1 - T_2)}{\gamma - 1} = 0$.
$$W_{\text{net}} = Q_1 - Q_2 = R T_1 \ln\left(\frac{V_2}{V_1}\right) - R T_2 \ln\left(\frac{V_3}{V_4}\right)$$
From the adiabatic relations (steps 2 $\to$ 3 and 4 $\to$ 1):
$$T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1} \quad \text{and} \quad T_1 V_1^{\gamma-1} = T_2 V_4^{\gamma-1}$$
Dividing the two equations:
$$\left(\frac{V_2}{V_1}\right)^{\gamma-1} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} \implies \frac{V_2}{V_1} = \frac{V_3}{V_4}$$
Therefore, $\ln(V_2/V_1) = \ln(V_3/V_4)$. The heat ratio simplifies to:
$$\frac{Q_2}{Q_1} = \frac{R T_2 \ln(V_3/V_4)}{R T_1 \ln(V_2/V_1)} = \frac{T_2}{T_1}$$
Hence the thermal efficiency is:
$$\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$$
Hence derived.

Part (c): Carnot's Theorem:
  1. No real engine operating between two given constant temperatures can be more efficient than a reversible Carnot engine operating between the same two temperatures ($\eta_{\text{real}} \le \eta_{\text{Carnot}}$).
  2. All reversible engines operating between the same two temperatures have identical efficiency, regardless of the nature of the working substance.

Common Misconceptions & Examiner Traps

Common Misconception

Assuming that ΔU = n Cv ΔT applies only to isochoric (constant volume) processes.

Scientific Reality & Correction

For an ideal gas, internal energy U depends EXCLUSIVELY on temperature (Joule's Law). Therefore, the formula ΔU = n Cv ΔT holds universally for ANY thermodynamic process (isobaric, isothermal, adiabatic, or general polytropic) between two states with temperature difference ΔT.

Common Misconception

Believing isothermal and adiabatic curves have the same slope on a P-V diagram.

Scientific Reality & Correction

Differentiating shows that the adiabatic slope (dP/dV)_adi = -γ (P/V) is strictly γ times steeper than the isothermal slope (dP/dV)_iso = -(P/V). Because γ = Cp/Cv > 1 (e.g. 1.4 for air), adiabatic curves always drop more steeply than isothermal curves.

Common Misconception

Using Celsius temperatures instead of absolute Kelvin in Carnot efficiency calculations.

Scientific Reality & Correction

The Carnot efficiency formula η = 1 - T₂/T₁ is derived on the absolute thermodynamic scale and requires temperatures strictly in Kelvin (K = °C + 273.15). Using Celsius values leads to completely erroneous results.

Common Misconception

Confusing the direction of a thermodynamic cycle with the sign of net work done.

Scientific Reality & Correction

On a P-V diagram, a CLOCKWISE cycle encloses positive net work (W_net > 0) because expansion occurs at higher pressure than compression (Heat Engine). A COUNTER-CLOCKWISE cycle encloses negative net work (W_net < 0) because compression occurs at higher pressure, requiring external work input (Refrigerator / Heat Pump).

Common Misconception

Confusing the sign conventions for work between physics and chemistry.

Scientific Reality & Correction

In physics, dW = P dV, meaning work done BY the gas during expansion is POSITIVE (First Law: dQ = dU + dW). In chemistry (IUPAC), work done ON the system is defined as positive: w = -P ΔV (First Law: ΔU = q + w). Both describe the exact same physical reality, but mixing signs causes fatal calculation errors.

Thermodynamics & Carnot Cycle Vector Blueprint

θ Thermodynamics: Laws, Carnot Cycle & P-V Indicator Diagram WBCHSE Class 11 Physics • Chapter 8 • TargetExams Technical Master Blueprint 📈 Carnot Cycle on P-V Indicator Diagram P V 1 (P₁,V₁,T₁) 2 (P₂,V₂,T₁) 3 (P₃,V₃,T₂) 4 (P₄,V₄,T₂) Net Work W = ∮ P dV 1→2 Isothermal Exp (T₁): Q₁ = nRT₁·ln(V₂/V₁) 2→3 Adiabatic Exp: T₁V₂^(γ-1) = T₂V₃^(γ-1) 3→4 Isothermal Comp (T₂): Q₂ = nRT₂·ln(V₃/V₄) 4→1 Adiabatic Comp: T₂V₄^(γ-1) = T₁V₁^(γ-1) Slope Ratio: (dP/dV)_adi = γ · (dP/dV)_iso (γ = Cp/Cv > 1) ⚡ 1st Law of Thermodynamics: Energy Balance ΔQ = ΔU + ΔW ⟹ dQ = dU + P·dV • ΔU = n·Cv·ΔT (State function) • Work W = ∫ P dV (Path dependent) ⚙️ Heat Engine & Refrigerator Thermodynamics Hot Source T₁ (Q₁) Q₁ E W = Q₁ - Q₂ Q₂ Cold Sink T₂ (Q₂) Heat Engine: η = W / Q₁ = 1 - Q₂ / Q₁ Efficiency: η = 1 - T₂/T₁ Refrigerator / Heat Pump: β = Q₂ / W = Q₂ / (Q₁ - Q₂) COP: β = T₂/(T₁ - T₂) Relation: β = (1 - η) / η 📌 Core Thermodynamic Formulations Mayer's Law: Cp - Cv = R Cv = R/(γ-1), Cp = γR/(γ-1) Adiabatic Law: P V^γ = const TV^(γ-1) = const, P^(1-γ)T^γ = const Isothermal Work (ΔT=0): W = nRT·ln(V₂/V₁) = 2.303 nRT log(V₂/V₁) Adiabatic Work (Q=0): W = (P₁V₁ - P₂V₂) / (γ - 1) = nRΔT/(1-γ)

Chapter Summary & 10 Key Takeaways

Takeaway 1
The Zeroth Law of Thermodynamics establishes the concept of temperature as the condition for thermal equilibrium and forms the basis of thermometry.
Takeaway 2
Internal energy U of an ideal gas is a state function depending solely on absolute temperature (Joule's Law), so ΔU = n Cv ΔT for all processes.
Takeaway 3
The First Law of Thermodynamics (dQ = dU + dW = dU + P dV) expresses energy conservation, where dQ and dW are path-dependent while dU is a path-independent exact differential.
Takeaway 4
Work done during any quasi-static process equals the area under the P-V curve; a clockwise cycle produces net positive work (heat engine), whereas a counter-clockwise cycle requires net work input (refrigerator).
Takeaway 5
In an isothermal process (T = const), ΔU = 0 and work done is W = nRT ln(V₂/V₁); in an adiabatic process (dQ = 0), work done is W = (P₁V₁ - P₂V₂) / (γ - 1) at the expense of internal energy.
Takeaway 6
The slope of an adiabatic curve on a P-V diagram is γ times steeper than that of an isothermal curve: (dP/dV)_adi = γ · (dP/dV)_iso.
Takeaway 7
Mayer's relation Cp - Cv = R proves that molar heat capacity at constant pressure exceeds that at constant volume because heat must also perform external work P dV during expansion.
Takeaway 8
The Kelvin-Planck statement prohibits 100% efficient heat engines; the Clausius statement prohibits spontaneous heat transfer from colder to hotter bodies without external work.
Takeaway 9
A heat engine has thermal efficiency η = W/Q₁ = 1 - Q₂/Q₁, while a refrigerator has Coefficient of Performance β = Q₂/W = (1 - η)/η.
Takeaway 10
Carnot's reversible engine achieves maximum theoretical efficiency η = 1 - T₂/T₁, which depends solely on absolute reservoir temperatures and is independent of the working substance.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Why does an expanding gas cool down in an adiabatic expansion, whereas its temperature remains constant in an isothermal expansion?
Reveal Answer & Explanation
Answer:
Consider the First Law dQ = dU + dW with dQ = 0 for adiabatic and dT = 0 for isothermal.
2
Can a closed room be cooled down by leaving the door of an electric refrigerator open inside the room? Explain scientifically.
Reveal Answer & Explanation
Answer:
Apply Q₁ = Q₂ + W and treat the entire room as the system.
3
Why can the thermal efficiency of a heat engine never reach 100%, according to the Second Law of Thermodynamics?
Reveal Answer & Explanation
Answer:
Reference the Kelvin-Planck statement and Carnot efficiency η = 1 - T₂/T₁.
4
A Carnot engine has an efficiency of 50% when its sink is at 27°C. By how many degrees must the source temperature be increased to raise its efficiency to 60%?
Reveal Answer & Explanation
Answer:
Use η = 1 - T₂/T₁ to find T₁ and T₁', then find the difference ΔT₁.
5
What is the fundamental difference between a reversible and an irreversible process? Why are all spontaneous natural processes irreversible?
Reveal Answer & Explanation
Answer:
Consider dissipative forces like friction and the Second Law / entropy increase.
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