Correct Answer - 7
`1^(st)` process is isobaric volume become double so, `T` is also become double
So, for adiabatic `T_(i)=2xx300=600`
`V_(i)=40`
`T_(f)=?, V_(f)=113`
for adiabatic
`P_(1)V_(1)^(gamma)="constant"=TV^(t-1)="constant"`
`T_(i)V_(i)^(5//3-1)=T_(f)V_(f)^(2//3)`
`600xx(40)^(2//3)=T_(f)xx(40)^(2//3)=T_(f)xx(110)^(2//3)`
`T_(f)=300 K`
`(300)/(100)=3`