A monoatomic ideal gas undergoes a process ABC.The heat given to the gas is

If T is temperature of gas

at A, then ${\mathrm{T}}_{\mathrm{B}} = 3\mathrm{T} = {\mathrm{T}}_{\mathrm{C}}$
In process AB, $∆\mathrm{W} = \frac{5\mathrm{PV}}{2}$

and $∆\mathrm{U} = \frac{1}{2}\mathrm{nR}\left(2\mathrm{T}\right) = 3\times 2 \mathrm{PV}$
So, $∆\mathrm{Q} = ∆\mathrm{W} + ∆\mathrm{U} = 8.5 \mathrm{PV}$
ln process BC, $∆\mathrm{U} = 0 \mathrm{as} ∆\mathrm{T}= 0$
$∆\mathrm{W} = 4\mathrm{V}.2\mathrm{P} = 8 \mathrm{PV} \mathrm{Hence} ∆{\mathrm{Q}}_2 = 8 \mathrm{PV}$

So $∆{\mathrm{Q}}_{\mathrm{T}} = 16.5 \mathrm{PV}$