Thermodynamics | Carnot Cycle and Second Law of Thermodynamics#

Carnot Cycle#

Carnot cycle is based on first law of thermodynamics. It tells about conversion of heat energy into work.
Carnot cycle also helps us in defining second law of thermodynamics.
A Carnot cycle consists of following components: (a) sink, (b) source, (c) insulated stand, (d) gas cylinder (system).
Sink: A sink can absorb any amount of heat released by the system.
Source: A source has infinite thermal conductivity and any amount of heat can be taken from it.
Insulated stand: An insulated stand is a diathermic substance, ie, no heat can be transferred to or taken from it.
A carnot cycle consists of four steps: (a) isothermal reversible expansion (b) adiabatic reversible expansion (c) isothermal reversible compression (d) adiabatic reversible compression.
Carnot cycle is a cyclic process.

Graphical Representation of carnot cycle#

Image credit: https://en.wikipedia.org/wiki/Carnot_cycle

In the above graph, process 1→2 is reversible isothermal expansion with constant temperature T₁. Process 2→3 is reversible adiabatic expansion. Process 3→4 is reversible isothermal expansion with constant temperature T₂. Process 4→1 is reversible adiabatic compression. Let's assume Q₁ heat is absorbed from source during process 1→2 and Q₂ heat is released to sink during process 3→4.

Step 1. Reversible Isothermal Expansion (Process 1→2)#

Let pressure and volume at state 1 be (P₁, V₁) and at state 2 be (P₂, V₂).
Change in internal energy(ΔU) is zero because it is isothermal process.
Work done, W₁₂ is given by:

W_{12} = - n R T_{1} l n \frac{V_{2}}{V_{1}}

Heat absorbed Q₁ can be calculated from first law of thermodynamics:

Δ U = q + W_{12}

0 = Q_{1} - n R T_{1} l n \frac{V_{2}}{V_{1}}

Q_{1} = n R T_{1} l n \frac{V_{2}}{V_{1}}

Q₁ will be positive because V₂ > V₁, which also indicates that heat is absorbed during this process.

Step 2. Reversible Adiabatic Expansion (Process 2→3)#

Let pressure and volume at state 3 be (P₃, V₃).
Change in internal energy(ΔU) is given by:

Δ U = n C_{v} (T_{2} - T_{1})

Since, it is an adiabatic process, no heat is absorbed or released.
Work done, W₂₃ is calculated from first law of thermodynamics:

Δ U = q + W_{23}

n C_{v} (T_{2} - T_{1}) = 0 + W_{23}

W_{23} = n C_{v} (T_{2} - T_{1})

Step 3. Reversible Isothermal Compression (Process 3→4)#

Let pressure and volume at state 4 be (P₄, V₄).
Change in internal energy(ΔU) is zero because it is isothermal process.
Work done, W₃₄ is given by:

W_{34} = - n R T_{2} l n \frac{V_{4}}{V_{3}}

Heat absorbed Q₂ can be calculated from first law of thermodynamics:

Δ U = q + W_{34}

0 = Q_{2} - n R T_{2} l n \frac{V_{4}}{V_{3}}

Q_{2} = n R T_{2} l n \frac{V_{4}}{V_{3}}

Q₂ will be negative because V₄ < V₃, which also indicates that heat is released during this process.

Step 4. Reversible Adiabatic Compression (Process 4→1)#

Change in internal energy(ΔU) is given by:

Δ U = n C_{v} (T_{1} - T_{2})

Since, it is an adiabatic process, no heat is absorbed or released.
Work done, W₂₃ is calculated from first law of thermodynamics:

Δ U = q + W_{41}

n C_{v} (T_{1} - T_{2}) = 0 + W_{41}

W_{41} = n C_{v} (T_{1} - T_{2})

W_{41} = - n C_{v} (T_{2} - T_{1})

Total Work done in Carnot cycle#

Total work done is given by:

W = W_{12} + W_{23} + W_{34} + W_{41}

W = - n R T_{1} l n \frac{V_{2}}{V_{1}} + n C_{v} (T_{2} - T_{1}) - n R T_{2} l n \frac{V_{4}}{V_{3}} - n C_{v} (T_{2} - T_{1})

W = - n R (T_{1} l n \frac{V_{2}}{V_{1}} + T_{2} l n \frac{V_{4}}{V_{3}})

Efficiency of carnot cycle#

For adiabatic reversible expansion process 2->3, we can write:

T_{1} V_{2}^{γ - 1} = T_{2} V_{3}^{γ - 1}

For adiabatic reversible compression process 4->1, we can write:

T_{1} V_{1}^{γ - 1} = T_{2} V_{4}^{γ - 1}

Dividing above two equations, we get:

\frac{V_{2}}{V_{1}} = \frac{V_{3}}{V_{4}}

Efficiency, η of carnot cycle in terms of temperatures T₁ and T₂ can be calculated as:

η = \frac{| W o r k d o n e |}{H e a t a b s o r b e d}

η = \frac{n R (T_{1} l n \frac{V_{2}}{V_{1}} + T_{2} l n \frac{V_{4}}{V_{3}})}{n R T_{1} l n \frac{V_{2}}{V_{1}}}

η = \frac{T_{1} l n \frac{V_{2}}{V_{1}} - T_{2} l n \frac{V_{3}}{V_{4}}}{T_{1} l n \frac{V_{2}}{V_{1}}}

S i n c e, \frac{V_{2}}{V_{1}} = \frac{V_{3}}{V_{4}}, w e c a n w r i t e :

η = \frac{T_{1} - T_{2}}{T_{1}}

Here, T₁ = Temperature of source and T₂ = Temperature of sink

Efficiency, η of carnot cycle in terms of Q₁ and Q₂ can be calculated as:

η = \frac{| W o r k d o n e |}{H e a t a b s o r b e d}

η = \frac{n R (T_{1} l n \frac{V_{2}}{V_{1}} + T_{2} l n \frac{V_{4}}{V_{3}})}{n R T_{1} l n \frac{V_{2}}{V_{1}}}

η = \frac{Q_{1} - Q_{2}}{Q_{1}}

Let us estimate the value of efficiency, η:

η = \frac{T_{1} - T_{2}}{T_{1}}

η = 1 - \frac{T_{2}}{T_{1}}

Here, T₂ = Temperature of sink and T₁ = Temperature of source.

Also, temperature of source, T₁ is always greater than temperature of sink, T₂.

So, value of η lies between 0 and 1 but it is never 1.

0 < η < 1

It means only a part of heat absorbed is converted into work. This will help us in defining second law of thermodynamics.

Entropy change in Carnot Cycle#

Since entropy is a state function and carnot cycle is a cyclic process, so entropy change in carnot cycle will be zero.

Second Law of Thermodynamics#

Claussius Statement

It is impossible to convey heat from a cooler body to a hotter body without the help of any external agent.

Kelvin-Planck Statement

It is impossible to build an engine which can convert heat completely into work in a complete cycle.