Correct Answer - B
Under the given process, the final volume `V` is 32 times the initial volume `V_(0)`. ltbr. Thus, `W=int_(V_(0))^(32V_(0))(PdV)/(2)=int _(V_(0)) ^(32 V_(0))(RT)/(2V)dV`
`=int_(V_(0))^(32V_(0))(R)/(2V)TdV`
Now, `TV_((1)/(5))=T_(0)V_(0)^((1)/(5))` gives` T=(T_(0)V_(0)^((1)/(5)))/(V^((1)/(5)))`
Hence,
`W=int_(V_(0))^(32V_(0))(R)/(2V)(T_(0)V_(0)^((1)/(5)))/(V^(1//5))dV`
`=RT_(0)V_(0)^(1//5)int_(V_(0))^(32V_(0))(V^(-6//5))/(2)dV`
`=RT_(0)V_(0)^(1//5)[(-5V^(-1//5))/(2)]_(V_(0))^(32V_(0))`
`W=-(5RT_(0))/(2)`
Work done by the gas `=(5rt_(0))/( 2)`