### Q.4:- An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (*R *= 8.31 J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 O_{2} = 32μ).

**Answer:-**

Volume of oxygen, *V*_{1} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure, *P*_{1} = 15 atm = 15 × 1.013 × 10^{5} Pa

Temperature, *T*_{1} = 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 *n*_{1}.

The gas equation is given as:

*P*_{1}*V*_{1} = *n*_{1}*RT*_{1}

∴ *n*_{1} = *P*_{1}*V*_{1}/ *RT*_{1}

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

But *n*_{1} = *m*_{1} / *M*

Where,

*m*_{1} = Initial mass of oxygen

*M* = Molecular mass of oxygen = 32 g

∴*m*_{1} = *n*_{1}*M *= 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, *V*_{2} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure, *P*_{2} = 11 atm = 11 × 1.013 × 10^{5} Pa

Temperature, *T*_{2} = 17°C = 290 K

Let *n*_{2} be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

*P*_{2}*V*_{2} = *n*_{2}*RT*_{2}

∴ *n*_{2} = *P*_{2}*V*_{2}/ *RT*_{2}

= (11.143 × 10^{5} × 30 × 10^{-3}) / (8.314 × 290) = 13.86

But *n*_{2} = *m*_{2} / *M*

Where,

*m*_{2} is the mass of oxygen remaining in the cylinder

∴ *m*_{2} = *n*_{2}*M *= 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 – Final mass of oxygen in the cylinder

= *m*_{1} – *m*_{2}

= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

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