Two moles of a perfect gas undergoes the following process : (i) a reversible isobaric expansion from (1.0 atm, 20.0 L) to (1.0 atm, 40.0 L). (ii) a reversible isochoric change of state from (1.0 atm, 40.0 L ) to (0.5 atm, 40.0 L). (iii) a reversible isothermal expansion from (0.5 atm, 40.0 L )to (1.0 atm, 20.0 L)
(i) Sketch with labels each procsses on the same P- V diagram.
(ii) Calculate the total work (w) and the total heat (Q) involved in the above process.
(iii) What will be values of `Delta U, Delta H and Delta S ` for overall process ?

Explanation

(ii) Work done in step AB :
`w_(1)= - P_(ex)(V_(2)-V_(1))= - 1(40-20)L` atm = - 20 L atm
Work done in step `BC : w_(2)= 0` [as the volume remains constant]
Work done in step CA : `w_(3)= -2.303 nRT " log" (P_(1))/(P_(2))`
Now, `PV= nRT " or ", 0.5xx40= nRT`
`therefore nRT= 0.5xx40 L` atm
`therefore w_(3)= - 2.303xx0.5xx40 L " atm log " (0.5)/(1)`
= 13.86 L atm.
`therefore` Total work in the overall process `= w_(1)+w_(2)+w_(3)`
`= (-20+0+13.86) L.` atm = -6.14 L atm
= `-621.77 J`
We know, `Delta U = q+w`
Since the process is cyclic and U is a state function,
`Delta U= 0`
`therefore q+w =0, or , q = - w = 621.77 J`.
`therefore ` total heat involved in the process = 621.77 J.
(iii) As H and S are state function and the process is cyclic, `Delta H =0 and Delta = 0` for the process .