(a) One mole of an ideal gas expands isothermally and reversibly at `25^(@)C` from a volume of `10` litres to a volume of `20` litres.
(i) What is the change in entropy of the gas?
(ii) How much work is done by the gas?
(iii) What is `q` (surroundings) ?
(iv) What is the change in the entropy of the surroundings?
(v) What is the change in the entropy of the system plus the surroundings ?
(b) Also answer the questions opening a stopcock and allowing the gas to rush into an avacuated bulb of `10 L` volume.

Explanation

(i) `DeltaS=2.303nR log. (V_(2))/(V_(1))=2.303xx1xx8.314xxlog.(20)/(10)=5.76J//K`
(a) (ii) `W_(rev)=-2.303nRTlog.(V_(2))/(V_(1))`
`=2.303xx1xx8.314xx298xxlog.(20)/(10)=-1718J`
(iii) For isothermal process, `DeltalI=0` and heat is absorbed by the gas,
`q_(rev)=DeltalI-W=0-(-178)=1718J`
`:.q_(rev)=1718J` (`:.` process is reversible)
(iv) `DeltaS_(surr)=-(1718)/(298)=-5.76J//K`
As entropy of the system increases by `5.76J`, the entropy of the surrounding decreases by `5.76J`, since the process is carried out reversibly.
(v) `DeltaS_(sys)+DeltaS_(surr)=0`.......for reversible process.
(b) (i) `DeltaS=5.76J//K`, which is the same as above because S is a state function.
`(ii) W=0` (`:. p_(ext)=0`)
`(iii)` No heat is exchanged with the surroundings
`(iv)DeltaS_(surr)=0`
`(v)` The entropy of the system plus surrounding increases by `5.76 J//K` as we except entropy to increase in an irrevesible process.