91Ó°ÊÓ

a ) The net work done in the process,$W=15.0\text{J}$

b) The energy transferred during AB,$Q_{AB}=20.0\text{J}$

c) The energy transferred during BC,$Q_{BC}=0\text{J}$

Since the process is a complete cycle, i.e., beginning and ending in the same thermodynamic state, the change in the internal energy is zero, and the heat absorbed by the gas is equal to the work done by the gas,$Q=W$.

Formula:

The heat produced during the whole process is equal to the total work done in the process, $Q_{AB}+Q_{BC}+Q_{CA}=W$ …(¾±)

Where,$Q_{AB}$is the heat produced along path AB,

$Q_{BC}$is the heat produced along path AB,

$Q_{CA}$is the heat produced along path AB,

$W$is the work done by the system in the whole heat process

Since, $Q=W$,in terms of the contributions of the individual parts of the cycle, using equation (i), we get that the energy transferred during the process CA is given as:

$\begin{array}{c}{Q}_{CA}=W−Q_{AB}−Q_{BC}\\ =15\text{J}−20.0\text{J}−0\\ =−5.0\text{J}\end{array}$

Hence, the energy transferred during the process is$−5.0\text{J}$