Show that in a cyclic process, the area of the loop on the indicator diagram is equal to net work done by or on the gas which in turn is equal to net heat absorbed or rejected by the gas.

Solution

When a system undergoes a cyclic process, it is finally brought back to initial state in all respects.
Suppose, an ideal gas is contained in a cylinder fitted with a massless and frictionless piston.
Let the gas be initially in the state represented by point A on the P–V graph. Let it be taken from state A to C along path ABC and brought back to A via path CDA.
Case 1: Work done by gas is equal to area ABCKLA.
Case 2: Work done on the gas is equal to area CDALKA.

Now net work done by gas,
$space dW equals Area left parenthesis ABCKLA right parenthesis space minus space Area left parenthesis CDALKA right parenthesis space space space space space space space space equals space Area space of space closed space loop left parenthesis ABCDA right parenthesis$
According to first law of thermodynamics,
$space increment straight Q space equals space increment straight U plus increment straight W$
In cyclic process the change in internal energy is zero.
So, $increment straight Q space equals space increment straight W$
Net heat absorbed during a cyclic process is equal to work done by gas.
Therefore, the area of the loop on the indicator diagram is equal to both work done and the heat absorbed in the cyclic process.