Question

Liquid oxygen can be produced for use as a rocket fuel by cooling dry air to -183°C, where the o, condenses. How many liters of dry air at 25°C and 750 torr would need to be processed to produce 150 L of liquid O, at -183°C? The mole fraction of oxygen in dry air is 0.21, and the density of liquid oxygen is 1.14 g/mL.

Answer 1

Here is the Solution:

Given data : Pressure = 750 torr 1 atm = 750 tour x = 760 tort -0.9868 atm - 1000 mL 1504 - = 150x103 mL • Volume = 150 L = . Temperature -25°C = 25+ 273,15 = 298.15 K Density = 1.14 g/mL. mole fraction of oxygen x = 0.21 - first, we will calciclate the mass of oxygen needed. Mass = volume x Density mass= (150 × 103 L) x (1.14 g /rot) Mass= 171000 g Second we will calculate the moles of oxygen (02) needed. (molar mass = 32 g/mol) mass Moles - - - 171000 g = 5343.75 moles. molar mass 32 g/mol Third, we will find the no. of moles of air needed. moles of O2 5343.75 mol moles of air = - mole fraction or 0.21 = 25446.43 mol Now, simply we will use ideal gas law equation, PU= nRT - where ,op= Pressure (in atm) . V = Volume = ?L .n=no. of moles (in mol) solue the equation R = gas constant (0.0821 L.atm/mobk) for 'v . = Temperature (in kelvin) u nRT at - е " > (25446.43 hool)*(0.0821 L. gtm/motik) x (298.15 0.98 6 8 atm v=6312 12.646 L OR v= 6.3 x 103 L 7

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