Air that initially occupies 0.59 m 3 at a gauge pressure of 140 kPa is expanded

Question

Air that initially occupies 0.59 m3 at a gauge pressure of 140 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

I cannot figure out what I am doing wrong.

For the first Isothermal portion I got Work = 24.8897 J

and for the second Isobaric part i got Work = -22.833 J

Which should make the answer Work Net = 2.0567 J

I also tried entering 2056.7 J as the other answer does not convert out of kPa to Pa

Explanation / Answer

Work done by the air is given by the integral
W = ? p dV from initial final state

In first step we expand at constant temperature. i.e.
p?V = n?R?T = constant
hence:
p = p??V?/V
and
W? = ? p??V?/V dV from V? to V?
= p??V??ln( V?/V?)

because p??V? = p??V? <=> V?/V? = p?/p?

W? = p??V??ln(p?/p?)

Add standard atmospheric pressure to initial gauge pressure:

p? = 140kPa + 101.3kPa = 241.3kPa

W? = 241300Pa ? 0.59m

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