What is the value of P that makes the work of the two processes the same?
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.
Assume mechanical reversibility and treat air as an ideal gas with 
and 
The initial volume of 1 mol of air at 800 K and 4 bar (400000 Pa) is V = RT/P = 8.3145 * 800 / 400000 = 0.01663 
. When it is cooled at constant volume to 350 K, the pressure drops to 4 bar * 350 K/800 K = 1.75 bar = 175000 Pa. No work is done during this constant volume cooling step. When it is heated back to 800 K at constant pressure, the volume increases to is V = RT/P = 8.3145 * 800 / 175000 = 0.03801 
The work done on the gas is -P?T = -175000 Pa * 
= -3742 J. So, the work done by the gas is 3742 J.
For an isothermal expansion, the work will be 
or 
or
= 2.27 bar.
So, the isothermal expansion does the same amount of work with less expansion of the gas.
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