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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder ($$R = 8.31 J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 $$O_2 = 32 u$$).

Asked by Pragya Singh | 1 year ago |  122

##### Solution :-

Volume of gas, V1 = 30 litres

= 30 x 10-3 m3

Gauge pressure, P1 = 15 atm

= 15 x 1.013 x 105 P a

Temperature, T1 = 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 n1

The gas equation is given as follows:

P1V1 = n1RT1

Hence,

n1 = $$\dfrac{P_1V_1}{RT_1}$$

$$\dfrac{(15.195 \times 10^5 \times 30 \times 10^{-3)}}{(8.314 \times 300)}$$

= 18.276

But n1 = $$\dfrac{m_1}{M}$$

Where,

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Thus,

m1 = N1M = 18.276 x 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.

Volume, V2 = 30 litres = 30 x 10-3 m3

Gauge pressure, P2 = 11 atm

= 11 x 1.013 x 10P a

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

Hence,

n2 = $$\dfrac{P_2V_2}{RT_2}$$

$$\dfrac{(11.143 \times 10^5 \times 30 \times 10^{-30)}}{ (8.314 \times 290)}$$

= 13.86

But

n2 = $$\dfrac{m_2}{M}$$

Where,

mis the mass of oxygen remaining in the cylinder

Therefore,

m2 = n2 x M = 13.86 x 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

We get,

= 131.74 g

= 0.131 kg

Hence, 0.131 kg of oxygen is taken out of the cylinder

Answered by Pragya Singh | 1 year ago

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