Answer 1

Volume of oxygen, V₁ = 30 litres = 30 × 10^–3 m³

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:

`P_1V_1 = eta_1RT_1`

`:.n_1 = P_1V_1/RT_1`

`= (15.195 xx10^5 xx 30 xx 10^(-3))/((8.314) xx 300) = 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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.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 m³

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

`:. n_2= (P_2V_2)/(RT_2)`

`= (11.143 xx 10^5 xx 30 xx 10^(-3))/8.314 xx 290 = 13.86`

But `n_2 = 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 – Final 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.

Answer 2

Initial Volume, `V_1 = 30 litre = 30 xx 10^3 cm^3`

`= 30 xx 10^3 xx 10^(-6) m^3 = 30 xx 10^(-3) m^3`

Initla Pressure, P_1 = 15 atm

`= 15 xx1.013 xx 10^5 N m^(-2)`

Inital temperature, `T_1 = (27 + 273) K = 300 K`

Initial number of moles 

`mu_1 = (P_1V_1)/RT_1 = (15xx1.013 xx10^5 xx 30xx10^(-3))/(8.31 xx 300) = 18.3`

Final Pressure, `P_2 = 11 atm`

`= 11 xx 1.013 xx 10^5 N m^(-2)`

Final Volume, V`_2 = 30 litre = 30 xx 10^(-3) cm^3`

Final temperature, `T_2 = 17 + 273 = 290 K`

Final number of moless, `mu_2 = (P_2V_2)/(RT_2) = (11xx 1.013xx 10^5xx 30xx10^(-3))/(8.31 xx 290) = 13.9`

 Number of moles takesout of cyclinder = 18.3 - 13.9 = 4.4 

Mass of gas taken out of cylinder = 4.4 xx 32 g = 140.8 g = 0.141 kg