Question -
Answer -
Volume of gas, V1 =30 litres = 30 x 10-3 m3
Gauge pressure, P1 =15 atm = 15 x 1.013 x 105 P a
Temperature, T1 =270 C = 300 K
Universal gas constant, R = 8.314 J mol-1 K-1
Let the initial number of moles of oxygen gas in the cylinder ben1
The gas equation is given as follows:
P1V1 = n1RT1
Hence,
n1 =P1V1 / RT1
= (15.195 x 105 x30 x 10-3) / (8.314 x 300)
= 18.276
But n1 =m1 / M
Where,
m1 =Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
Thus,
m1 =N1M = 18.276 x 32 =584.84 g
After some oxygen is withdrawn from the cylinder, the pressureand temperature reduce.
Volume, V2 =30 litres = 30 x 10-3 m3
Gauge pressure, P2 =11 atm
= 11 x 1.013 x 105 Pa
Temperature, T2 =170 C = 290 K
Let n2 bethe number of moles of oxygen left in the cylinder
The gas equation is given as:
P2V2 = n2RT2
Hence,
n2 =P2V2 / RT2
= (11.143 x 105 x30 x 10-30) / (8.314 x 290)
= 13.86
But
n2 =m2 / M
Where,
m2 isthe mass of oxygen remaining in the cylinder
Therefore,
m2 =n2 x M = 13.86 x32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by therelation:
Initial mass of oxygen in the cylinder – Final mass of oxygen inthe cylinder
= m1 –m2
= 584.84 g – 453.1 g
We get,
= 131.74 g
= 0.131 kg
Hence, 0.131 kg of oxygen is taken out of the cylinder