Determine the mass of 5 liters of oxygen at atmospheric pressure and a temperature of 20 C

Under conditions that differ from standard ones, Mendeleev Clapeyron’s law is used to determine the parameters of gases:
p * V = n * R * T, where
p – gas pressure, in Pa
V – gas volume, in m3
n is the amount of gas substance, in mol
T – temperature, in K
R – universal gas constant, R = 8.31 J / (mol * K)
Let’s bring our parameters to the units of the formula.
p = 101325 Pa, V = 5 l = 0.005 m3, T = 20 ° C = 293 K.
Calculate n = p * V / (R * T) = 101325 * 0.005 / (8.31 * 293) = 0.208 mol
Find the mass of oxygen m = n * M, where M = 32 g / mol is the molar mass of oxygen
m = 0.208 * 32 = 6.66 g



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