Question

An oxygen cylinder of volume 30 litres has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given: R = `100/12`Jmol⁻¹ K⁻¹,and molecular mass of 12 O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m]

पर्याय

• 0.125 kg

• 0.144 kg

• 0.116 kg

• 0.156 kg

Answer 1

0.116 kg

Explanation:

Given: R = `100/12`Jmol⁻¹ K⁻¹

Molecular mass of 12 O₂ = 32

1 atm pressure = 1.01 × 10⁵ N/m

From the general gas equation

pV = nRT

n = `(pV)/(RT)`

where p = absolute (final) pressure = gauge pressure + atmospheric pressure

p = 11 + 1

= 12 atm

∴ n = no. of moles remaining

= `(12 xx 1.01 xx 10^5 xx 30 xx 10^-3)/(100/12 xx 300)`

= `(36.36 xx 10^2)/(100 xx 25)`

= `36.36/2500 xx 10^2`

= 0.01454 × 10²

= 14.54

No. of moles removed = 18.20 −14.54

= 3.66

Mass removed = 3.66 × Molar mass

= `3.66 xx 32/1000`

= `117.1/1000`

= 0.1171 kg