Explanation:
As it is given that,
[tex]P_{1} = C_{2}V^{\frac{1}{3}}[/tex]
[tex]P_{1}V^{-\frac{1}{3}}_{1} = C_{2}[/tex]
As the system is not gaining or losing heat. So, it is an adiabatic process in an assumed ideal gas. The polytropic extent n is [tex]\frac{-1}{3}[/tex].
[tex]P_{1} = C_{1}V\frac{1}{3}_{1}[/tex]
= [tex](100)(1)^{\frac{1}{3}}[/tex]
= 100 kpa
[tex]P_{2} = C_{2}V^{\frac{1}{3}}_{2}[/tex]
= [tex]100(4)^{\frac{1}{3}}[/tex]
= 158.74 kpa
Now, work done by the air is as follows.
[tex]W_{2} = \int PdV[/tex]
= [tex]\frac{P_{2}V_{2} - P_{1}V_{1}}{1 - n}[/tex]
= [tex]\frac{158.74 \times 4 - 100 \times 1}{1 - (\frac{-1}{3})}[/tex]
= 401.22 kJ
Work done by the air is as follows.
[tex]P_{2}V_{2} = m_{2}RT_{2}[/tex]
[tex]m_{2} = \frac{P_{2}V_{2}}{RT_{2}}[/tex]
= [tex]\frac{158.74 \times 4}{0.287 \times (273 + 25)}[/tex]
= 7.424 kg
Thus, we can conclude that final mass of air is 7.424 kg and work done by the air is 401.22 kJ.