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cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.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.