Answer:
the total work W = 29.05 kJ
the change in total internal energy is [tex]\mathbf{\Delta U = - 27.19 \ kJ}[/tex]
the total heat transferred in [kJ] is Q = 1.860 kJ
Explanation:
Given that
mass of carbon dioxide in the closed system = 1 kg
Temperature [tex]T_1= 30 ^0 C[/tex] = (273+30 ) K = 303 K
Pressure [tex]P_1 = \ 200 \ kPa[/tex]
Pressure [tex]P_2 = 100 \ kPa[/tex]
polytropic expansion n = 1.27
Note that we are also given the following data set:
R = 188.9 J/kg.K
c_v = 655 J/kg.K
So; for a polytropic process ; [tex]PV^{1.27} = c[/tex]
[tex]\dfrac{T_2}{T_1}= ( \dfrac{V_1}{V_2})^{n-1} = (\dfrac{P_2}{P_1})^{\frac{n-1}{n}[/tex]
[tex]T_2 = T_1 [\dfrac{P_2}{P_1}]^{\frac{n-1}{n}[/tex]
[tex]T_2 = 303 [\dfrac{100}{200}]^{\frac{1.27-1}{1.27}[/tex]
[tex]T_2 = 261.48 \ K[/tex]
Since the system does not follow the first order of thermodynamics; To calculate the total work by using the expression:
[tex]W = \dfrac{P_1V_1-P_2V_2}{n-1} = \dfrac{mR(T_1-T_2)}{n-1}[/tex]
[tex]W = \dfrac{1*188.9(303-261.48)}{1.27-1}[/tex]
W = 29048.62222 J
W = 29.05 kJ
Thus, the total work W = 29.05 kJ
The change in internal energy can be expressed by the formula:
[tex]\Delta U = mc_v (T_2-T_1)[/tex]
[tex]\Delta U = 1*655(261.48-303)[/tex]
[tex]\Delta U = -27195.6 \ J[/tex]
[tex]\mathbf{\Delta U = - 27.19 \ kJ}[/tex]
Hence; the change in total internal energy is [tex]\mathbf{\Delta U = - 27.19 \ kJ}[/tex]
Finally; to determine the total heat transferred in [kJ]; we go by the expression for the first order of thermodynamics which say:
Total Heat Q = ΔU + W
Q = (-27.19 + 29.05)kJ
Q = 1.860 kJ
Hence; the total heat transferred in [kJ] is Q = 1.860 kJ
