Astream 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

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 =

or

n =

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 =

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