Ncert Exercises & Solutions

13.4 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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J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, the molecular mass of $O_{2}$ = 32 u).

Initial volume of oxygen,

V_{1}

= 30 litres = 30 ×

10^{- 3} m^{3}

Initial gauge pressure,

P_{1}

= 15 atm = 15 × 1.013 ×

10^{5}

Initial temperature,

T_{1}

= 27°C = 300 K

Universal gas constant, R = 8.314 J mol

^{- 1} k^{- 1}

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

n_{1}

According to the ideal gas equation:

P_{1} V_{1} = P_{2} V_{2}

n_{1} = \frac{P_{1} V_{1}}{{RT}_{1}}

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

= 18.276

Initial mass of

O_{2}

m_{1} = n_{1} M

= 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.

Let

n_{2}

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

Volume,

V_{2}

= 30 litres = 30 × 1

0^{- 3} m^{3}

Gauge pressure,

P_{2}

= 11 atm = 11 × 1.013 ×

10^{5}

Temperature,

T_{2}

= 17°C = 290 K

Again according to the ideal gas equation:

n_{2} = \frac{P_{2} V_{2}}{{RT}_{2}}

= \frac{11.143 \times 10^{5} \times 30 \times 10^{- 3}}{8.314 \times 290}

= 13.86

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

= 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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