Physics

Chapter 4 The Second Law of Thermodynamics

Imron Rosyadi

Introduction

The Second Law of Thermodynamics explains the direction of natural processes.

It helps us understand why some events spontaneously occur and others don’t.

This law introduces entropy as a fundamental concept related to disorder and energy dispersal.

4.1 Reversible and Irreversible Processes

Learning Objectives

• Define reversible and irreversible processes.
• State the second law of thermodynamics via an irreversible process.

4.1 Reversible and Irreversible Processes: Free Expansion

Imagine an ideal gas in one half of a thermally insulated container.

The other half is a vacuum.

If the wall is removed, the gas expands to fill the entire container.

Figure 4.2 A gas expanding from half of a container to the entire container (a) before and (b) after the wall in the middle is removed.

• Work done (W): Since the other half is a vacuum, no force is exerted, so \(W=0\).
• Heat transfer (Q): The container is thermally insulated, so \(Q=0\).
• First Law: \(\Delta E_{int} = Q - W = 0\).
  • For an ideal gas, \(\Delta E_{int} = 0\) means temperature (T) is constant.
  • Pressure halves as volume doubles: \(p = p_0/2\).

Note

Can the gas spontaneously return to its original half?

Our intuition says no. This process is irreversible.

• Start by clearly defining the setup for free expansion: gas in one half, vacuum in the other, insulated container.
• Explain why work done and heat transfer are zero.
• Connect this to the first law of thermodynamics and its implications for internal energy and temperature of an ideal gas.
• Highlight the change in pressure and volume.
• Pose the question about the gas returning to its original state to introduce the concept of reversibility.
• Emphasize that while the first law doesn’t explicitly forbid it, our intuition and observation point to it being unlikely.

4.1 Reversible and Irreversible Processes: Definitions

• Reversible Process:
  • A process where both the system and its environment can be restored to their exact initial states.
  • Requires the process to be quasi-static (system always near equilibrium).
  • Idealized processes, rarely seen in reality.
  • Example: Slow, isothermal expansion against a piston, where heat is continuously exchanged to maintain constant temperature.
• Irreversible Process:
  • A process where the system and/or its environment cannot be restored to their original states simultaneously.
  • Almost all real-world processes are irreversible (e.g., friction, heat transfer across finite temperature differences).
  • Often called “natural processes.”
  • Characterized by finite gradients (e.g., temperature difference for heat flow).

Tip

Think of breaking a glass: easy to break (irreversible), impossible to put back exactly as it was without external changes.

• Clearly define a reversible process, emphasizing the “system AND environment” aspect and the quasi-static requirement.
• Provide a simple example of a nearly reversible process.
• Define an irreversible process, highlighting that it’s what we encounter in reality and the inability to restore both system and environment.
• Explain the role of finite gradients in irreversibility (e.g., heat flowing from hot to cold).
• Use the analogy of breaking glass to make the concept more relatable and memorable.

4.1 Reversible and Irreversible Processes: Microscopic View

• From a microscopic view, individual particle motions are reversible (Newton’s Laws).
• However, in macroscopic systems (\(>10^{23}\) particles), numerous collisions erase memory of initial states.
• The probability of all particles spontaneously reversing their paths is astronomically small.
  • Example: For gas expanded into vacuum, the chance of all molecules returning to the original half is virtually zero.
  • This is why free expansion is irreversible.

Note

Irreversibility arises from the overwhelming number of possible microscopic arrangements, making the “ordered” initial state incredibly improbable to spontaneously reoccur.

• Discuss the contrast between microscopic reversibility (Newton’s laws) and macroscopic irreversibility.
• Explain that the sheer number of particles in macroscopic systems leads to this effective irreversibility due to the vast phase space.
• Reiterate the free expansion example, connecting it to the microscopic probability argument.
• Emphasize that irreversibility isn’t about forbidden individual particle motion, but the statistical improbability of collective reversal.

4.1 Reversible and Irreversible Processes: Heat Flow

Consider two objects in thermal contact:

• One at temperature \(T_2\)
• The other at temperature \(T_1\), where \(T_2 > T_1\).

Figure 4.3 Spontaneous heat flow from an object at higher temperature T2 to another at lower temperature T1.

• Observation: Heat always flows spontaneously from the hotter object to the colder one.
  • Example: Holding ice in your hand (heat leaves hand, enters ice).
  • Example: Heating a metal rod (heat flows from fire to hand).

Second Law of Thermodynamics (Clausius Statement)

Heat never flows spontaneously from a colder object to a hotter object.

• “Spontaneously” means without any external effort.
• This statement summarizes empirical observations and is one of several equivalent formulations of the Second Law.

• Introduce the example of heat transfer between two objects at different temperatures.
• Relate it to everyday experiences (ice, hot rod) to make it intuitive.
• Formally state the Clausius statement of the Second Law, highlighting the word “spontaneously.”
• Emphasize that this is an empirical observation and a fundamental principle.
• Briefly mention that there are other equivalent statements, which will be covered later.

4.1 Reversible and Irreversible Processes: Summary of Processes

Process	Constant Quantity and Resulting Fact
Isobaric	Constant pressure \(W = p\Delta V\)
Isochoric	Constant volume \(W = 0\)
Isothermal	Constant temperature \(\Delta T = 0\)
Adiabatic	No heat transfer \(Q = 0\)

Table 4.1 Summary of Simple Thermodynamic Processes

Tip

These idealized processes are reversible in principle because the system can always be considered at equilibrium.

• Review the four basic thermodynamic processes.
• For each, state the constant quantity and a key resulting fact (e.g., work done).
• Emphasize that these are “idealized” and “reversible in principle,” connecting back to the quasi-static requirement.

4.2 Heat Engines

Learning Objectives

• Describe the function and components of a heat engine.
• Explain the efficiency of an engine.
• Calculate the efficiency of an engine for a given cycle of an ideal gas.

4.2 Heat Engines: Function and Components

• Heat Engine: A device that converts thermal energy (heat) into mechanical work.
  • Examples: Steam engines, internal combustion engines, power plants.
• Components:
  1. Heat Source (Hot Reservoir): Supplies heat (\(Q_h\)) at high temperature (\(T_h\)).
  2. Working Substance: Carries out the cycle (e.g., water in a steam engine, gas-air mixture in a car engine).
  3. Engine: Performs work (\(W\)) from absorbed heat.
  4. Heat Sink (Cold Reservoir): Receives discarded heat (\(Q_c\)) at low temperature (\(T_c\)).

Figure 4.4 A schematic representation of a heat engine. Energy flows from the hot reservoir to the cold reservoir while doing work.

• Start with a clear definition of a heat engine and its purpose.
• List the essential components: hot reservoir, cold reservoir, working substance, and the engine itself.
• Explain the role of each component, relating them to the schematic diagram.
• Emphasize that work is done from absorbed heat, and some heat is always discarded.
• Briefly mention the sign convention for heat and work in engine diagrams (amounts, directions shown by arrows).

4.2 Heat Engines: Energy Conservation and Efficiency

• Heat engines operate in cycles.

  • After one cycle, the working substance returns to its initial state.
  • Therefore, \(\Delta E_{int} = 0\) for a complete cycle.

• First Law of Thermodynamics for a Cycle:

  \(W = Q_{net} - \Delta E_{int}\)

  \(W = Q_h - Q_c - 0\)

  \[W = Q_h - Q_c\]

  • \(Q_h\): Heat absorbed from hot reservoir.
  • \(Q_c\): Heat discarded to cold reservoir.
  • \(W\): Net work done by the engine.

• Efficiency (\(e\)): “What we get out” divided by “what we put in.” \[e = \frac{W_{out}}{Q_{in}} = \frac{W}{Q_h}\] Substituting \(W = Q_h - Q_c\): \[e = \frac{Q_h - Q_c}{Q_h} = 1 - \frac{Q_c}{Q_h}\]

Warning

A heat engine cannot be 100% efficient (\(Q_c\) can never be zero). This is a consequence of the Second Law.

• Explain the significance of cyclic operation for a heat engine, leading to \(\Delta E_{int} = 0\).
• Apply the First Law to derive the relationship between work, \(Q_h\), and \(Q_c\).
• Define thermal efficiency intuitively (“what you get out / what you put in”).
• Show the derivation of the efficiency formula in terms of \(Q_h\) and \(Q_c\).
• Pre-emptively state the limitation on 100% efficiency, linking it to the Second Law.

4.2 Heat Engines: Example - A Lawn Mower

A lawn mower has an efficiency of 25.0% and an average power of 3.00 kW. What are (a) the average work done and (b) the minimum heat discharge into the air by the lawn mower in one minute of use?

Given:

\(e = 0.250\)

\(P = 3.00 \text{ kW} = 3000 \text{ J/s}\)

\(\Delta t = 1 \text{ min} = 60 \text{ s}\)

Solution:

(a) Average work delivered (W):

\(W = P \times \Delta t\)

\(W = (3000 \text{ J/s}) \times (60 \text{ s})\)

\(W = 180,000 \text{ J} = 180 \text{ kJ}\)

(b) Minimum heat discharged (\(Q_c\)):

We know \(e = \frac{W}{Q_h}\), so \(Q_h = \frac{W}{e}\).

\(Q_h = \frac{180 \text{ kJ}}{0.250} = 720 \text{ kJ}\)

From \(W = Q_h - Q_c\), we get \(Q_c = Q_h - W\).

\(Q_c = 720 \text{ kJ} - 180 \text{ kJ}\)

\(Q_c = 540 \text{ kJ}\)

Note

Alternatively, use \(Q_c = Q_h(1-e)\).

\(Q_c = (720 \text{ kJ})(1 - 0.250) = (720 \text{ kJ})(0.750) = 540 \text{ kJ}\)

The minimum heat discharge is \(540 \text{ kJ}\).

• Clearly state the problem and list the given values.
• Break down the solution into two parts, (a) and (b).
• For part (a), show the calculation of work using power and time.
• For part (b), demonstrate two methods to calculate \(Q_c\):
  • Method 1: Calculate \(Q_h\) first, then use \(Q_c = Q_h - W\).
  • Method 2: Use the efficiency definition \(e = W/Q_h\) to relate \(Q_c\) directly.
• Emphasize the significance: higher efficiency means less waste heat, which is better for the environment.

4.3 Refrigerators and Heat Pumps

Learning Objectives

• Describe a refrigerator and a heat pump and list their differences.
• Calculate the performance coefficients of simple refrigerators and heat pumps.

4.3 Refrigerators and Heat Pumps: Overview

• Reversed Heat Engines: Refrigerators and heat pumps are essentially heat engines operating in reverse.
  • They require work input (\(W\)) to transfer heat from a colder region to a hotter region.

Figure 4.6 A schematic representation of a refrigerator (or a heat pump). The arrow next to work (W) indicates work being put into the system.

• Refrigerator:
  • Purpose: To remove heat (\(Q_c\)) from a cold reservoir (e.g., inside a fridge, air-conditioned room).
  • Work is done to move heat against its natural flow (from cold to hot).
• Heat Pump:
  • Purpose: To dump heat (\(Q_h\)) into a hot reservoir (e.g., inside a house during winter).
  • Also uses work to move heat from a colder outside to a warmer inside.

• Introduce refrigerators and heat pumps as reversed heat engines.
• Highlight the key difference from heat engines: they require work input.
• Explain the distinct primary purpose of each:
  • Refrigerator: focusing on removing heat from the cold side.
  • Heat pump: focusing on delivering heat to the hot side.
• Refer to the schematic to show the direction of heat flows (\(Q_c\) absorbed, \(Q_h\) discarded) and work input (\(W\)).

4.3 Refrigerators and Heat Pumps: Performance Coefficients

• Since the focus is on “how much heat is moved for a given work input,” efficiency for these devices is called the Coefficient of Performance (COP).

• For a Refrigerator (\(K_R\)):

  • “What we get out” (heat removed from cold reservoir) divided by “what we put in” (work done). \[K_R = \frac{Q_c}{W}\]
  • Using \(W = Q_h - Q_c\) (from the First Law for a cycle, with \(W\) now being input): \[K_R = \frac{Q_c}{Q_h - Q_c}\]
  • Typical \(K_R\) values are between 2 and 6.

• For a Heat Pump (\(K_P\)):

  • “What we get out” (heat delivered to hot reservoir) divided by “what we put in” (work done). \[K_P = \frac{Q_h}{W}\]
  • Using \(W = Q_h - Q_c\): \[K_P = \frac{Q_h}{Q_h - Q_c}\]
  • Typical \(K_P\) values are between 3 and 7.

Tip

Note that \(K_P = K_R + 1\). This means a heat pump always delivers more heat to the hot reservoir than it removes from the cold reservoir, because the work input also becomes part of the heat delivered.

• Explain why “efficiency” isn’t the best term for these devices and introduce “Coefficient of Performance” (COP).
• Define \(K_R\) for a refrigerator, clearly stating the “get out” and “put in” terms.
• Show the derivation of \(K_R\) using the First Law.
• Define \(K_P\) for a heat pump, similarly.
• Show the derivation of \(K_P\).
• Highlight the relationship \(K_P = K_R + 1\) as an important takeaway.
• Mention typical COP values to give a sense of scale.

4.4 Statements of the Second Law of Thermodynamics

Learning Objectives

• Contrast the second law of thermodynamics statements according to Kelvin and Clausius formulations.
• Interpret the second of thermodynamics via irreversibility.

4.4 Statements of the Second Law: Clausius and Kelvin

The Second Law of Thermodynamics can be stated in several equivalent ways.

• Clausius Statement (Heat Flow):
  • “Heat never flows spontaneously from a colder object to a hotter object.”
  • Implies that a “perfect refrigerator” (one that transfers heat from cold to hot without work input) is impossible.
• Kelvin Statement (Heat Engines):
  • “It is impossible to convert the heat from a single source into work without any other effect.”
  • Implies that a “perfect heat engine” (one that is 100% efficient, converting all heat absorbed into work in a cycle) is impossible.

Figure 4.8 (a) A “perfect heat engine” converts all input heat into work. (b) A “perfect refrigerator” transports heat from a cold reservoir to a hot reservoir without work input. Neither of these devices is achievable in reality.

• Reiterate that there are multiple equivalent statements of the Second Law.
• Present the Clausius statement again, linking it to the impossibility of a perfect refrigerator.
• Present the Kelvin statement, linking it to the impossibility of a perfect heat engine (100% efficiency).
• Use Figure 4.8 to visually illustrate these “impossible” devices.
• Explain the importance of “without any other effect” in the Kelvin statement, emphasizing that it applies to cyclic processes.

4.4 Statements of the Second Law: Equivalence

The Kelvin and Clausius statements are equivalent. This means if one is false, the other must also be false.

Proof (by contradiction): 1. Assume Clausius statement is FALSE: A perfect refrigerator exists (transfers \(Q\) from \(T_c\) to \(T_h\) with \(W=0\)). 2. Combine with a real heat engine: - Real engine takes \(Q_h' = Q + \Delta Q\) from \(T_h\). - Does work \(W = \Delta Q\). - Discards \(Q_c' = Q\) to \(T_c\). 3. Net effect of combined device: - Net heat removed from \(T_h\): \((Q + \Delta Q) - Q = \Delta Q\). - Net heat transfer to/from \(T_c\): \(Q - Q = 0\). - Net work done: \(W = \Delta Q\).

Figure 4.9 Combining a perfect refrigerator and a real heat engine yields a perfect heat engine because W = \(\Delta\)Q.

4. This combined device is a perfect heat engine (takes \(\Delta Q\) from a single reservoir \(T_h\) and converts it all to work \(W = \Delta Q\)).
5. This contradicts the Kelvin statement.
6. Therefore, if the Clausius statement is false, the Kelvin statement is also false. This proves their equivalence.

• Explain the concept of proving equivalence by showing that if one statement is false, the other must also be false.
• Walk through the steps of the proof by contradiction:
  • Start by assuming the Clausius statement is false (perfect refrigerator exists).
  • Introduce a real heat engine.
  • Combine these two devices conceptually.
  • Analyze the net heat flows and work done by the combined system.
  • Show that this combination results in a “perfect heat engine,” which directly contradicts the Kelvin statement.
• Conclude that the initial assumption (Clausius statement is false) must be wrong, thus proving the equivalence.

4.4 Statements of the Second Law: Efficiency Limits

Two important properties of heat engines operating between two heat reservoirs:

1. Reversible engines are more efficient than irreversible engines.
   • Any reversible engine operating between two reservoirs has a greater efficiency than any irreversible engine operating between the same two reservoirs.
   • Proof involves showing that if an irreversible engine were more efficient, it could be combined with a reversible engine (run in reverse as a refrigerator) to violate the Clausius statement.
2. All reversible engines have the same efficiency.
   • All reversible engines operating between the same two reservoirs have the same efficiency.
   • Proof involves showing that if two reversible engines had different efficiencies, one could be combined with the other (run in reverse) to violate the Clausius statement.

Important

These properties are fundamental to understanding the Carnot efficiency, which sets the upper limit for heat engine performance.

• State the two key properties regarding heat engine efficiencies.
• Briefly explain the reasoning behind each property, mentioning the use of contradiction and the Clausius statement. Avoid going into full proof detail, as it’s implied the undergraduate audience might not need the full mathematical rigor here.
• Connect these properties to the upcoming concept of Carnot efficiency as a “preview.”

4.5 The Carnot Cycle

Learning Objectives

• Describe the Carnot cycle with the roles of all four processes involved.
• Outline the Carnot principle and its implications.
• Demonstrate the equivalence of the Carnot principle and the second law of thermodynamics.

4.5 The Carnot Cycle: Overview

• Proposed by Sadi Carnot in 1824.
• A hypothetical reversible cycle that achieves the highest possible efficiency between two given heat reservoirs (\(T_h\) and \(T_c\)).
• Crucial for theoretical understanding and defining an absolute temperature scale.
• Consists of four reversible processes:

Figure 4.11 The four processes of the Carnot cycle.

• Introduce Sadi Carnot and the significance of his cycle.
• Emphasize its theoretical importance as the most efficient possible cycle.
• List the four types of processes involved in the cycle.
• Refer to Figure 4.11 to show the conceptual steps.

4.5 The Carnot Cycle: Four Processes

Let’s trace the Carnot cycle on a \(pV\) diagram, assuming an ideal gas as the working substance.

Figure 4.12 The total work done by the gas in the Carnot cycle is shown and given by the area enclosed by the loop MNOPM.

1. Isothermal Expansion (M to N):
   • Gas absorbs heat \(Q_h\) from hot reservoir (\(T_h\)).
   • Expands at constant \(T_h\), doing work \(W_1\).
   • \(\Delta E_{int} = 0\), so \(Q_h = W_1 = nRT_h \ln(\frac{V_N}{V_M})\).
2. Adiabatic Expansion (N to O):
   • Gas thermally isolated (\(Q=0\)).
   • Expands, doing work \(W_2\), and its temperature drops from \(T_h\) to \(T_c\).
   • \(T V^{\gamma-1} = \text{constant}\).
3. Isothermal Compression (O to P):
   • Gas releases heat \(Q_c\) to cold reservoir (\(T_c\)).
   • Compressed at constant \(T_c\), work \(W_3\) is done on the gas.
   • \(\Delta E_{int} = 0\), so \(Q_c = nRT_c \ln(\frac{V_O}{V_P})\).
4. Adiabatic Compression (P to M):
   • Gas thermally isolated (\(Q=0\)).
   • Compressed, work \(W_4\) is done on the gas, and its temperature rises from \(T_c\) to \(T_h\).

• Use Figure 4.12 as the visual aid for the \(pV\) diagram.
• Go through each of the four steps in detail:
  • Name of the process.
  • What happens to the gas (heat exchange, work, temperature, volume).
  • Relevant thermodynamic relationships for ideal gas (e.g., for isothermal, \(\Delta E_{int} = 0\); for adiabatic, \(TV^{\gamma-1} = \text{const}\)).
  • Clearly indicate where \(Q_h\) is absorbed and \(Q_c\) is rejected.
  • Mention work done by or on the gas.
• Emphasize that the area enclosed by the loop represents the net work done.

4.5 The Carnot Cycle: Efficiency

• For a complete Carnot cycle, \(\Delta E_{int} = 0\).

  • From the First Law: \(W = Q_h - Q_c\).

• The efficiency of a Carnot engine (\(e_C\)) is: \[e_C = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}\]

• Through analysis of the adiabatic steps, it can be shown that for a Carnot cycle: \[\frac{Q_c}{Q_h} = \frac{T_c}{T_h}\] (where temperatures must be in Kelvin).

• Therefore, the Carnot Efficiency is: \[e_C = 1 - \frac{T_c}{T_h}\]

Carnot’s Principle

No engine working between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine. This means the Carnot efficiency \(e_C\) is the maximum possible efficiency for any heat engine operating between \(T_h\) and \(T_c\). Real engines always have \(e < e_C\).

• Reiterate the First Law for a cycle (\(W = Q_h - Q_c\)).
• Present the general efficiency formula.
• State the crucial relationship \(Q_c/Q_h = T_c/T_h\) for a Carnot cycle (without re-deriving it fully, as it’s complex for a single slide, but acknowledge its derivation comes from the adiabatic steps).
• Provide the final Carnot efficiency formula \(e_C = 1 - T_c/T_h\).
• Formally state Carnot’s Principle.
• Emphasize its implication: \(e_C\) is the absolute upper limit for efficiency, and real engines are always less efficient.

4.5 The Carnot Cycle: Refrigerator/Heat Pump COPs

• A Carnot refrigerator or heat pump is a Carnot engine running in reverse.

  • The relationships \(W = Q_h - Q_c\) and \(Q_c/Q_h = T_c/T_h\) still hold.

• Coefficient of Performance for a Carnot Refrigerator (\(K_{R,C}\)): \[K_{R,C} = \frac{Q_c}{W} = \frac{Q_c}{Q_h - Q_c}\] Substituting \(Q_h = Q_c \frac{T_h}{T_c}\): \[K_{R,C} = \frac{Q_c}{Q_c \frac{T_h}{T_c} - Q_c} = \frac{1}{\frac{T_h}{T_c} - 1}\] \[K_{R,C} = \frac{T_c}{T_h - T_c}\]

• Coefficient of Performance for a Carnot Heat Pump (\(K_{P,C}\)): \[K_{P,C} = \frac{Q_h}{W} = \frac{Q_h}{Q_h - Q_c}\] Substituting \(Q_c = Q_h \frac{T_c}{T_h}\): \[K_{P,C} = \frac{Q_h}{Q_h - Q_h \frac{T_c}{T_h}} = \frac{1}{1 - \frac{T_c}{T_h}}\] \[K_{P,C} = \frac{T_h}{T_h - T_c}\]

• Explain that Carnot refrigerator/heat pump are simply reversed Carnot engines.
• Derive the \(K_{R,C}\) formula, showing the steps for substitution.
• Derive the \(K_{P,C}\) formula, showing the steps for substitution.
• Emphasize that these are the maximum possible COPs for devices operating between \(T_h\) and \(T_c\).

4.5 The Carnot Cycle: Example - A Carnot Engine

A Carnot engine has an efficiency of 0.60 and the temperature of its cold reservoir is 300 K.

1. What is the temperature of the hot reservoir?
2. If the engine does 300 J of work per cycle, how much heat is removed from the high-temperature reservoir per cycle?
3. How much heat is exhausted to the low-temperature reservoir per cycle?

Given:

\(e_C = 0.60\)

\(T_c = 300 \text{ K}\)

\(W = 300 \text{ J}\)

Solution:

(a) Temperature of the hot reservoir (\(T_h\)):

\(e_C = 1 - \frac{T_c}{T_h}\)

\(0.60 = 1 - \frac{300 \text{ K}}{T_h}\)

\(\frac{300 \text{ K}}{T_h} = 1 - 0.60 = 0.40\)

\(T_h = \frac{300 \text{ K}}{0.40} = 750 \text{ K}\)

(b) Heat removed from hot reservoir (\(Q_h\)):

\(e_C = \frac{W}{Q_h} \implies Q_h = \frac{W}{e_C}\)

\(Q_h = \frac{300 \text{ J}}{0.60} = 500 \text{ J}\)

(c) Heat exhausted to cold reservoir (\(Q_c\)):

\(W = Q_h - Q_c \implies Q_c = Q_h - W\)

\(Q_c = 500 \text{ J} - 300 \text{ J} = 200 \text{ J}\)

Note

Always use Kelvin for temperature in Carnot efficiency/COP calculations.

• Walk through the problem step-by-step.
• Clearly show the formula used for each calculation.
• Emphasize the importance of using Kelvin for temperatures.
• Highlight the relationship between the three quantities (\(W\), \(Q_h\), \(Q_c\)) via the First Law.

4.6 Entropy

Learning Objectives

• Describe the meaning of entropy.
• Calculate the change of entropy for some simple processes.

4.6 Entropy: Definition and State Function

• Entropy (\(S\)): A thermodynamic variable that quantifies the disorder or randomness of a system.
  • Related to how energy is distributed among particles.
• State Function:
  • Like internal energy (\(E_{int}\)), entropy is a state function.
  • The change in entropy (\(\Delta S\)) between two states depends only on the initial and final states, not on the path taken.
• Entropy Change for Reversible Process at Constant Temperature:
  • If a system exchanges heat \(Q\) reversibly at a constant temperature \(T\) (in Kelvin): \[\Delta S = \frac{Q}{T}\]
  • If \(Q > 0\) (heat absorbed), \(\Delta S > 0\) (entropy increases).
  • If \(Q < 0\) (heat lost), \(\Delta S < 0\) (entropy decreases).

Note

For phase changes (e.g., melting ice), temperature remains constant, so \(\Delta S = Q/T\) is directly applicable.

• Start with a concise definition of entropy, linking it to disorder/randomness.
• Emphasize that entropy is a state function, meaning path independence.
• Introduce the formula for entropy change in a reversible process at constant temperature.
• Explain the sign convention for \(Q\) and its effect on \(\Delta S\).
• Give phase change as a direct application example.

4.6 Entropy: General Reversible Processes

• For an arbitrary reversible transition where temperature is not constant:
  • Imagine the process as a series of infinitesimal steps, each exchanging heat \(dQ_{rev}\) at temperature \(T\).
  • The total change in entropy is found by integrating: \[\Delta S = \int_{A}^{B} \frac{dQ_{rev}}{T}\]
  • This integral is path-independent for reversible processes, reinforcing that \(S\) is a state function.
• Entropy Change in a Reversible Cyclic Process:
  • For any complete reversible cyclic process (like the Carnot cycle), the net change in entropy of the system is zero. \[\oint \frac{dQ_{rev}}{T} = 0\]
  • This is because the system returns to its initial state.

• Extend the entropy change calculation to variable-temperature reversible processes using the integral form.
• Reiterate path independence from the integral definition.
• Explain the entropy change for any reversible cyclic process, which must be zero because entropy is a state function.
• Connect this back to the Carnot cycle, where \(\Delta S_{engine} = 0\).

4.6 Entropy: Second Law (Entropy Statement)

• What about irreversible processes?
  • The entropy of a closed system (or the universe) always increases for an irreversible process.
  • For a reversible process, the entropy of a closed system remains constant (\(\Delta S = 0\)).

Second Law of Thermodynamics (Entropy Statement)

The entropy of a closed system and the entire universe never decreases.

\[\Delta S_{universe} \ge 0\]

• The equality (\(= 0\)) holds for reversible processes.
• The inequality (\(> 0\)) holds for irreversible (natural) processes.

• This statement is consistent with the Clausius and Kelvin statements and Carnot’s principle.
  • Natural processes always proceed in a direction that increases the total entropy of the universe.
  • This is why spontaneous heat flow is from hot to cold, and why gas expands to fill a vacuum.

• Transition from reversible to irreversible processes regarding entropy.
• State the entropy formulation of the Second Law clearly.
• Explain the meaning of the equality and inequality signs.
• Emphasize that this statement underpins why natural processes occur in one direction.
• Connect it back to the earlier examples of free expansion and heat flow.

4.6 Entropy: Example - Entropy Change of Melting Ice

Heat is slowly added to a 50-g chunk of ice at 0°C until it completely melts into water at the same temperature. What is the entropy change of the ice?

Given:

\(m = 50 \text{ g}\)

\(T = 0^\circ\text{C} = 273 \text{ K}\)

Latent heat of fusion for water, \(L_f = 335 \text{ J/g}\)

Strategy:

• This is a phase change at constant temperature, so the process can be considered reversible for the ice.
• Use \(\Delta S = Q/T\).
• First, calculate the heat absorbed during melting, \(Q = mL_f\).

Solution:

1. Calculate heat absorbed (\(Q\)):

   \(Q = m L_f = (50 \text{ g}) \times (335 \text{ J/g})\)

   \(Q = 16,750 \text{ J} = 16.75 \text{ kJ}\)

2. Calculate entropy change (\(\Delta S\)):

   \(\Delta S = \frac{Q}{T} = \frac{16,750 \text{ J}}{273 \text{ K}}\)

   \(\Delta S \approx 61.36 \text{ J/K}\)

Note

The entropy of the ice increases because it absorbs heat and becomes more disordered (liquid water molecules have more freedom than solid ice molecules).

• Clearly state the problem and list the given values.
• Outline the strategy, identifying it as a constant temperature reversible process for the system.
• Walk through the calculation of heat absorbed (\(Q\)).
• Then, calculate the entropy change (\(\Delta S\)).
• Emphasize the physical interpretation of the positive entropy change (increased disorder).

4.7 Entropy on a Microscopic Scale

Learning Objectives

• Interpret the meaning of entropy at a microscopic scale.
• Calculate a change in entropy for an irreversible process of a system and contrast with the change in entropy of the universe.
• Explain the third law of thermodynamics.

4.7 Entropy on a Microscopic Scale: Disorder and Probability

• On a microscopic scale, entropy is related to the number of possible microscopic arrangements (microstates) that correspond to a given macroscopic state (macrostate).
  • A system with more microstates has higher entropy.
  • More microstates generally means more disorder or randomness.
• Boltzmann’s Entropy Formula: \[S = k_B \ln W\]
  • \(k_B\): Boltzmann constant (\(1.38 \times 10^{-23} \text{ J/K}\)).
  • \(W\): Number of microstates corresponding to the macrostate.
• Second Law (in terms of disorder): In any irreversible process, the universe becomes more disordered.
  • Example: Shuffling cards: A new deck (ordered) has low entropy. Shuffling increases the number of possible arrangements, increasing disorder and entropy.
  • Example: Free expansion: Gas molecules can occupy a larger volume, meaning more possible positions (microstates), thus increasing entropy.

• Introduce the microscopic interpretation of entropy, linking it to microstates and disorder.
• Present Boltzmann’s formula (\(S = k_B \ln W\)) as the mathematical representation, explaining \(k_B\) and \(W\).
• Rephrase the Second Law in terms of disorder, providing intuitive examples like shuffling cards and free expansion.
• Emphasize that natural processes tend towards states of higher probability (more microstates/disorder).

4.7 Entropy on a Microscopic Scale: Example - Entropy Change of the Universe

50 g of ice at 0°C is placed in contact with a heat reservoir at 20°C. Heat flows spontaneously from the reservoir to the ice, which melts and eventually reaches 20°C. Find the change in entropy of (a) the ice and (b) the universe.

Given:

\(m_{ice} = 50 \text{ g} = 0.050 \text{ kg}\)

\(T_{ice, initial} = 0^\circ\text{C} = 273 \text{ K}\)

\(T_{reservoir} = 20^\circ\text{C} = 293 \text{ K}\)

\(L_f = 335 \text{ J/g} = 335,000 \text{ J/kg}\)

\(c_{water} = 4186 \text{ J/(kg} \cdot \text{K)}\)

Solution:

(a) Change in entropy of the ice (\(\Delta S_{ice}\)):

This involves two reversible steps for the ice: 1. Melting at 0°C (273 K):

$Q_1 = m L_f = (0.050 \text{ kg})(335,000 \text{ J/kg}) = 16,750 \text{ J}$

$\Delta S_1 = \frac{Q_1}{T_1} = \frac{16,750 \text{ J}}{273 \text{ K}} = 61.36 \text{ J/K}$

2. Warming water from 0°C to 20°C (273 K to 293 K):

   \(\Delta S_2 = \int_{T_1}^{T_2} \frac{dQ}{T} = \int_{T_1}^{T_2} \frac{mc dT}{T} = mc \ln(\frac{T_2}{T_1})\)

   \(\Delta S_2 = (0.050 \text{ kg})(4186 \text{ J/(kg} \cdot \text{K)}) \ln(\frac{293 \text{ K}}{273 \text{ K}})\)

   \(\Delta S_2 = (209.3 \text{ J/K}) \ln(1.073) \approx 14.65 \text{ J/K}\)

   Total \(\Delta S_{ice} = \Delta S_1 + \Delta S_2 = 61.36 \text{ J/K} + 14.65 \text{ J/K} = 76.01 \text{ J/K}\)

(b) Change in entropy of the universe (\(\Delta S_{universe}\)): The reservoir provides all the heat.

Total heat transferred from reservoir:

\(Q_{total} = Q_1 + Q_2 = m L_f + m c \Delta T = 16,750 \text{ J} + (0.050 \text{ kg})(4186 \text{ J/(kg} \cdot \text{K)})(20 \text{ K})\)

\(Q_{total} = 16,750 \text{ J} + 4186 \text{ J} = 20,936 \text{ J}\)

Change in entropy of the reservoir (at constant \(T_R = 293 \text{ K}\)):

\(\Delta S_{reservoir} = \frac{-Q_{total}}{T_{reservoir}} = \frac{-20,936 \text{ J}}{293 \text{ K}} = -71.45 \text{ J/K}\)

Total \(\Delta S_{universe} = \Delta S_{ice} + \Delta S_{reservoir} = 76.01 \text{ J/K} - 71.45 \text{ J/K} = 4.56 \text{ J/K}\)

Important

Since \(\Delta S_{universe} > 0\), the irreversible process (spontaneous heat flow and melting) indeed increases the total entropy of the universe, confirming the Second Law.

• This is a multi-step calculation. Guide students through each part.
• Break down the entropy change for the ice into two phases: melting (constant T) and warming (variable T, requiring integration).
• Clearly show the calculation for each part.
• Then, calculate the total heat provided by the reservoir.
• Calculate the entropy change of the reservoir (it’s negative as heat is lost).
• Finally, sum the entropy changes to get the universe’s entropy change.
• Emphasize that \(\Delta S_{universe} > 0\) is the key result confirming the Second Law.
• Discuss the increase in disorder of the ice (solid to liquid) vs. the slight decrease in disorder of the reservoir (losing heat), with the net result being an increase in overall disorder.

4.7 Entropy on a Microscopic Scale: Third Law of Thermodynamics

• The Second Law states that \(\Delta S_{universe} \ge 0\).
• But what happens as temperature approaches absolute zero (\(T \to 0 \text{ K}\))?

Third Law of Thermodynamics

The absolute zero temperature (\(0 \text{ K}\)) cannot be reached through any finite number of cooling steps. (Alternatively: A system becomes perfectly ordered when its temperature approaches absolute zero, and its entropy approaches its absolute minimum (often taken as zero for a pure, crystalline substance)). \[\lim_{T \to 0 \text{ K}} \Delta S = 0\]

• Implications:
  • It’s impossible to build a “perfect engine” that uses a reservoir at \(0 \text{ K}\) to achieve 100% efficiency.
  • At \(0 \text{ K}\), all thermal motion ceases, and a system is in its lowest energy state, ideally with only one possible microstate (\(W=1 \implies S = k_B \ln 1 = 0\)).

• Introduce the Third Law of Thermodynamics as an extension beyond the Second Law regarding low temperatures.
• State the two common formulations of the Third Law.
• Explain the implication for reaching absolute zero and for engine efficiency (connecting back to the Second Law).
• Briefly connect it to the microscopic view: at 0 K, ideally, there’s only one microstate, meaning zero entropy (for a perfect crystal).

Key Takeaways

• Reversible vs. Irreversible: Reversible processes are idealized and can be reversed without any change in the environment; irreversible processes (natural processes) cannot.
• Second Law (Clausius): Heat never flows spontaneously from cold to hot.
• Second Law (Kelvin): It’s impossible for an engine in a cycle to convert all absorbed heat into work.
• Heat Engine Efficiency: \(e = 1 - \frac{Q_c}{Q_h}\).
• Refrigerator/Heat Pump COP: \(K_R = \frac{Q_c}{W}\), \(K_P = \frac{Q_h}{W}\).
• Carnot Cycle: The most efficient reversible cycle between two temperatures.
• Carnot Efficiency: \(e_C = 1 - \frac{T_c}{T_h}\) (maximum possible efficiency).
• Carnot COPs: \(K_{R,C} = \frac{T_c}{T_h - T_c}\), \(K_{P,C} = \frac{T_h}{T_h - T_c}\) (maximum possible COPs).
• Entropy (\(S\)): A measure of disorder/randomness; a state function.
• Entropy Change (reversible, constant T): \(\Delta S = \frac{Q_{rev}}{T}\).
• Entropy Change (reversible, variable T): \(\Delta S = \int \frac{dQ_{rev}}{T}\).
• Second Law (Entropy): \(\Delta S_{universe} \ge 0\) (increases for irreversible processes, constant for reversible).
• Third Law: Absolute zero temperature cannot be reached; entropy approaches its minimum (zero for perfect crystals) as \(T \to 0 \text{ K}\).

• This slide acts as a comprehensive summary of all the main concepts covered.
• Each bullet point should be a concise reminder of a key definition, principle, or formula.
• Encourage students to review these points as a quick reference for the chapter.

Key Equations

Equation	Description
\(W = Q_h - Q_c\)	Work done by a heat engine in a cycle (First Law)
\(e = 1 - \frac{Q_c}{Q_h}\)	Efficiency of a heat engine
\(K_R = \frac{Q_c}{W}\)	Coefficient of Performance for a Refrigerator
\(K_P = \frac{Q_h}{W}\)	Coefficient of Performance for a Heat Pump
\(e_C = 1 - \frac{T_c}{T_h}\)	Carnot efficiency (maximum possible)
\(K_{R,C} = \frac{T_c}{T_h - T_c}\)	COP of a Carnot Refrigerator (maximum possible)
\(K_{P,C} = \frac{T_h}{T_h - T_c}\)	COP of a Carnot Heat Pump (maximum possible)
\(\Delta S = \frac{Q_{rev}}{T}\)	Entropy change for reversible process at constant T
\(\Delta S = \int \frac{dQ_{rev}}{T}\)	Entropy change for general reversible process
\(\Delta S_{universe} \ge 0\)	Second Law of Thermodynamics (Entropy Statement)
\(\lim_{T \to 0 \text{ K}} \Delta S = 0\)	Third Law of Thermodynamics

• This slide provides a focused reference for the most important formulas.
• Students can quickly consult this for problem-solving.
• Ensure all variables are implicitly understood from previous slides or briefly explained if ambiguity exists.

Key Terms

Term	Definition
Reversible Process	A thermodynamic process where both the system and its environment can be restored to their exact initial states.
Irreversible Process	A thermodynamic process where the system and/or its environment cannot be restored to their original states simultaneously.
Heat Engine	A device that converts thermal energy (heat) into mechanical work by operating between a hot and a cold reservoir.
Efficiency (\(e\))	The ratio of work output to heat input for a heat engine.
Refrigerator	A device that uses work input to transfer heat from a cold reservoir to a hot reservoir.
Heat Pump	A device that uses work input to transfer heat from a cold reservoir to a hot reservoir, with the goal of heating the hot reservoir.
Coefficient of Performance (COP)	A measure of the effectiveness of a refrigerator or heat pump, defined as the ratio of desired heat transfer to work input.
Carnot Cycle	A theoretical reversible thermodynamic cycle that represents the most efficient possible cycle for converting heat into work between two temperatures.
Carnot’s Principle	States that no engine working between two constant temperature reservoirs can have a greater efficiency than a reversible engine.
Entropy (\(S\))	A thermodynamic state function that quantifies the disorder or randomness of a system and the dispersal of energy.
Clausius Statement	States that heat never flows spontaneously from a colder object to a hotter object.
Kelvin Statement	States that it is impossible to convert the heat from a single source entirely into work without any other effect.
Third Law of Thermodynamics	States that absolute zero temperature cannot be reached, and a system’s entropy approaches its minimum as temperature approaches absolute zero.

• This slide defines the essential vocabulary for the chapter.
• Each term should have a concise, clear definition.
• Encourage students to understand these terms thoroughly as they form the foundation of the topic.