A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bar

Question

A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bar is cooled at constant volume to T = 350 K. (2) The air is then heated at constant pressure until its temperature reaches 800 K. If this two-step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to some final pressure P, what is the value of P that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with C_p = (7/2)K and C_v = (5/2)/R.

Explanation / Answer

Original process consisting of two steps:

For the 1st step : The process at which the volume is constant is called Isochoric process

Where, work done (W) = 0

One mole of air at Ti=800 K and Pi = 4 bar is cooled to Tf =350 K (i = initial, f= Final)

consider Air is an ideal gas, Thus PiVi = RTi

Vi = R x 800 / 4 = 200R

And PfVf = RTf

Since it is a isochoric process Vi = Vf,

Pf = R x 350 / (200R) = 1.75 bar

For the 2nd step: The process at which the pressure is constant is called Isobaric process:

Pi = 1.75 bar; Ti = 350 K; Vi = 200R

Since pressure is constant

Pf = 1.75 bar; Tf = 800 K

PfVf = RTf

Vf = R x 800 / 1.75 = 457.14R

Work done W = ò PdV = Pf(Vf – Vi) = 1.75 x (457.14R - 200R) = 257.14R

Total work done in the two step process = 257.14R

For the single step isothermal process where temperature is constant:

T=800 K

Work done W = RT ln (Pi /Pf) = 257.14 R

=> R x 800 x ln (4/Pf) = 257.14 R

=> ln (4/Pf) = 257.14 R/ 800 R = 0.32

=> 4/P = e(0.32)

=> 4/P = 1.37

=> P = 4/1.37 = 2.91 bar

Thus P= 2091 bar makes the work of the proposed process equal to that of the existing process

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