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Physics · Ch 11 — Thermodynamics

Cyclic Process

11.8.6

Cyclic Process

The Meaning of a Cyclic Process

A thermodynamic process is called cyclic when the system, after undergoing a series of changes, returns exactly to its initial state. This means every state variable — pressure, volume, temperature, internal energy — comes back to its original value.

Because the system returns to its starting point, the net change in any state function over one complete cycle is zero. The most important consequence is for internal energy UU, a state function:

ΔUcycle=0\Delta U_{\text{cycle}} = 0

This single fact drives all the work and heat relations for a cyclic process.

Important

For any cyclic process, ΔU=0\Delta U = 0. This is not an approximation — it follows directly from the definition of a state function.

Work Done in a Cycle

On a PP–VV diagram, a cyclic process is represented by a closed curve. The system may expand (doing work on the surroundings) and then contract (work being done on the system). The net work done by the system in one complete cycle equals the area enclosed by the closed curve on the PP–VV diagram.

If the cycle is traversed clockwise, the net work done by the system is positive. If traversed anticlockwise, the net work is negative (work is done on the system).

The First Law for a Cyclic Process

Apply the first law of thermodynamics to one complete cycle:

ΔU=Q−W\Delta U = Q - W

Since ΔU=0\Delta U = 0, we get:

0=Q−W⇒Q=W0 = Q - W \quad \Rightarrow \quad Q = W

Here QQ is the net heat supplied to the system over the cycle, and WW is the net work done by the system over the cycle.

Qnet=WnetQ_{\text{net}} = W_{\text{net}}

This is a powerful result: in a cyclic process, the net heat absorbed by the system is entirely converted into net work output. No net change in internal energy occurs.

Sign Convention in a Cycle

Heat may be added to the system during some parts of the cycle and rejected during others. Similarly, work may be done by the system during expansion and on the system during compression. The equation Q=WQ = W refers to the algebraic sum of all heat transfers and all work transfers over the complete cycle.

Let Q1Q_1 be the heat absorbed, Q2Q_2 the heat rejected (so Q2Q_2 is negative). Then:

Q=Q1+Q2Q = Q_1 + Q_2

And the net work done by the system is:

W=Q1+Q2W = Q_1 + Q_2

Since Q2Q_2 is negative, WW is less than Q1Q_1. This is the basis of heat engine efficiency.

Efficiency of a Cyclic Process (Heat Engine)

For a cyclic process that operates as a heat engine:

  • Heat Q1Q_1 is absorbed from a hot reservoir.
  • Heat Q2Q_2 is rejected to a cold reservoir (Q2Q_2 is negative in the sign convention where QQ is heat added to the system).
  • Net work done by the system: W=Q1−∣Q2∣W = Q_1 - |Q_2|.

The thermal efficiency η\eta is defined as:

η=Net work outputHeat input=WQ1\eta = \frac{\text{Net work output}}{\text{Heat input}} = \frac{W}{Q_1}

Substituting W=Q1−∣Q2∣W = Q_1 - |Q_2|:

η=1−∣Q2∣Q1\eta = 1 - \frac{|Q_2|}{Q_1}

η=1−∣Q2∣Q1\eta = 1 - \frac{|Q_2|}{Q_1}

Efficiency is always less than 1 because some heat must always be rejected (a consequence of the second law of thermodynamics).

Example: A Simple Cyclic Process

Consider a gas taken through the following cycle:

  1. Isothermal expansion at temperature T1T_1 from volume V1V_1 to V2V_2, absorbing heat Q1Q_1.
  2. Adiabatic expansion from V2V_2 to V3V_3, temperature falling to T2T_2.
  3. Isothermal compression at T2T_2 from V3V_3 to V4V_4, rejecting heat Q2Q_2.
  4. Adiabatic compression from V4V_4 back to V1V_1, temperature rising to T1T_1.

This is the Carnot cycle. For an ideal gas, one can show:

Q1Q2=T1T2\frac{Q_1}{Q_2} = \frac{T_1}{T_2}

Hence the efficiency becomes:

η=1−T2T1\eta = 1 - \frac{T_2}{T_1} …