Work Done in Cyclic Process

In thermodynamics, a cyclic process refers to a sequence of thermodynamic changes that return a system to its initial state. During such processes, the system undergoes a series of expansions and compressions, absorbing and releasing heat and performing work. In a practical sense, if a system, like a heat engine, completes a cycle, the work done during each part of the cycle accumulates. For example, in an idealized engine cycle like the Carnot cycle, the system performs work during the expansion phase (when it absorbs heat), and the work is recovered during the compression phase (when it releases heat). Understanding cyclic processes helps optimise these systems for better performance, efficiency, and sustainability.

1.0Definition of Cyclic Process

• A process in which a system returns to its original state after undergoing a series of changes is known as a cyclic process.
• A cyclic process in thermodynamics is a sequence of thermodynamic changes that returns a system to its original state. This means that after completing the cycle, the system's properties—such as pressure, volume, and temperature—are precisely as they were at the beginning. These processes are fundamental to various practical applications, including engines and refrigerators.

2.0Work Done During a Cyclic Process

Let us assume that expansion of a gas starts from the initial point A to the final point B after undergoing a series of pressure and volume changes along the path AXB. Work done by the gas during the expansion is 

$W_1=+\text{ Area }AXBCDA$

• In the following sequence, the gas is subjected to a different series of pressure changes and volume to return to its initial state A via the path BYA.
• Work done by the gas during the compression is 

$W_2=-\text{ Area }BYADCB$

Work is done on the gas during compression, which is taken negative.

• Net work done during the cyclic process is

$\mathrm{W}=W_1+W_2$

W= Area AXBCDA - Area BYADCB

W= +Area AXBYA

3.0Conclusion for a Cyclic Process

1. The work done per cycle numerically equals the area enclosed by the loop representing the cycle on a pressure-volume (P-V) diagram.
2. If the closed loop is traced clockwise, the expansion curve lies above the compression curve $\left(W_1>W_2\right)$. The positive area of the loop implies that the system does work.
3. If the closed loop is traced anticlockwise, the expansion curve lies below the compression curve $\left(W_1<W_2\right)$. The area of the loop is negative, implying that net work is done on the system.
4. Net change in internal energy in cyclic process is zero, so according to first law of thermodynamics 

$\Delta Q=\text{ Work done }=\text{ Area of loop }$

4.0Examples of Cyclic Process

Carnot Cycle:

• Description: An idealized thermodynamic cycle consisting of two isothermal and two adiabatic processes defines the maximum possible efficiency for a heat engine.
• Components: Isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.
• Applications: Used as a benchmark to compare the efficiency of real-world heat engines.

5.0Graph in Cyclic Process

In P-V Graph

$\Delta W$= Area enclosed between the loop

$\Rightarrow$ Clockwise direction $\to \Delta W=\text{ Positive }$

$\Rightarrow$ Anticlockwise direction $\to \Delta W=\text{ Negative }$

In V-P graph

$\Delta W$ =Area enclosed between the loop

$\Rightarrow$ Clockwise direction $\to \Delta W=\text{ Negative }$

$\Rightarrow$ Anticlockwise direction $\to \Delta W=\text{ Positive }$

6.0Efficiency of cyclic process

Efficiency $(\eta )$ is defined as 

$\eta =\frac{\text{ net work done }}{\text{ Heat supplied to the system }}$

$\%\eta =\frac{\text{ net work done }}{\text{ Heat supplied to the system }}\times 100$

Note that here we are taking net work done(both positive and negative) but only +ve heat is considered.

Example: Given plot is for a monatomic gas undergoing cyclic process. We have to find its efficiency

Net work done = area of loop

= 2P × 3V = 6PV(+ve as clockwise cycle)

Calculation of heat

AB : Isobaric expansion

$\Delta Q=\frac{5}{2}nR\Delta T=\frac{5}{2}V\Delta P=\left(\frac{5}{2}\right)\times 3P\times 3V=\frac{45}{2}PV(+ve)$

BC : Isochoric cooling

$\Delta Q=\frac{3}{2}nR\Delta T=\frac{3}{2}V\Delta P=\left(\frac{3}{2}\right)\times -2P\times 4V=-12PV(-ve)$

CD :Isobaric compression

$\Delta Q=\frac{5}{2}nR\Delta T=\frac{5}{2}V\Delta P=\left(\frac{5}{2}\right)\times P\times -3V=-\frac{15}{2}PV(-ve)$

DA :isochoric heating

$\Delta Q=\frac{3}{2}nR\Delta T=\frac{3}{2}V\Delta P=\left(\frac{3}{2}\right)\times 2P\times V=3PV(+ve)$

Only positive heat is to be considered for efficiency calculation,

$\Delta Q=45\frac{PV}{2}+3PV=51\frac{PV}{2}$

$\Rightarrow \%\eta =\frac{\text{ net work done }}{\text{ Heat supplied to the system }}\times 100$

$=\frac{6PV}{51\frac{PV}{2}}\times 100=\frac{1200}{51}=23.53\%$

7.0Solved Questions on Work Done In Cyclic Process

Q-1. Calculate

1. Area 
2. Work done  
3. change in internal energy  
4. Heat

Solution:

1. $\text{ Area }=2P_1\times 3V_1=6P_1{V}_1$
2. $\text{ Work }=-6P_1{V}_1\mathrm{J}$
3. In cyclic process change in internal energy U=O
4. Heat, by using The First Law Of Thermodynamics

$\begin{array}{rl}& \Delta Q=\Delta U+\Delta W\\ & (\therefore \Delta U=0)\\ & \Delta Q=\Delta W\end{array}$

Q-2. An ideal gas is taken through the cycle ABCD as shown in fig. If the heat added to the gas in the cycle is 5J,what does the gas do the total work  in process $\left(W_{CA}\right)$

Solution:

For the cyclic process

$\begin{array}{rl}& \Delta Q=\Delta W\\ & \mathrm{Q}=W_{AB}+W_{BC}+W_{CA}\\ & 5=10(2-1)+0+W_{CA}\end{array}$

($\therefore W_{BC}=0$, work done in isochoric process is zero)

$W_{CA}=-5\mathrm{J}$

Q-3. The diagram shows a thermodynamic system undergoing a cyclic process ABCDA. Find the system's work in the cycle.

Solution:

W = area BCO + area ADO

$\mathrm{W}=-W_0+W_0=0$

Net work done by the system in the cycle is zero.

Q-4.A monoatomic ideal gas with two moles undergoes a cyclic process starting from state A, as depicted in the diagram. The volume ratios are $\frac{V_B}{V_A}=2\text{ and }\frac{V_D}{V_A}=4$ If the temperature T_A  at A is 27⁰C.Calculate the temperature of the gas at point B, the heat exchanged (absorbed or released) during each stage of the process, and the total work performed by the gas over the entire cycle. Provide the total work done in terms of the gas constant R.

Solution:

The process AB is 

Solution:

• The process AB is isobaric

$\begin{array}{rl}& \left(\therefore \frac{V}{T}=\text{ Constant }\right)\\ & \frac{V_A}{T_A}=\frac{V_B}{T_B}\\ & T_B=\frac{V_B}{V_A}\times T_A\\ & T_B=2\times 300=600\mathrm{K}=27^0\mathrm{C}\end{array}$

• Heat absorbed in process $A\to B$

$Q_1=n{C}_{p}dT=2\times \frac{5}{2}R\times (600-300)=1500R$

Heat absorbed in the process B C isothermal process)

$Q_2=W_2=nR{T}_B\log \left(\frac{V_D}{V_B}\right)=2\times R\times 600{\log }_{e}2=1200R\times 0.693$

$Q_2=831.6R$

Heat absorbed in process CD (isochoric process)

$Q_3=n{C}_{V}dT=2\times \frac{3}{2}R\times (300-600)=-900R$

Heat absorbed in process DA (isothermal process)

$Q_4=nR{T}_A{\log }_e\frac{V_A}{V_D}=2\times R\times 300{\log }_e\frac{1}{4}=-1200R{\log }_{e}2=-831.6R$

• The total work done by the gas during the complete cycle

$\mathrm{W}=Q_1+Q_2+Q_3+Q_4$

$W=1500R+831.6R-900R-831.6R=600R$