Answer:
Q = -68.859 kJ
Explanation:
given details
mass [tex]co_2 = 1 kg[/tex]
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of [tex]co_2 = 44[/tex]
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT
[tex]V_1 = \frac{m RT_1}{P_1}[/tex]
[tex]=\frac{1*0.189*298}{104}[/tex]
[tex]V_1 = 0.5415 m3[/tex]
[tex]V_2 = \frac{m RT_2}{P_2}[/tex]
[tex] =\frac{1*0.189*584}{1068}[/tex]
[tex]V_1 = 0.1033 m3[/tex]
WORK DONE
[tex]W =P_{avg}*{V_2-V_1}[/tex]
w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS
[tex]\Delta U = m *c_v*{V_2-V_1}[/tex]
[tex]\Delta U = 1*0.657*(584-298)[/tex]
[tex]\Delta U =187.902 kJ[/tex]
HEAT TRANSFER
[tex]Q = \Delta U +W[/tex]
= 187.902 +(-256.46)
Q = -68.859 kJ