Question

A 1.0  L container of liquid nitrogen is kept in a closet measuring 1.2  m by 1.1  m by 1.9  m . Assuming that the container is completely full, that the temperature is 29.9 ∘C, and that the atmospheric pressure is 1.0  bar , calculate the percent (by volume) of air that is displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 gmL−1.)

Answer 1

Given that Pressure = 1.0 bar or 0.987 atm density of liquid nitrogen; N2=0.807 gmL-1 the molar mass of nitrogen=28.0 g/mol volume of nitrogen = 1.0 L So: 1.0 L*1L/1000 mL = 1000 ml Density = mass / volume Mass = density * volume 1000 mL * 0.807 g/mL = 807 g Number of moles of N2 = amount in g/molar mass = 807 g/28.0 (g/mol) = 28.82 mol Ideal Gas Law for volume: PV = nRT V=nRT/P For R, we use R=0.082 atm*L/(mol*K), and for T we use (29.9+273) K. V=28.82 mol * 0.082 atm*L/(mol*K) * (302.9) K/(0.987 atm) V = 725 L for sig figs. the total volume of the chamber in liters. 1.2 m * 1.1 m * 1.9 m = 2.508 m^3 Now convert this to liters, we use the fact that 1 m^3 = 1000 L. 2.508 m^3 * 1000L /1 m^3 = 2508 L. Now, to find the percent volume displaced, as followed: Volume of nitrogen / total volume *100 725 L * 100%/2508 L = 28.9 %