(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

(a) (i) `DeltaS= 2.303 nR "log"(V_(2))/(V_(1))=2.303xx1xx8.314xx"log"(20)/(10)=5.76J//K`.
(ii) `W_(rev)= -2.303 nRT "log"(V_(2))/(V_(1))`
`= -2.303 xx1xx8.314xx298xx"log"(20)/(10)= -1781 J`.
(iii) For isothermal process, `DeltaU=0` and heat is absorbed by the gas,
`q_(rev)=DeltaU-W=0-(-1718)=1718J`
`:. q_(rev)=1718J`. ( `:.` Process is reversible)
(iv) `DeltaS_(surr)=(1718)/(298)= - 5.76 J//K`
As entropy of the system increased by `5.76 J`, the entropy of the surroundings decreases by `5.76 J`. Since the process is carried out reversibly.
(v) `DeltaS_(sys)+DeltaS_(srr)=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 ( because p_(ext)=0)`
(iii) No heat is exchanged withe the surroundings.
(iv) `Delta S_(surr)=0`
(v) The entropy of the system plus surroundings increases by `5.76 J//K`, as we expect entropy to increase in an irreversible process.