Respuesta :
Answer:
Percentage of error is 67.2%
Explanation:
from table A-1 of thermodynamics " gas constant and critical point properties" obtain CO_2 properties
from table
critical temperature [tex]T_{CR} = 304.2 k[/tex]
critical pressure[tex] P_{cr} = 7.39 MPa[/tex]
Gas constant R = 0.188 kJ/kg -K
From ideal gas equation we have
PV = RT
solving of v we have
[tex]v = \frac{0.188(380)}{7\times 10^{3}}[/tex]
v = 0.01020 m^3/kg
Reduced Pressure
[tex]P_R = \frac{P}{P_{CR}} = \frac{7}{7.39} = 0.947[/tex]
reduced temperature
[tex]T_R = \frac{T}{T_{CR}} = \frac{380}{304.2} = 1.249[/tex]
obtained compressibility factor (z) from chart between reduced temperature and reduced pressure.
from[tex]P_R = 0.947 and T_R = 1.249, Z value is 0.6[/tex]
so we have
Pv = ZRT
[tex]v = \frac{0.6\times 0.1889\times 380}{7\times 10^3}[/tex]
v = 0.00615 m^3/kg
percentage of error is
[tex]= \frac{0.01020- 0.0061}{0.0061} \times 100 = 67.21[/tex]%
