05_2nd-law

1. The Carnot Cycle (Sadi Carnot, 1824)

The Carnot cycle is an idealized thermodynamic cycle consisting of four reversible processes: two isothermal and two adiabatic.

Deriving Carnot Efficiency (\(\varepsilon_{\text{Carnot}}\)):

For the isothermal expansion (\(A \rightarrow B\)), the change in internal energy \(\Delta U = 0\), so the heat absorbed equals the work done:

\[ Q_H = W = nRT_H \ln\left(\frac{V_B}{V_A}\right) \]

Similarly, for the isothermal compression (\(C \rightarrow D\)), the heat released is:

\[ Q_L = nRT_L \ln\left(\frac{V_C}{V_D}\right) \]

The efficiency of the cycle is defined as:

\[ \varepsilon_{\text{Carnot}} = 1 - \frac{Q_L}{Q_H} = 1 - \frac{T_L \ln(V_C/V_D)}{T_H \ln(V_B/V_A)} \]

To simplify, we use the adiabatic relations for processes \(D \rightarrow A\) and \(B \rightarrow C\):

\[ T_L V_D^{\gamma-1} = T_H V_A^{\gamma-1} \Rightarrow \frac{T_L}{T_H} = \left(\frac{V_A}{V_D}\right)^{\gamma-1} \]

\[ T_H V_B^{\gamma-1} = T_L V_C^{\gamma-1} \Rightarrow \frac{T_L}{T_H} = \left(\frac{V_B}{V_C}\right)^{\gamma-1} \]

Equating these gives \(\frac{V_A}{V_D} = \frac{V_B}{V_C}\), which rearranges to \(\frac{V_C}{V_D} = \frac{V_B}{V_A}\). This cancels out the logarithmic terms in the efficiency equation.

Final Carnot Efficiency:

\[ \varepsilon_{\text{Carnot}} = 1 - \frac{T_L}{T_H} \]

Note: Efficiency depends only on the ratio of the reservoir temperatures.

• Example 1: If \(T_L =\) 303 K (30°C) and \(T_H =\) 503 K (230°C), then \(\varepsilon_{\text{Carnot}} \approx\) 40%.

• Example 2: If \(T_L / T_H \approx\) 0, then \(\varepsilon_{\text{Carnot}} \approx\) 1 (100% efficiency, which is practically unattainable).

2. Reverse Carnot Cycle (Refrigerator)

Operating the Carnot cycle in reverse creates an ideal refrigerator.

• Work (\(W\)) is done TO the system.

• The system absorbs heat (\(Q_L\)) FROM the low-\(T\) reservoir and releases heat (\(Q_H\)) TO the high-\(T\) reservoir.

Coefficient of Performance (COP / \(\varepsilon^{(r)}\)):

\[ \varepsilon^{(r)} \equiv \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L} = \frac{1}{1 - Q_L/Q_H} - 1 = \frac{1}{\varepsilon} - 1 \]

For a reverse Carnot cycle specifically:

\[ \varepsilon^{(r)}_{\text{Carnot}} = \frac{1}{1 - T_L/T_H} - 1 = \frac{T_L}{T_H - T_L} \]

Reversible vs. Irreversible Processes:

• Reversible process: A process where the direction of both the system and its surroundings can be exactly reversed (e.g., a frictionless quasi-static process).

• Irreversible process: All natural processes are irreversible (e.g., quasi-static processes with friction, adiabatic free expansion, a cooling cup of hot water).

3. The Second Law of Thermodynamics

The Second Law can be expressed through two functionally equivalent statements:

1. Clausius Statement (No perfect refrigerator):

Heat can never pass from a colder to a warmer body without some other change, connected with it, occurring at the same time.

2. Kelvin-Planck Statement (No perfect heat engine):

It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work.

Equivalence of the Statements:

• If Kelvin-Planck is violated: A "perfect" engine converts heat \(Q_0\) entirely into work (\(W = Q_0\)). This work can be used to drive a regular refrigerator that moves heat \(Q\) from \(T_L\) and dumps \(Q_0 + Q\) into \(T_H\). The net effect is heat \(Q\) moving from \(T_L\) to \(T_H\) with no external work input—violating the Clausius statement.

• If Clausius is violated: A "perfect" refrigerator moves heat \(Q_0\) from \(T_L\) to \(T_H\) without external work. A regular heat engine can then take \(Q_0 + Q\) from \(T_H\), do work \(W=Q\), and dump \(Q_0\) back to \(T_L\). The net effect is converting heat \(Q\) from \(T_H\) entirely into work \(W\) without any other effect—violating the Kelvin-Planck statement.

4. Carnot's Theorem

Carnot's theorem dictates the upper limits of thermodynamic efficiency based on two core postulates:

1. For heat engines working between two constant-\(T\) reservoirs, the efficiency of an irreversible engine (\(\varepsilon_I\)) is always lower than that of a reversible engine (\(\varepsilon_R\)).

2. For all reversible engines working between the same two constant-\(T\) reservoirs, the efficiencies are identical.

Proof of Postulate 1 (By Contradiction):

Assume an irreversible engine (\(I\)) is more efficient than a reversible engine (\(R\)), meaning \(\varepsilon_I > \varepsilon_R\).

• Given \(\varepsilon_I = \frac{W'}{Q_1}\) and \(\varepsilon_R = \frac{W}{Q_1}\), this assumption implies \(W' > W\).

• Design a combined process: Run engine \(I\) forward and run engine \(R\) in reverse (as a refrigerator) using a portion of the work from \(I\).

• Engine \(I\) absorbs \(Q_1\) from \(T_H\) and produces work \(W'\). Engine \(R\) uses work \(W\) to return \(Q_1\) to \(T_H\).

• Net Effect: The high-temp reservoir (\(T_H\)) experiences zero net heat change (\(Q_1 - Q_1 = 0\)). The combined system absorbs net heat (\(Q_2 - Q_2'\)) from the low-temp reservoir (\(T_L\)) and converts it entirely into net work (\(W' - W\)).

• This directly violates the Kelvin-Planck statement of the Second Law. Therefore, the assumption is false, proving that \(\varepsilon_I \leq \varepsilon_R\).

(Postulate 2 is proven using identical logic by pitting two reversible engines against each other, yielding \(\varepsilon_{R1} \leq \varepsilon_{R2}\) and \(\varepsilon_{R2} \leq \varepsilon_{R1}\), dictating they must be equal).

Corollary of Carnot's Theorem:

For heat engines working between two constant-temperature reservoirs, the efficiency of any irreversible engine is strictly bounded by the Carnot efficiency:

\[ \varepsilon_I \leq \varepsilon_{\text{Carnot}} \]