Question

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32μ).

Answer 1

use
PV=nRT
(15atm)(30litres)=nR (300k)
n=15/10R
finally
( 11atm)(30litres)=n' R (290)
n'=33/29R
hence mole of oxygen taken out
=15/10R-33/29R=105/290R=21/58R
now
mole =weight /molar weight
21/58R=weight/(0.032kg)
(21/58R)(0.032kg)=weight of oxygen taken out

Answer 2

vol. of oxygen V1=30 litre 30×10^- 3 m^3
gauge pressure P1=15atm 15×1.013×10^5 Pa
temperature T1=27 degree Celsius =300K
universal gas constant R=8.314I mole^-1K^- 1
let initial no. of moles of oxygen gas be n1

gas equation is
P1V1=n1RT1

n1=P1V1/RT1
15.195×10^5×30×10^-3/(8.314)×300
18.276

but,
n1=m1/M
where m1 is initial mass of gas
M is molecular mass of oxygen =32g
m1=n1M =18.276×32
=584.84g

after some oxygen is withdrawn from cylinder
pressure and temperature reduces

V2=30litre 30×10^- 3m^3
gauge pressure P2=11atm 11×1.013×10^5 Pa
temperature T2=17 degree Celsius =290K

let n2 be no. of moles of oxygen gas left
the gas equation is
aP2V2=n2RT2

n2=P2V2/RT2
11.143×10^5×30×10^- 3/8.314×290
13.86
but,
n2=m2/M
where m2 is mass of oxygen gas left in cylinder
m2=n2M
13.86×32=453.1g
mass of oxygen gas taken out from cylinder is given by following relation
initial mass of oxygen in cylinder- final mass of oxygen in cylinder
m1- m2

584.84- 453.1 g
131.74g
0.131kg
therefore 0.131 kg of oxygen gas is taken out of
cylinder