Helium weighing `16 g` is expanded from `1atm` to one-tenth of its original pressure at `30^(@)C`. Calculate the change in entropy assuming it to be an ideal gas.

Explanation

Since the process is an isothermal expansion of an idela gas.
`DeltaS =- nRT "In" (P_(2))/(P_(1)) or =nRT "In"(P_(1))/(P_(2))`
`n= (16g)/(4g mol^(-1)) = 4 mol, R = 8.314 J K^(-1) mol^(-1), p_(1) = 1 atm`
`p_(2) = (1)/(10)atm`
`DeltaS = (4mol) xx (8.314 J K^(-1) mol^(-1)) xx 2.303 "log" (1)/(1//10)`
`= 76.59 J K^(-1)`