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Asked by Kunal Last Modified
Answer 
Harshit Bhardwaj
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:
P1V1 = n1RT1
= 18.276
But, 
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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.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
= 13.86
But, 
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.
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