Four grams of helium is expanded from 1 atm to one-tenth of its origiinal pressure at `30^(@)C`. Change in entropy (assuming ideal gas behaviour) is :

Options

A. `38.3 JK^(-1)`
B. `76.6 JK^(-1)`
C. `19.15 JK^(-1)`
D. `100 JK^(-1)`

Answer

C

Explanation

For isothermal expansion of gas:
`Delta S=nR "in" (P_(1))/(P_(2))`
`n= (4.0)/(4) = 1.0` mol, R= 8.314 `p_(1) =1` atm
`p_(2) = 1//10` atm
`Delta S= (1.0) xx(8.314) "log" (1)/(1//10)`
`=19.15JK^(-1)`