Question

A steel tank contains 377 g of ammonia gs (NH₃ with molar mass 17.0 g/mol) at a pressure of 1.33 x 10⁶ and a temperature of 75.0^oC. (a) What is the volume of the tank in liters? (b) Later the temperature is 60.5^oC and the pressure is 0.831 x 10⁶ Pa. How many grams of gas have leaked out of the tank? (Ignore the thermal contraction of the tank.)

Answer 1

Universal gas constant = R = 8.314 J/(mol.K)

Molar mass of ammonia gas = M = 17 g/mol

Initial mass of ammonia gas = m₁ = 377 g

Initial number of moles of ammonia gas = n₁

m₁ = n₁M

377 = n₁(17)

n₁ = 22.176 mol

Initial temperature of the ammonia gas = T₁ = 75 ^oC = 75 + 273 K = 348 K

Initial pressure of the ammonia gas = P₁ = 1.33 x 10⁶ Pa

Initial volume of the ammonia gas = V₁

By Ideal Gas Law,

P₁V₁ = n₁RT₁

(1.33x10⁶)V₁ = (22.176)(8.314)(348)

V₁ = 4.824 x 10^-2 m³

Converting from m³ to liters,

V₁ = 4.824x10^-2 x 1000 L

V₁ = 48.24 L

Some amount of the ammonia gas has leaked out from the tank.

Final number of moles of ammonia gas in the tank = n₂

Final temperature of the ammonia gas = T₂ = 60.5 ^oC = 60.5 + 273 K = 333.5 K

Final pressure of the ammonia gas = P₂ = 0.831 x 10⁶ Pa

Final volume of the ammonia gas = V₂

The volume of the tank remains constant.

V₂ = V₁ = 4.824 x 10^-2 m³

P₂V₂ = n₂RT₂

(0.831x10⁶)(4.824x10^-2) = n₂(8.314)(333.5)

n₂ = 14.458 mol

Mass of ammonia gas left inside the tank = m₂

m₂ = n₂M

m₂ = (14.458)(17)

m₂ = 245.8 g

Mass of the gas leaked out = m

m = m₁ - m₂

m = 377 - 245.8

m = 131.2 g

a) Volume of the tank = 48.24 L

b) Mass of the gas leaked out of the tank = 131.2 g