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Two moles of a perfect gas undergo the f...

Two moles of a perfect gas undergo the following process: (a) A reversible isobaric expansion from (`1.0` atm, `20.0` litre) to (`1.0` atm,`40.0`litre). (b) A reversible isochoric change of state from (`1.0` atm, `40.0` litre) to (`1.0` atm, `40.0` litre). (b) A reversible isochoric change of state from (`1.0` atm, `40.0` litre) to (`0.5` atm, `40.0` litre). (c) A reversible isothermal compression from (`0.5`atm, `40.0` litre) to (`1.0` atm, `20.0` litre). (i) Sketch with labels each of the process on the same `P-V` diagram. (ii) Calculate the total work `(W)` and the total heat change `(q)` involved in the above process. (iii) What will be the value of `DeltaU, DeltaH` and `DeltaS` for the overall process?

Text Solution

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(i)
`T = (PV)/(nR) = (1 xx 20)/(2 xx 0.082) = 121.95` K
(ii) Total work (W) `=W_(1) + W_(2) + W_(3)`
`= -1 xx 20 + 2.303 xx 2 xx 0.082 xx 121.95 log2`
`=-20 + 13.86 = -6.13` L atm.
Since the system has returned to its initial state, i.e. the process is cyclic. So, `DeltaU=0`
`DeltaU = q+W=0`
`q=-W = 6.13 L atm = 620.7` J
In a cyclic process heat absorbed is completely converted into work.
Entropy is a state function and since the system has returned to its initial state so `DeltaS=0`. Similarly, `DeltaH=0` and `DeltaU=0` for the same reason, i.e. U and H are also state functions having definite values in a given state of a system.
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