A 1.3-L container of liquid nitrogen is kept in a closetmeasuring 1.0 m by 1.0 m
ID: 687350 • Letter: A
Question
A 1.3-L container of liquid nitrogen is kept in a closetmeasuring 1.0 m by 1.0 m by 2.0 m. Assuming that the container is completely full, that thetemperature is 28.9 degrees C, and that the atmospheric pressure is1.2 atm, calculate the percent (by volume) of air that wouldbe displaced if all of the liquid nitrogen evaporated. (Liquidnitrogen has a density of 0.807 g/mL.) A 1.3-L container of liquid nitrogen is kept in a closetmeasuring 1.0 m by 1.0 m by 2.0 m. Assuming that the container is completely full, that thetemperature is 28.9 degrees C, and that the atmospheric pressure is1.2 atm, calculate the percent (by volume) of air that wouldbe displaced if all of the liquid nitrogen evaporated. (Liquidnitrogen has a density of 0.807 g/mL.)Explanation / Answer
We know, mass = density * volume =0.807g / mL * 1.1L (1000mL / 1L) =887.7g Moles of N2 = 887.7g / 28g/mol = 31.7mol Volume of closet = 1.0m * 1.0m *2.0m = 2m^3 (1dm / 0.1m)^3 = 2*10^3 m^3 (1dm^3 / m^3) = 2*10^3dm^3 =2*10^3L [Since 1dm^3 = 1L] Volume occupied by N2 gas = nRT / p = 31.7mol * 0.0821Latmmol^-1K^-1 * 298K / 1atm = 775.57L So percent of volume occupied by N2 = (775.57L /2*10^3L )*100 = 38.79% We know, mass = density * volume =0.807g / mL * 1.1L (1000mL / 1L) =887.7g Moles of N2 = 887.7g / 28g/mol = 31.7mol Volume of closet = 1.0m * 1.0m *2.0m = 2m^3 (1dm / 0.1m)^3 = 2*10^3 m^3 (1dm^3 / m^3) = 2*10^3dm^3 =2*10^3L [Since 1dm^3 = 1L] Volume occupied by N2 gas = nRT / p = 31.7mol * 0.0821Latmmol^-1K^-1 * 298K / 1atm = 775.57L So percent of volume occupied by N2 = (775.57L /2*10^3L )*100 = 38.79%Related Questions
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