An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder.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= 8.3 J/mol/ K, molecular mass of O₂=32].

Solution

Ideal gas equation is given by,

PV = nRT

∴

n = \frac{PV}{RT}

... (1)

Volume of oxygen gas cylinder, V = 30

l

30 \times 10^{- 3} m^{3}

Pressure, P = 15 atm + 1 atm = 16 atm =

16 \times 1.01 \times 10^{5}

Pa
Temperature, T = 27^o C = 300 K

Putting this in equation (1), we have

n = \frac{16 \times 1.01 \times 10^{5} \times 30 \times 10^{- 3}}{300 \times 8.3} = 19 . 5 moles

After some amount of oxygen is withdrawn from the cylinder,

New pressure, P' = 11 atm + 1 atm = 12 atm
Temperature, T' = 17^o C = 290 K

So, number of moles =

\frac{12 \times 1.01 \times 10^{5} \times 30 \times 10^{- 3}}{290 \times 8.3}

= 15.1 moles

So, number of moles withdrawn from the cylinder is given by,

∆ n = 19.5 - 15.1 = 4.4 moles

So, mass of O₂ taken out is given by,

∆ m = 4.4 \times 32 gm = 140.8 gm

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