One mole of an ideal mono-atomic gas is taken round cyclic process `ABCD` as shown in figure below. Calculate

Explanation

Path `AB` is isochoric `(w_(1)=0)`, path `BC` is isothermal `(w_(2)=-ve)`, path `CA` is isobaric `(w_(3)=+ve)`
Total work done by gas `(w)`
`=w_(1)+w_(2)+w_(3)`
`=0+2.303nRT"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))`
`=0+2.303P_(B)V_(B)"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))`
`=2.303xx3P_(0)V_(0)"log"(V_(0))/(2V_(0))+P_(0)(2V_(0)-V_(0))`
`=-2P_(0)V_(0)+P_(0)V_(0)=-P_(0)V_(0)`
Also, `w_(2)=-2P_(0)V_(0)` and `w_(3)=P_(0)V_(0)`
also, For the path `AB`, ie. isochoric
`q_(1)=nxxC_(P)xx(T_(B)-T_(A))=1xx3/2R[(P_(B)V_(B)-P_(A)V_(A))/R]`
`=3/2[3P_(0)V_(0)-P_(0)V_(0)]=+3P_(0)V_(0)`
For the path `CA`, i.e., isochoric:
`q_(3)=nxxC_(P)xx(T_(A)-T_(B))`
`=1xx5/2R[(P_(A)V_(A)-P_(B)V_(B))/R]`
`=5/2[P_(0)V_(0)-2P_(0)V_(0)]`
`q_(3)=-5/2 P_(0)V_(0)`
Also net heat absorbed `=3P_(0)V_(0)-5/2P_(0)V_(0)-=(P_(0)V_(0))/2`
`:. q_("net")=(P_(0)V_(0))/2`
Also, `(P_(0)V_(0))/(T_(1))=(3P_(0)V_(0))/(T_(2)):. T_(2)=3T_(1)=(3P_(0)V_(0))/R`