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Sagot :
Answer: (a) Mole fraction of [tex]H_{2}[/tex] is 0.66.
Mole fraction of [tex]N_{2}[/tex] is 0.33
(b) The partial pressure of [tex]H_{2}[/tex] is 1.98 atm.
The partial pressure of [tex]N_{2}[/tex] is 0.99 atm.
(c) The total pressure is 3.0 atm
Explanation:
Given: Volume = [tex]22.4 dm^{3}[/tex] (1 [tex]dm^{3}[/tex] = 1 L) = 22.4 L
Moles of [tex]H_{2}[/tex] = 2.0 mol
Moles of [tex]N_{2}[/tex] = 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.
[tex]PV = nRT\\[/tex]
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.
[tex]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[/tex]
- The mole fractions of each component:
The mole fraction of [tex]H_{2}[/tex] is calculated as follows.
[tex]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[/tex]
The mole fraction of [tex]N_{2}[/tex] is as follows.
[tex]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[/tex]
- The partial pressures of each component:
Partial pressure of [tex]H_{2}[/tex] are as follows.
[tex]P_{H_{2}} = P_{total} \times mole fraction of H_{2}\\= 3.0 atm \times 0.66\\= 1.98 atm[/tex]
Partial pressure of [tex]N_{2}[/tex] are as follows.
[tex]P_{N_{2}} = P_{total} \times mola fraction of N_{2}\\= 3.0 atm \times 0.33\\= 0.99 atm[/tex]
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