###
Isothermal followed by Isochoric Process

Home ->
ChE Learning Resources
->
Solved Problems
->
Thermodynamics->

An existing process consists of two steps:

(a) One mole of air T_{1} = 900 K and P_{1} = 3 bar is cooled at constant volume to T_{2} = 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 C_{p} = 7R/2 and C_{v} = 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 (P_{2}):

P_{1}V_{1} = RT_{1}

V1 = R x 900 / 3 = 300R

P_{2}V_{2} = RT_{2}

Since V_{1} = V_{2},

P_{2} = R x 300 / (300R) = 1 bar

(b) For the constant pressure process:

P_{2} = 1 bar; T_{2} = 300 K; V_{2} = 300R

P_{3} = 1 bar; T_{3} = 900 K

P_{3}V_{3} = RT_{3}

V_{3} = R x 900 / 1 = 900R

W = ∫PdV = P_{2}(V_{3} - V_{2}) = 1 x (900R - 300R) = 600R

Total work done in the two step process = 600R

For the single step isothermal process:

W = 600R = RT ln (P_{1}/P)

i.e., 600R = R x 900 x ln (3/P)

ln(3/P_{2}) = 600/900 = 2/3

e^{(2/3)} = 3/P

1.9477 = 3/P

P = 3/1.9477 = 1.54
bar

[Index]

Last Modified on: 01-May-2024

Chemical Engineering Learning Resources - msubbu

e-mail: learn[AT]msubbu.academy

www.msubbu.in