Air that initially occupies 0.22 m3 at a gauge pressure of 86 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.)

total work done = -5960.8 J

Explanation:

given data

initial volume v1 = 0.22 m³

initial pressure  p1 = 86 kPa

final pressue p2 = 101.3 kPa

solution

we apply here  isothermal expansion that is express as

p1 × v1 = p2 × v2    ......................1

put here value

86 × 0.22 = 101.3 × v2

v2 = 0.1867 m³  

and

work done will be here

w1 = p1 × v1  × ln()      ....................2

w1 = 86 × 10³ × 0.22 ×

w1 = -3.097  × 10³ J

and

it is cooled to initial volume at constant pressure  so here work done will be

w2 = p(v2 - v1)    .................3

w2 =  86 × 10³ × ( 0.1867 - 0.22 )

w2 = -2863.8 J

so

total work done is

total work done = w1 + w2

total work done = -3097 +  -2863.8

total work done = -5960.8 J