Correct Answer - A
Entropy change for `n` moles of isothenal expansion of an ideal gas from volume `V_(1)` to volume `V_(2)` is
`DeltaS=2.303nR"log"(V_(2))/(V_(1))` ltbr. Hence, `n=2` , `V_(2)=100dm^(3) , V_(1)=10dm^(3)`
`2.303xx2xx8.3143"log"(100)/(10)`
`=38.296Jmol^(-1)K^(-1)`