Question

PLZ HELP "NO LINKS"

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

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.

$PV = nRT\\P \times 22.4 L = 3.0 mol \times 0.0821 L atm/mol K \times 273.15 K\\P = 3.0 atm$

• The mole fractions of each component:

The mole fraction of $H_{2}$ is calculated as follows.

$Mole fraction = \frac{moles of H_{2}}{moles of H_{2} + moles of N_{2}}\\= \frac{2.0 mol}{(2.0 + 1.0) mol}\\= 0.66$

The mole fraction of $N_{2}$ is as follows.

$Mole fraction = \frac{moles of N_{2}}{moles of H_{2} + moles of N_{2}}\\= \frac{1.0 mol}{(2.0 + 1.0) mol}\\= 0.33$

• The partial pressures of each component:

Partial pressure of $H_{2}$ are as follows.

$P_{H_{2}} = P_{total} \times mole fraction of H_{2}\\= 3.0 atm \times 0.66\\= 1.98 atm$

Partial pressure of $N_{2}$ are as follows.

$P_{N_{2}} = P_{total} \times mola fraction of N_{2}\\= 3.0 atm \times 0.33\\= 0.99 atm$