Laws of Thermodynamics
The four fundamental laws that govern energy, heat, work, and entropy
Historical Context
The laws of thermodynamics emerged from the practical study of steam engines in the early 19th century. Sadi Carnot's 1824 work on the motive power of fire laid the groundwork, while Rudolf Clausius and Lord Kelvin formalized the first and second laws in the 1850s. The zeroth law was articulated later (1930s) by Ralph Fowler, despite being logically prior to the others. Walther Nernst proposed the third law in 1906, completing the theoretical framework.
These laws are remarkable for their universality: they apply to all physical systems regardless of their microscopic details. As Arthur Eddington wrote, "If your theory is found to be against the second law of thermodynamics, I can give you no hope; there is nothing for it but to collapse in deepest humiliation."
Key contributors: Sadi Carnot (1824), Rudolf Clausius (1850), Lord Kelvin (1851), Ludwig Boltzmann (1870s), Walther Nernst (1906), Ralph Fowler (1930s).
Derivation 1: The Zeroth Law and Temperature
The zeroth law establishes the concept of thermal equilibrium and provides the logical foundation for temperature as a measurable quantity.
Statement of the Zeroth Law
If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A is in thermal equilibrium with B.
This seemingly obvious statement has profound consequences. It guarantees that we can define a property called temperature such that two systems in thermal equilibrium share the same temperature. Mathematically, thermal equilibrium defines an equivalence relation on the set of thermodynamic states, and temperature labels the equivalence classes.
Consider three systems with equations of state relating their thermodynamic coordinates:
Solving for $P_C$ from each and equating, we can eliminate the properties of C entirely, yielding a relation between A and B alone. This shows that a single scalar function$\theta$ (temperature) characterizes thermal equilibrium:
Derivation 2: The First Law of Thermodynamics
The first law is a statement of energy conservation extended to include heat transfer. It defines internal energy U as a state function and relates it to heat and work.
First Law (Differential Form)
where $dU$ is an exact differential (state function), while $\delta Q$ and$\delta W$ are inexact differentials (path-dependent).
Proof that U is a State Function
Consider a cyclic process returning to the initial state. Joule's experiments showed that the adiabatic work required to change the state of a system depends only on the initial and final states, not the path:
For a general (non-adiabatic) process between the same states:
For reversible work on an ideal gas ($\delta W = P\,dV$):
Applications
Isochoric Process (constant V)
Isobaric Process (constant P)
Isothermal Process (ideal gas)
Adiabatic Process (ideal gas)
Derivation 3: The Second Law and Entropy
The second law introduces entropy and establishes the directionality of natural processes. It can be stated in multiple equivalent forms.
Clausius Statement
Heat cannot spontaneously flow from a colder body to a hotter body without external work.
Kelvin-Planck Statement
No cyclic process can convert heat entirely into work; some heat must be rejected to a cold reservoir.
Clausius Inequality and Entropy
For any cyclic process, Clausius proved:
with equality holding for reversible processes. This allows us to define entropy S:
For an irreversible process between the same endpoints:
For an isolated system ($\delta Q = 0$):
This is the law of entropy increase: the entropy of an isolated system never decreases, and it increases for any irreversible process.
Derivation 4: The Third Law of Thermodynamics
The third law addresses the behavior of entropy at absolute zero and provides an absolute reference point for entropy calculations.
Nernst Heat Theorem (1906)
As temperature approaches absolute zero, the entropy change for any isothermal process approaches zero:
Planck's Statement
The entropy of a perfect crystalline substance approaches zero as temperature approaches absolute zero:
Consequences
1. Heat capacity vanishes at T = 0
Since $S(T) = \int_0^T \frac{C_P}{T'} dT'$ must be finite,$C_P \to 0$ as $T \to 0$.
2. Unattainability of absolute zero
No finite sequence of processes can bring a system to T = 0. Each cooling step produces diminishing returns as the entropy approaches its minimum.
3. Absolute entropy
The third law allows us to calculate absolute entropies by integrating heat capacity data from T = 0: $S(T) = \int_0^T \frac{C_P(T')}{T'} dT'$
Derivation 5: Carnot Efficiency
The Carnot cycle represents the most efficient heat engine operating between two thermal reservoirs. Its efficiency depends only on the reservoir temperatures.
The Carnot Cycle Steps
Efficiency Derivation
From the adiabatic relations, dividing step 2 by step 4:
Therefore $|Q_C| / Q_H = T_C / T_H$, and the efficiency is:
Key result: No heat engine operating between two reservoirs can exceed the Carnot efficiency. This is a direct consequence of the second law. For the total entropy change of a Carnot cycle: $\Delta S_{\text{total}} = Q_H/T_H - |Q_C|/T_C = 0$.
Applications
Power Generation
Coal, natural gas, and nuclear power plants are all governed by the Carnot limit. Modern combined-cycle gas turbines achieve 60%+ efficiency by cascading hot and cold cycles, though they remain bound by $\eta \leq 1 - T_C/T_H$.
Refrigeration
The Carnot cycle run in reverse defines the upper limit for refrigeration performance. The coefficient of performance $\text{COP} = T_C / (T_H - T_C)$sets the theoretical maximum for heat pumps and refrigerators.
Chemical Equilibrium
The second law determines whether a chemical reaction proceeds spontaneously via the Gibbs energy criterion $\Delta G < 0$. This connects thermodynamics directly to chemistry and materials science.
Cryogenics
The third law explains why reaching absolute zero requires infinite effort. Techniques like adiabatic demagnetization approach millikelvin temperatures but can never reach T = 0 exactly.
Simulation: Carnot Cycle PV Diagram
This simulation computes all four steps of the Carnot cycle for an ideal monatomic gas, generating the PV diagram and performing a complete energy balance. Modify the reservoir temperatures to explore how efficiency depends on the temperature ratio.
Carnot Cycle PV Diagram and Energy Balance
PythonClick Run to execute the Python code
Code will be executed with Python 3 on the server
Simulation: Entropy of Mixing
This simulation visualizes the entropy and Gibbs energy of mixing for an ideal binary solution. The entropy of mixing is always positive (mixing is entropically favorable), and the Gibbs energy of mixing becomes more negative at higher temperatures.
Entropy and Gibbs Energy of Mixing
PythonClick Run to execute the Python code
Code will be executed with Python 3 on the server
Chapter Summary
• Zeroth Law: Thermal equilibrium is transitive, defining temperature as a state variable.
• First Law: Energy is conserved: $dU = \delta Q - \delta W$. Internal energy is a state function.
• Second Law: Entropy never decreases in an isolated system: $dS \geq \delta Q / T$.
• Third Law: Entropy approaches zero for a perfect crystal as $T \to 0$.
• Carnot Efficiency: $\eta = 1 - T_C/T_H$ sets the upper bound for all heat engines.
Practice Problems
Problem 1:A Carnot engine operates between a hot reservoir at $T_H = 600\,\text{K}$ and a cold reservoir at $T_C = 300\,\text{K}$. If the engine absorbs $Q_H = 1000\,\text{J}$ per cycle, find the work output, heat rejected, and efficiency.
Solution:
1. Carnot efficiency: $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$ (50%).
2. Work output: $W = \eta Q_H = 0.50 \times 1000 = 500\,\text{J}$.
3. Heat rejected: $|Q_C| = Q_H - W = 1000 - 500 = 500\,\text{J}$.
4. Verify entropy balance: $\Delta S = \frac{Q_H}{T_H} - \frac{|Q_C|}{T_C} = \frac{1000}{600} - \frac{500}{300} = 1.667 - 1.667 = 0$. Reversible, as expected.
5. If the cold reservoir were at $T_C = 200\,\text{K}$: $\eta = 1 - 200/600 = 66.7\%$, $W = 667\,\text{J}$.
6. Key insight: efficiency depends only on the temperature ratio, not on the working substance or the amount of heat.
Problem 2:Calculate the total entropy change when 1 kg of water at $90°\text{C}$ is mixed with 1 kg of water at $10°\text{C}$ in an insulated container. ($c_p = 4186\,\text{J/(kg·K)}$)
Solution:
1. Final temperature: $T_f = \frac{T_1 + T_2}{2} = \frac{363.15 + 283.15}{2} = 323.15\,\text{K}$ (50 degrees C).
2. Entropy change of hot water: $\Delta S_H = mc_p \ln\frac{T_f}{T_H} = 1 \times 4186 \times \ln\frac{323.15}{363.15} = 4186 \times (-0.1171) = -490.1\,\text{J/K}$.
3. Entropy change of cold water: $\Delta S_C = mc_p \ln\frac{T_f}{T_C} = 4186 \times \ln\frac{323.15}{283.15} = 4186 \times 0.1328 = 555.9\,\text{J/K}$.
4. Total entropy change: $\Delta S_{\text{total}} = -490.1 + 555.9 = 65.8\,\text{J/K} > 0$.
5. The positive total entropy change confirms this is an irreversible process (heat flows spontaneously from hot to cold).
6. Note: $\Delta S > 0$ always for mixing at different temperatures. Equality holds only if $T_1 = T_2$ (no entropy production if already in equilibrium).
Problem 3:An ideal gas undergoes free expansion from volume $V_i$ to $V_f = 2V_i$ in an insulated rigid container. Find $\Delta T$, $\Delta U$, $W$, $Q$, and $\Delta S$.
Solution:
1. Free expansion into vacuum: no external pressure, so $W = \int P_{\text{ext}}\,dV = 0$.
2. Insulated container: $Q = 0$.
3. By the first law: $\Delta U = Q - W = 0$. For an ideal gas, $U$ depends only on $T$, so $\Delta T = 0$.
4. To find $\Delta S$, use a reversible path between the same states (isothermal expansion): $\Delta S = nR\ln\frac{V_f}{V_i} = nR\ln 2$.
5. For 1 mol: $\Delta S = 8.314 \times 0.693 = 5.76\,\text{J/K}$.
6. Even though $Q = 0$, $\Delta S > 0$. This is because $dS \geq \delta Q / T$ with strict inequality for irreversible processes. Free expansion is irreversible because the gas cannot spontaneously compress back.
Problem 4:Prove the Clausius inequality $\oint \frac{\delta Q}{T} \leq 0$ for any cyclic process using the second law.
Solution:
1. Consider a cyclic process where the system exchanges heat $\delta Q$ with a reservoir at temperature $T_R$ at each step.
2. Replace each heat exchange with a Carnot engine operating between $T_R$ and a universal reservoir at $T_0$. The Carnot engine absorbs $\delta Q_0 = T_0 \cdot \delta Q / T_R$ from the universal reservoir.
3. After one complete cycle, the system returns to its initial state. The total heat drawn from the universal reservoir is $Q_0 = T_0 \oint \frac{\delta Q}{T_R}$.
4. By the Kelvin-Planck statement, no cyclic process can convert heat entirely into work. So the net work $W \leq 0$ (can't extract net work from a single reservoir).
5. Since the combined system is cyclic: $W = Q_0$ (first law for the cycle). So $Q_0 \leq 0$.
6. Therefore $T_0 \oint \frac{\delta Q}{T_R} \leq 0$, and since $T_0 > 0$: $\oint \frac{\delta Q}{T} \leq 0$. Equality holds for reversible processes. $\square$
Problem 5:A heat pump maintains a building at $T_H = 293\,\text{K}$ (20 degrees C) when the outside temperature is $T_C = 263\,\text{K}$ (-10 degrees C). Find the maximum COP and the minimum electrical power needed for a heating rate of 10 kW.
Solution:
1. The coefficient of performance for a heat pump: $\text{COP}_{\text{HP}} = \frac{Q_H}{W} = \frac{T_H}{T_H - T_C}$.
2. Maximum COP: $\text{COP}_{\text{HP}} = \frac{293}{293 - 263} = \frac{293}{30} = 9.77$.
3. For a heating rate of $\dot{Q}_H = 10\,\text{kW}$: minimum power $\dot{W} = \dot{Q}_H / \text{COP} = 10/9.77 = 1.02\,\text{kW}$.
4. Heat extracted from outside: $\dot{Q}_C = \dot{Q}_H - \dot{W} = 10 - 1.02 = 8.98\,\text{kW}$.
5. Compared to direct electric heating (COP = 1 requiring 10 kW), the heat pump uses only 1/10 the power.
6. Note: if $T_C$ drops to $243\,\text{K}$ (-30 degrees C), $\text{COP} = 293/50 = 5.86$ and $\dot{W} = 1.71\,\text{kW}$. The COP degrades as the temperature difference increases, consistent with Carnot limits.