Answer :

__Explanation: We have to consider 2 cases here and find the masses of O _{2}__

__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

V_{1} = 30 litres = 0.03 m^{3}

P_{1} = 15 atm = 15 × 1.01 × 10^{5} Pa

T_{1} = 27+273 = 300 K

If n_{1} is the moles O_{2} in the cylinder, then by using

P_{1}V_{1} = n_{1}RT_{1} ---( Ideal Gas Law )

⇒ n_{1} = =

⇒ n_{1} = 18.276 moles

Molecular weight of O_{2} (M) = 32 g

Initial mass of cylinder (m_{1}) = n_{1}M

⇒ m_{1} = n_{1}M

⇒ m_{1} = (18.276 moles) × ( 32 g)

⇒ m_{1} = 584.84 g

Now,

Let n_{2} be the moles of O_{2} left in the cylinder

V_{2} = 30 litres = 0.03 m^{3}

P_{2} = 11 atm = 11 × 1.01 × 10^{5} Pa

T_{2} = 17+273 = 290 K

Thus n_{2} = =

⇒ n_{1} = 13.86 moles

Therefore, m_{2} be the final mass of O_{2} in the cylinder

⇒ m_{2} = n_{2}M

⇒ m_{2} = (13.86 moles) × ( 32 g)

⇒ m_{2} = 453.1 g

∴ the mass of oxygen taken out of the cylinder

Δm = m_{1}-m_{2}

⇒ 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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