Answer to Question #125255 in General Chemistry for Jay Shendurnikar

Question

A mixture of oxygen and nitrogen (equal proportions by mass) is contained in a rigid
cylinder at a temperature of 30 oC and 2 bar gauge pressure. If the volume of the cylinder is
10 m3
calculate the mass of oxygen present in the cylinder. If the temperature of the cylinder
is changed to 100 oC, calculate the new partial pressure of oxygen gas.

Answer 1

Let, x g of N₂ and O₂ mixed

Moles of N₂ = x/28

Moles of O₂ = x/32

Total Moles = x/28 + x/32 = 0.066964285x

Total pressure (P) = atmospheric pressure + gauge pressure

                      = (1.013+2) bar

                     = 3.013 bar =3.013× 0.987 atm   =   2.974 atm

T = (273+30) K  = 303 K

V= 10 m³ = 10000 L

R = 0.082 L atm K^-1 mol^-1

 PV = nRT                n= Total number of mole

n = PV/RT = (2.974 × 10000 ) / ( 0.082 × 303) mol = 1196.97 mol

0.066964285x = 1196.97

x = 17874.75 g (Answer)

The mass of oxygen present in cylinder is 17874.75 g.

Moles of O₂ = 17874.75 / 32 = 558.58

At 100 °C = 373 K,     P_O2 = Partial pressure of O₂

P_O2 = n_O2RT / V = (558.58 × 0.082 × 373) / 10000 atm = 1.708 atm (Answer)

The new partial pressure of oxygen gas is 1.708 atm.