The steel tank with volume of 438 L filled with certain amou

The steel tank with volume of 438 L filled with certain amount of O2 develops a slow leak that is discovered and sealed. A pressure gauge fitted on the tank shows the new pressure is 1.37 atm at 25° C.
How many grams of O2 remain?

Solution

V =438 L

P= 1.37 atm

T= 25^oC

PV = nRT (ideal gas equation)

n= PV/RT

n =(1.37 atm * 438 L)/(0.0821 L atm mol^-1 K^-1 * 298 K)

=600.0600/24.4658 = 24.5265 mol

mass of O₂ remaining= 24.5265*16 =392.424 g