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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27ºC. The mass of the oxygen withdrawn from the cylinder is nearly equal to :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 O 2 = 32, 1 atm pressure = 1.01 × 10 5 N m -2 ]

Options

A

0.125 kg

B

0.144 kg

C

0.116 kg

D

0.156 kg

Show Answer

Correct Answer :

Option C

0.116 kg

0.116 kg

Solution :

To find the mass of the oxygen withdrawn from the cylinder, we can determine the initial number of moles of oxygen and the final number of moles of oxygen remaining in the cylinder after some gas is withdrawn.


1. Initial State:
The volume of the cylinder, V=30 litres=30×10-3 m3.
The initial number of moles of oxygen, n1=18.20 moles.


2. Final State:
Let the final pressure inside the cylinder be P. The question states that the gauge pressure drops to 11 atm. Assuming the atmospheric pressure is 1 atm, the absolute final pressure P inside the cylinder is:
P=Pgauge+Patm=11 atm+1 atm=12 atm.
Converting this pressure to SI units (N m-2):
P=12×1.01×105 N m-2=12.12×105 N m-2.


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


3. Calculating the final number of moles (n2):
Using the ideal gas equation, PV=n2RT:
n2=PVRT


Substituting the values:
n2=(12.12×105)×(30×10-3)(10012)×300
n2=363602500=14.544 moles.


4. Calculating the number of moles withdrawn (Δn):
Δn=n1-n2=18.20-14.544=3.656 moles.


5. Converting moles to mass:
The molecular mass of O2 is 32 g mol-1=32×10-3 kg mol-1.
The mass of oxygen withdrawn is:
m=Δn×M=3.656×32×10-3 kg
m0.11699 kg0.116 kg.


Therefore, the mass of oxygen withdrawn is approximately 0.116 kg.

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