adiabatic process, $\mathrm{Q}=0$ For isothermal process
${Q_{rejected}}\, = - W = nR{T_0}\ell n\left( {{V_0}/\frac{{{V_0}}}{2}} \right) = nR{T_0}\ell n\,2$
$=0.693 \mathrm{nRT}_{0}$
For isobaric process
$Q_{\text {rejected }}=-n C_{p} \Delta T=-n\left(\frac{f}{2}+1\right) n R\left(-\frac{T_{0}}{2}\right)$
$=\left(0.5+\frac{f}{4}\right) n R T_{0}>0.693 n R T_{0}$
It is clear that more heat is rejected in isobaric process.


