Answer to Question #238162 in Physical Chemistry for deejay

Question

Liquid oxygen for use as a rocket fuel can be produced by cooling dry air to

           –183ºC, where the O2 condenses. How many liters of dry air at 25ºC and 750 torr would need to be processed to produce 150 L of liquid O2 at –183 ºC? (The mole fraction of oxygen in dry air is 0.21, and the density of liquid oxygen is 1.14 g/mL).

Answer 1

Molar mass of $O_2$ $2(15.9994)=31.9988g/mol$

Moles of $O_2=\frac{150×1000×1.14}{31.9988}=5343.95moles$

Using the mole fraction to determine the total number of moles of $O_2$

$n_{o_2}=Xo_2n_{tot}$

$n_{tot}=\frac {n_{o_2}}{Xo_2}$

$n_{tot}=\frac{5343.95}{0.21}=2.5447×10^4mol$

Converting the pressure into atmospheres

$750torr=\frac{750}{760}=0.98684atm$

Converting the temperature into Kelvins

$T=(25+273.15)=298.15K$

Use the ideal gas law to calculate the volume

$PV=n_{tot}RT$

$V=\frac{n_{tot}RT}{P}$

$V=\frac{2.5447×10^4×0.08206×298.15}{0.98684}=6.3×10^5L$