An existing process consists of two steps:
(a) One mole of air T1 = 900 K and P1 = 3 bar is cooled at constant volume to T2 = 300 K.
(b) The air is then heated at constant pressure until its temperature reaches 900 K.
It is proposed to replace this two-step process by a single isothermal expansion of the air from 900 K and 3 bar to some final pressure P. What is the value of P that makes the work of the proposed process equal to that of the existing process? Assume mechanical reversibility and treat air as an ideal gas with Cp = 7R/2 and Cv = 5R/2
Calculations:
Work in the original process consisting of two steps:
(a) For the constant volume process:
W = 0
Pressure at the end of the constant volume process (P2):
P1V1 = RT1
V1 = R x 900 / 3 = 300R
P2V2 = RT2
Since V1 = V2,
P2 = R x 300 / (300R) = 1 bar
(b) For the constant pressure process:
P2 = 1 bar; T2 = 300 K; V2 = 300R
P3 = 1 bar; T3 = 900 K
P3V3 = RT3
V3 = R x 900 / 1 = 900R
W = ∫PdV = P2(V3 - V2) = 1 x (900R - 300R) = 600R
Total work done in the two step process = 600R
For the single step isothermal process:
W = 600R = RT ln (P1/P)
i.e., 600R = R x 900 x ln (3/P)
ln(3/P2) = 600/900 = 2/3
e(2/3) = 3/P
1.9477 = 3/P
P
= 3/1.9477 = 1.54 barLast Modified on: 04-Feb-2022
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