Respuesta :
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant [tex]\gamma =1.4[/tex]
specific heat is given as [tex]=\frac{\gamma R}{\gamma -1}[/tex]
gas constant =287 J⋅kg−1⋅K−1
[tex]Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}[/tex]
specific heat at constant volume
[tex]Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}[/tex]
change in internal energy [tex]= Cv(T_2 -T_1)[/tex]
[tex] \Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}[/tex]
change in enthalapy [tex]\Delta H = Cp(T_2 -T_1)[/tex]
[tex] \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}[/tex]
change in entropy
[tex]\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})[/tex]
[tex]\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})[/tex]
[tex]\Delta S = 382.79 J kg^{-1} K^{-1}[/tex]
