Answer:
The answer is "[tex]W= 100.44 \ KJ\ \ \ \W_{win}=7.23 \ KJ[/tex]"
Explanation:
Please find the complete question in the attached file.
From of the ideal gas relation that initial and the last temperatures were determined: Â
[tex]T_1 = \frac{P_1 V}{m R}[/tex]
  [tex]= \frac{100 \times 0.8}{1.54 \times 0.1889} \\\\ = 275 \ K[/tex]
[tex]T_2 = \frac{P_2 V}{m R}[/tex]
  [tex]= \frac{135 \times 0.8}{1.54 \times 0.1889} \\\\ = 317 \ K[/tex]
In the initial and final states, the internal energies for given temperatures are described from A-20 by means of intelmpolation and divided by the carlxon molar mass. Â
[tex]u_1 = 141.56 \frac{KJ}{kg}\\\\u_2 = 206.78 \frac{KJ}{kg}[/tex]
The real job is just the difference between internal energies: Â
[tex]W = m(u_2 - u_1) \\\\[/tex]
  [tex]= 1.54(206.78 -141.56) \ kJ \\\\ =100.44 \ kJ[/tex]
In the initial and final states, the zero entries are as determined as internal energies:
[tex]S_1^{\circ} =4.788 \frac{KJ}{kg K}\\\\S_2^{\circ} =5.0478 \frac{KJ}{kg K}[/tex]
From its energy increase, the minimum work required is determined:
[tex]W_{min} = m(u_2-u_1 - T_0(s_2 -S_1))\\\\=W-mT_0(S_2^{\circ}- S_1^{\circ} -R \In \frac{P_2}{P_1})\\\\= 100.44kJ -1.54 \times 298( 5.0478-4.788-0.1889 \In \frac{135}{100})\\\\=7.23\ KJ\\[/tex]
