Answer:
[tex]T_2=604.70K=331.55\°C[/tex]
[tex]W=220.10\frac{J}{g}[/tex]
Explanation:
Hello!
In this case, for this compression of air from 101 kPa to 1200 kPa, we should first realize that equation relating P and T is:
[tex]T_1*P_1^{\frac{1-\gamma}{\gamma} }=T_2*P_2^{\frac{1-\gamma}{\gamma} }[/tex]
Whereas:
[tex]\gamma=\frac{Cp_{air}}{Cv_{air}} =\frac{1.00J/(g\°C)}{0.718J/(g\°C)}=1.40[/tex]
That is why the final temperature is:
[tex]T_2=T_1(\frac{P_1}{P_2})^{\frac{1-\gamma}{\gamma} }\\\\T_2=298.15K(\frac{101kPa}{1200kPa})^{\frac{1-1.40}{1.40} }\\\\T_2=604.70K=331.55\°C[/tex]
Moreover, for the work done, since the process is adiabatic, no heat is in the equation:
[tex]Q-W=\Delta U[/tex]
But as the work is done on the system we can write:
[tex]-(-W)=\Delta U=Cv(T_2-T_1)\\\\W=0.718\frac{J}{g*K}(604.70K-298.15K)\\ \\W=220.10\frac{J}{g}[/tex]
Best regards!
