Respuesta :
Answer : The mole fraction and partial pressure of [tex]CH_4,CO_2[/tex] and [tex]He[/tex] gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of [tex]CH_4[/tex] = 1.79 mole
Moles of [tex]CO_2[/tex] = 1.20 mole
Moles of [tex]He[/tex] = 3.71 mole
Now we have to calculate the mole fraction of [tex]CH_4,CO_2[/tex] and [tex]He[/tex] gases.
[tex]\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}[/tex]
[tex]\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267[/tex]
and,
[tex]\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}[/tex]
[tex]\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179[/tex]
and,
[tex]\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}[/tex]
[tex]\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554[/tex]
Thus, the mole fraction of [tex]CH_4,CO_2[/tex] and [tex]He[/tex] gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of [tex]CH_4,CO_2[/tex] and [tex]He[/tex] gases.
According to the Raoult's law,
[tex]p_i=X_i\times p_T[/tex]
where,
[tex]p_i[/tex] = partial pressure of gas
[tex]p_T[/tex] = total pressure of gas = 5.78 atm
[tex]X_i[/tex] = mole fraction of gas
[tex]p_{CH_4}=X_{CH_4}\times p_T[/tex]
[tex]p_{CH_4}=0.267\times 5.78atm=1.54atm[/tex]
and,
[tex]p_{CO_2}=X_{CO_2}\times p_T[/tex]
[tex]p_{CO_2}=0.179\times 5.78atm=1.03atm[/tex]
and,
[tex]p_{He}=X_{He}\times p_T[/tex]
[tex]p_{He}=0.554\times 5.78atm=3.20atm[/tex]
Thus, the partial pressure of [tex]CH_4,CO_2[/tex] and [tex]He[/tex] gases are, 1.54, 1.03 and 3.20 atm respectively.
