Respuesta :
Answer: The partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
We know that:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Rearranging the above equation:
[tex]M=\frac{dRT}{P}[/tex]
where,
d = density of gas mixture = 2.2 g/L
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]26^oC=[26+273]K=299K[/tex]
P = pressure of the mixture = 0.80 atm
M = average molar mass of mixture
Putting values in above equation, we get:
[tex]M_{avg}=\frac{2.2g/L\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 299}{0.80atm}\\\\M_{avg}=67.5g/mol[/tex]
We know that:
Molar mass of [tex]NO_2[/tex] = 46 g/mol
Molar mass of [tex]N_2O_4[/tex] = 92 g/mol
Let the mole fraction of [tex]NO_2[/tex] be 'x' and that of [tex]N_2O_4[/tex] be '(1-x)'
For average molar mass calculation:
[tex]M_{avg}=M_{NO_2}\chi_{NO_2}+M_{N_2O_4}\chi_{N_2O_4}[/tex]
Putting values in above equation:
[tex]67.5=46x+92(1-x)\\\\x=0.533[/tex]
Mole fraction of [tex]N_2O_4[/tex] = (1 - x) = (1 - 0.533) = 0.467
To calculate the partial pressure, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex]
- For [tex]NO_2[/tex] :
We are given:
[tex]p_T=0.80atm\\\chi_{NO_2}=0.533[/tex]
Putting values in above equation, we get:
[tex]p_{NO_2}=0.80atm\times 0.533\\\\p_{NO_2}=0.426atm[/tex]
- For [tex]N_2O_4[/tex] :
We are given:
[tex]p_T=0.80atm\\\chi_{N_2O_4}=0.467[/tex]
Putting values in above equation, we get:
[tex]p_{N_2O_4}=0.80atm\times 0.467\\\\p_{N_2O_4}=0.374atm[/tex]
Hence, the partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
