# An oxygen c

Explanation: We have to consider 2 cases here and find the masses of O2 in each case and then by calculating the difference of those masses gives the the mass of oxygen taken out of the cylinder.

Given:

Initially, Case1

V1 = 30 litres = 0.03 m3

P1 = 15 atm = 15 × 1.01 × 105 Pa

T1 = 27+273 = 300 K

If n1 is the moles O2 in the cylinder, then by using

P1V1 = n1RT1 ---( Ideal Gas Law )

n1 = =

n1 = 18.276 moles

Molecular weight of O2 (M) = 32 g

Initial mass of cylinder (m1) = n1M

m1 = n1M

m1 = (18.276 moles) × ( 32 g)

m1 = 584.84 g

Now,

Let n2 be the moles of O2 left in the cylinder

V2 = 30 litres = 0.03 m3

P2 = 11 atm = 11 × 1.01 × 105 Pa

T2 = 17+273 = 290 K

Thus n2 = =

n1 = 13.86 moles

Therefore, m2 be the final mass of O2 in the cylinder

m2 = n2M

m2 = (13.86 moles) × ( 32 g)

m2 = 453.1 g

the mass of oxygen taken out of the cylinder

Δm = m1-m2

Δm = 584.84 g - 453.1 g

Δm = 131.74 g

The mass of oxygen taken out of the cylinder is 131.74 g.

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