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ChemistryEdu Logo 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#

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)#

Carnot cycle - Reversible isothermal expansion

  • 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:

\[W_{12} = -nRT_1ln{V_2 \over V_1}\]
  • Heat absorbed Q1 can be calculated from first law of thermodynamics:
\[ΔU = q + W_{12}\]
\[0 = Q_1 - nRT_1ln{V_2 \over V_1}\]
\[Q_1 = nRT_1ln{V_2 \over V_1}\]

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)#

Carnot cycle - Reversible adiabatic expansion

  • Let pressure and volume at state 3 be (P3, V3).

  • Change in internal energy(ΔU) is given by:

\[ΔU = nC_v(T_2 - T_1)\]

  • Since, it is an adiabatic process, no heat is absorbed or released.

  • Work done, W23 is calculated from first law of thermodynamics:

\[ΔU = q + W_{23}\]
\[nC_v(T_2 - T_1) = 0 + W_{23}\]
\[W_{23} = nC_v(T_2 - T_1)\]

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

Carnot cycle - Reversible isothermal compression

  • 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:

\[W_{34} = -nRT_2ln{V_4 \over V_3}\]
  • Heat absorbed Q2 can be calculated from first law of thermodynamics:
\[ΔU = q + W_{34}\]
\[0 = Q_2 - nRT_2ln{V_4 \over V_3}\]
\[Q_2 = nRT_2ln{V_4 \over V_3}\]

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)#

Carnot cycle - Reversible adiabatic compression

  • Change in internal energy(ΔU) is given by:
\[ΔU = nC_v(T_1 - T_2)\]

  • Since, it is an adiabatic process, no heat is absorbed or released.

  • Work done, W23 is calculated from first law of thermodynamics:

\[ΔU = q + W_{41}\]
\[nC_v(T_1 - T_2) = 0 + W_{41}\]
\[W_{41} = nC_v(T_1 - T_2) \]
\[W_{41} = -nC_v(T_2 - T_1)\]

Total Work done in Carnot cycle#

  • Total work done is given by:
\[W = W_{12} + W_{23} + W_{34} + W_{41}\]
\[W = -nRT_1ln{V_2 \over V_1} + nC_v(T_2 - T_1) - nRT_2ln{V_4 \over V_3} - nC_v(T_2 - T_1)\]
\[W = -nR(T_1ln{V_2 \over V_1} + T_2ln{V_4 \over V_3})\]

Efficiency of carnot cycle#

  • For adiabatic reversible expansion process 2->3, we can write:
\[T_1V_2^{γ-1} = T_2V_3^{γ-1}\]
  • For adiabatic reversible compression process 4->1, we can write:
\[T_1V_1^{γ-1} = T_2V_4^{γ-1}\]
  • Dividing above two equations, we get:
\[ {V_2 \over V_1} = {V_3 \over V_4}\]
  • Efficiency, η of carnot cycle in terms of temperatures T1 and T2 can be calculated as:
\[η = {|Work\ done| \over Heat\ absorbed}\]
\[η = {nR(T_1ln{V_2 \over V_1} + T_2ln{V_4 \over V_3}) \over nRT_1ln{V_2 \over V_1}}\]
\[η = {T_1ln{V_2 \over V_1} - T_2ln{V_3 \over V_4} \over T_1ln{V_2 \over V_1}}\]
\[Since,\ {V_2 \over V_1} = {V_3 \over V_4},\ we\ can\ write:\]
\[η = {T_1 - T_2 \over T_1}\]

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:
\[η = {|Work\ done| \over Heat\ absorbed}\]
\[η = {nR(T_1ln{V_2 \over V_1} + T_2ln{V_4 \over V_3}) \over nRT_1ln{V_2 \over V_1}}\]
\[η = {Q_1 - Q_2 \over Q_1}\]
  • Let us estimate the value of efficiency, η:
\[η = {T_1 - T_2 \over T_1}\]
\[η = 1 - {T_2 \over T_1}\]

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.