Question

you fill a rigid steel container that has a volume of 20 L with nitrogen gas to a final pressure of 2 x 10^4 kpa at 23 Celsius. how mny kilorams of n2 does this cylinder contai

Answer 1

4.549 kg.

Step-by-step explanation:

• We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).

V is the volume of the gas in L (V = 20.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 23° C + 273 = 296 K).

∴ n = PV/RT = (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = 162.5 mol.

• To find the mass of N₂ in the cylinder, we can use the relation:

mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = (162.5 mol)*(28.0 g/mol) = 4549 g = 4.549 kg.