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blsea [12.9K]
3 years ago
10

What is the percentage of error involved in treating carbon dioxide at 7 MPa and 380 K as an ideal gas?

Engineering
1 answer:
Lorico [155]3 years ago
6 0

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 T_{CR} = 304.2 k

critical pressureP_{cr} = 7.39 MPa

Gas constant R = 0.188 kJ/kg -K

From ideal gas equation we have

PV = RT

solving of v we have

v = \frac{0.188(380)}{7\times 10^{3}}

v = 0.01020 m^3/kg

Reduced Pressure

P_R = \frac{P}{P_{CR}} = \frac{7}{7.39} = 0.947

reduced temperature

T_R = \frac{T}{T_{CR}} = \frac{380}{304.2} = 1.249

obtained compressibility factor (z) from chart between reduced temperature and reduced pressure.

fromP_R = 0.947 and T_R = 1.249, Z value is 0.6

so we have

Pv = ZRT

v = \frac{0.6\times 0.1889\times 380}{7\times 10^3}

v = 0.00615 m^3/kg

percentage of error is

= \frac{0.01020- 0.0061}{0.0061} \times 100 = 67.21%

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