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(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 (i) to (v) if the expansion of the gas occurs irreversibly by simply opening a stopcock and allowing the gas to rush into an evacuated bulb of 10-L volume. |
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Answer» Solution :(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. |
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