__Question 13.4__:*R*= 8.31 J mol

^{–1}K

^{–1}, molecular mass of O

_{2}= 32μ).

__Solution__:*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 X 10

^{5}X 30 X 10

^{-3}) / (8.314 X 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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.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 X 10

^{5}X 30 X 10

^{-3}) / (8.314 X 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.

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