I got 7.02 bar but the answer is 2.279 bar. Please show me what I\'m doing wrong

Question

I got 7.02 bar but the answer is 2.279 bar. Please show me what I'm doing wrong. I'm using Q=-W=-RTln(p2/4bar)=-R(800-350)+W2(no change in volume means no W=0 for this term

A process consists of two steps: (1) One mole of air at T = 800 K and 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 make the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with Cp = (7/2)R and Cy = (5/2) R.

Explanation / Answer

from the 1st process using Q = n*Cv*(T2-T1) = -450R

now work done in an isothermal expansion = - nRTln(4/P2) = Q

450R = 1*R*T*ln(4/P2)
ln(4/P2) = 450/800

P2 = 4/e^0.5625
P2 = 2.279 bar

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