Question

An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^–1 K^–1, molecular mass of O₂ = 32 u).

Answer 1

Volume of oxygen, V₁ = 30 litres = 30 × 10^–3 m ³ 

Gauge pressure, P₁ = 15 atm = 15 × 1.013 × 10⁵ Pa 

Temperature, T₁ = 27°C = 300 K 

Universal gas constant, R = 8.314 J mole^–1 K^–1 

Let the initial number of moles of oxygen gas in the cylinder be n₁. 

The gas equation is given as: 

P₁V₁ = n₁RT₁

\(∴  n_1=\frac{P_1V_1}{RT}\)

\(=\frac{15.195×10^5×30c10^{-3}}{(8.314)×300}=18.276\)

But, \(n_1=\frac{m_1}{M}\)

P₁V₁ = n₁RT₁ = 18.276 

Where, 

m₁ = Initial mass of oxygen 

M = Molecular mass of oxygen = 32 g 

∴ m₁ = n₁M = 18.276 × 32 = 584.84 g 

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces. 

Volume, V₂ = 30 litres = 30 × 10^–3 m ³ 

Gauge pressure, P₂ = 11 atm = 11 × 1.013 × 10⁵ Pa 

Temperature, T₂ = 17°C = 290 K 

Let n₂ be the number of moles of oxygen left in the cylinder. 

The gas equation is given as: 

P₂V₂ = n₂RT₂ = 13.86 

But,  \(∴  n_2=\frac{m_2}{M}\)

Where, 

m₂ is the mass of oxygen remaining in the cylinder 

∴ m₂ = n₂M = 13.86 × 32 = 453.1 g 

The mass of oxygen taken out of the cylinder is given by the relation: 

Initial mass of oxygen in the cylinder 

= m₁ – m₂ 

= 584.84 g – 453.1 g 

= 131.74 g 

= 0.131 kg 

Therefore, 0.131 kg of oxygen is taken out of the cylinder.