$Q =\Delta U + W$
For isothermal process $\Delta U =0$
$\therefore$ Internal energy remains constant
$Q = W$
Positive work done by external agent is completely lost in the form of heat.
As volume gets reduced, pressure increases
$( PV =$ constant $)$

$I.$ Area $ABCD =$ Work done on the gas
$II.$ Area $ABCD =$ Net heat absorbed
$III.$ Change in the internal energy in cycle $= 0$
Which of these are correct

