The pressure of an oxygen containing cylinder is 10.3atm.Under same temperature, another oxygen cylinder having 1/3rd volume and has a pressure 1.1 atm Keeping the temperature same, two cylinders are connected, find the total pressure of the system
Answers
Answer:
Volume of oxygen, V
1
= 30 litres =30×10
−3
m
3
Gauge pressure, P
1
= 15 atm =15×1.013×10
5
Pa
Temperature, T
1
=27
o
C=300K
Universal gas constant, R=8.314Jmol
−1
K
−1
Let the initial number of moles of oxygen gas in the cylinder be n1 .
The gas equation is given as:
P
1
V
1
=n
1
RT
1
∴n
1
=P
1
V
1
/RT
1
=(15.195×10
5
×30×10
−3
)/(8.314×300)=18.276
But n
1
=m
1
/M
Where,
m
1
= Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m
1
=N
1
M=18.276×32=584.84g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V
2
= 30 litres =30×10
−3
m
3
Gauge pressure, P
2
= 11 atm =11×1.013×10
5
Pa
Temperature, T
2
=17
o
C=290K
Let n
2
be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P
2
V
2
=n
2
RT
2
∴n
2
=P
2
V
2
/RT
2
=(11.143×10
5
×30×10
−30
)/(8.314×290)=13.86
But n
2
=m
2
/M
Where,
m
2
is the mass of oxygen remaining in the cylinder
∴m
2
=n
2
×M=13.86×32=453.1g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
=m
1
−m
2
=584.84g–453.1g
=131.74g
=0.131kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
Explanation:
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We know that
Pi Vi = Pf Vf
Now
Since volume becomes 1/3
Pi Vi = 10.3V + 1.1V/3
Pf Vf = 4PV/3
Now
10.3 V + 1.1V/3 = 4PV/3
V and P will be cancelled
10.3 + 1.1/3 = 4P/3
30.9 + 1.1/3 = 4P/3
32/3 = 4P/3
P = 32/3 \times× 3/4
P = 32/8
P = 8 atom