Question

a mixture of gases consists of 0.1 kg of oxygen, 1 kg of carbon dioxide, and 0.5 kg of helium is compressed to 17,500 kPa and 20 C. determine volume occupied by the mixture?

Answer 1

Molar mass of O2 = 32 g/mol

mass of O2 = 0.1 Kg

= 100.0 g [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(100.0 g)/(32 g/mol)

= 3.125 mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = 1 Kg

= 1000.0 g [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(1000.0 g)/(44.01 g/mol)

= 22.72 mol

Molar mass of He = 4.003 g/mol

mass of He = 0.5 Kg

= 500.0 g [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol of He,

n = mass of He/molar mass of He

=(500.0 g)/(4.003 g/mol)

= 124.9 mol

so, total number of moles,

n = 3.125 + 22.72 + 124.9 = 150.7 mol

we have:

P= 17500.0 kPa

= (17500.0*1000) pa

= 1.75E7 Pa

= 1.75E7 Pa

= (1.75E7/101325) atm

= 172.7116 atm

n = 150.7 mol

T = 20.0 oC

= (20.0+273) K

= 293 K

we have below equation to be used:

P * V = n*R*T

172.7116 atm * V = 150.7 mol* 0.0821 atm.L/mol.K * 293 K

V = 21.0 L

Answer: 21.0 L