4g of oxygen gas initially at STP is adiabatically compressed to a third of its original volume,find the values of,the pressure and temperature
Answers
Answer:
Volume of oxygen, V1= 30 litres =30×10−3m3
Gauge pressure, P1= 15 atm =15×1.013×105Pa
Temperature, T1=27oC=300K
Universal gas constant, R=8.314Jmol−1K−1
Let the initial number of moles of oxygen gas in the cylinder be n1 .
The gas equation is given as:
P1V1=n1RT1
∴n1=P1V1/RT1
=(15.195×105×30×10−3)/(8.314×300)=18.276
But n1=m1/M
Where,
m1= Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m1=N1M=18.276×32=584.84g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V2= 30 litres =30×10−3m3
Gauge pressure, P2= 11 atm =11×1.013×105Pa
Temperature, T2=17oC=290K
Let n2 be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P2V2=n2RT2
∴n2=P2V2/RT2
=(11.143×105×30×10−30)/(8.314×290)=13.86
But n2=m2/M
Where,
m2 is the mass of oxygen remaining in the cylinder
∴m2=n2×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
=m1−m2
=584.84g–453.1g
=131.74g
=0.131kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.