10 gm of helium at 127^(@) C is expanded isothermally from 100 atm to 1 atm.Calculate the work done when the expansion is carried out (i) in single step, (ii) In three steps, the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.

10 gm of helium at 127^(@) C is expanded isothermally from 100 atm to

1.	10 gm of helium at 127^(@) C is expanded isothermally from 100 atm to 1 atm.Calculate the work done when the expansion is carried out (i) in single step, (ii) In three steps, the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.
Answer» Solution :(i) Work DONE = `V.DeltaP` `V = 10/4 xx (8.314 xx 400)/(100 xx 10^(5)) = 83.14 xx 10^(-5) m^(3)` So, `W = (83.14)/10^(5) (100-1) xx 10^(5) = 8230.86` J (ii) Ini three steps, `V_(i) = 83.14 xx 10^(5) m^(3)` `w_(i)= (83.14 xx 10^(-5)) xx (100-60) xx 10^(5)` `=3325.6` J `V_(3) = (2.5 xx 8.314 xx 400)/(60 xx 10^(5)) = 138.56 xx 10^(-5) m^(3)` `w_(a) =V.DeltaP` `w_(1) = 138.56 xx 10^(-5)(60-30) xx 10^(5)` `=4156.99 = 4157` J `V_(\|\|\|) = (2.5 xx 8.314 xx 400)/(30 xx 10^(5)) = 277. 1.3 xx 10^(-5) m^(3)` `w_(\|\|\|) = 8036.88` J `W_("total") =w_(1) + w_(2) + w_(3)` `=3325.6 + 4156.909 + 8036.86 = 15519.45` J (iii) For reversible process. `w = 2.303 nRT LOG P_(1)/P_(2)` `=2.303 xx (10/4) xx 8.314 xx 400 xx log(100/1)` `w = 38294.28` JULES.