Question

Air that initially occupies 0.144 m3 at a gauge pressure of 211.0 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.)

Answer 1

Using Gas equation

For isothermal process

$V_{2} = \frac{P_{1}V_{1} }{P_{2}}$

$V_{2} = \frac{211.0 kPa\times 0.144 m^{3} }{101.3 kPa} = 0.29994 m^{3}$

Work done during isothermal expansion

$W_{1} = P_{1}V_{1}\ln \left ( \frac{P_{1}}{P_{2}} \right )$

$W_{1} = 211.0\times 10^{3}Pa\times 0.144 m^{3}\times \ln \left ( \frac{211.0}{101.3} \right )= 22294.92 J$

Now work done during isobaric process

$W_{2} = P_{2}(V_{1}-V_{2}) = 101.3\times 10^{3} Pa\times (0.144 m^{3}-0.29994 m^{3})= -15796.72 J$

Total Work done