Question Details

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:


[Given, R=100/12 J mol-1 k-1 and molecular mass of O2 = 32, 1 atm pressure = 1.01 × 105 N/m]

Options

A

0.144 kg

B

0.116 kg

C

0.156 kg

D

0.125 kg

Show Answer

Correct Answer :

Option B

0.116 kg

0.116 kg

Solution :

To find the mass of oxygen withdrawn from the cylinder, we first need to determine the number of moles of oxygen remaining in the cylinder after some of it has been withdrawn.

Let the initial state of the oxygen in the cylinder be represented by:
Initial volume, V=30 litres=30×10-3 m3
Initial number of moles, n1=18.20 moles

After some oxygen is withdrawn, the gauge pressure drops to 11 atm. The pressure inside the cylinder is the absolute pressure, which is the sum of gauge pressure and atmospheric pressure:
Absolute pressure, P=Pgauge+Patm=11 atm+1 atm=12 atm

Given that 1 atm=1.01×105 N/m2, the final absolute pressure in SI units is:
P=12×1.01×105 N/m2=12.12×105 N/m2

The final temperature is T=27C=27+273=300 K.
The gas constant is given as R=10012 J mol-1 K-1.

Using the ideal gas equation, PV=n2RT, where n2 is the final number of moles of oxygen remaining in the cylinder:
n2=PVRT

Substituting the given values:
n2=(12.12×105)×(30×10-3)(10012)×300
n2=12.12×30×10210012×300
n2=3636002500=14.544 moles

The number of moles of oxygen withdrawn from the cylinder is:
Δn=n1-n2=18.20-14.544=3.656 moles

The molecular mass of O2 is 32 g/mol=32×10-3 kg/mol.
The mass of oxygen withdrawn is:
m=Δn×M=3.656×32×10-3 kg
m0.11699 kg0.116 kg

Therefore, the mass of the oxygen withdrawn is nearly equal to 0.116 kg.

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