Question

PLZ HELP "NO LINKS"

A vessel of volume 22.4 dm3 contains 2.0 mol H2(g) and 1.0 mol N2(g) at 273.15 K.
(a) Calculate the mole fractions of each component.
H2:

N2:

(b) Calculate the partial pressures of each component.
H2:

N2:

(c) Calculate the total pressure.

Thanks!

Answer 1

Answer: (a) Mole fraction of
`H_(2)` is 0.66.

Mole fraction of
`N_(2)` is 0.33

(b) The partial pressure of
`H_(2)` is 1.98 atm.

The partial pressure of
`N_(2)` is 0.99 atm.

(c) The total pressure is 3.0 atm

Step-by-step explanation:

Given: Volume =
`22.4 dm^(3)` (1
`dm^(3)` = 1 L) = 22.4 L

Moles of
`H_(2)` = 2.0 mol

Moles of
`N_(2)` = 1.0 mol

Total moles = (2.0 + 1.0) mol = 3.0 mol

Temperature = 273.15 K

• Now, using ideal gas equation the total pressure is calculated as follows.

`PV = nRT\\`

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

`PV = nRT\\P * 22.4 L = 3.0 mol * 0.0821 L atm/mol K * 273.15 K\\P = 3.0 atm`

• The mole fractions of each component:

The mole fraction of
`H_(2)` is calculated as follows.

`Mole fraction = (moles of H_(2))/(moles of H_(2) + moles of N_(2))\\= (2.0 mol)/((2.0 + 1.0) mol)\\= 0.66`

The mole fraction of
`N_(2)` is as follows.

`Mole fraction = (moles of N_(2))/(moles of H_(2) + moles of N_(2))\\= (1.0 mol)/((2.0 + 1.0) mol)\\= 0.33`

• The partial pressures of each component:

Partial pressure of
`H_(2)` are as follows.

`P_{H_(2)} = P_(total) * mole fraction of H_(2)\\= 3.0 atm * 0.66\\= 1.98 atm`

Partial pressure of
`N_(2)` are as follows.

`P_{N_(2)} = P_(total) * mola fraction of N_(2)\\= 3.0 atm * 0.33\\= 0.99 atm`