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A stream of oxygen at -65°C and 8.3 atm flows at a rate of 250 kg/h. Use the SRK equation of state to estimate the volumetric flow rate (L/hr) of this stream. (See Example 5.3-3.) The ideal gas equation of state is an approximation. Under which conditions, it is suggested that the ideal gas equation be used for? Select one: O a. Temperatures above about 0°C and pressures below about 1 atm O b. Temperatures below about 0°C and pressures below about 1 atm c. Temperatures above about 0°C and pressures above about 1 atm d. Under any condition e. standard condition of 25°C and 1 atm O o

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Answer:

16064 m³/h

Option a) " Temperatures above about 0°C and pressures below about 1 atm "

Explanation:

Given:

Temperature of oxygen = -65° C = 273 - 65 = 208 K

Pressure = 8.3 atm = 8.3 × 101325 Pa

Mass flow rate = 250 kg/h

also,

Molecular weight of oxygen, O₂ = 2 × 16 = 32 grams/mol = 0.032 kg/mol

now,

number of moles of oxygen, n = [tex]\frac{\textup{Mass flow rate}}{\textup{Molecular weight of oxygen}}[/tex]

or

n = [tex]\frac{\textup{250}}{\textup{0.032}}[/tex]

or

n = 7812.5 moles/h

Now,

from the ideal gas equation

PV = nRT

here,

V is the volume flow rate

P is the pressure

R is the ideal gas constant = 8.314 Pa.m³/mol.K

T is the temperature

thus,

V = [tex]\frac{\textup{7812.5}\times8.314\times208}{\textup{8.3}\times101325}[/tex]

or

V = 16.064 m³/h

also,

1 m³ = 1000 L

thus,

V = 16.064 × 1000 m³/h = 16064 m³/h

Option a) " Temperatures above about 0°C and pressures below about 1 atm " is correct because the ideal gas law does not work at very low temperature and as well as at very high pressure

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