Respuesta :
Answer : The partial pressure of [tex]O_2\text{ and }CH_4[/tex] is, 0.281 atm and 0.839 atm respectively.
The total pressure in vessel is, 1.12 atm.
Explanation :
First we have to calculate the moles of oxygen and methane.
[tex]\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{6.00g}{32g/mol}=0.188mol[/tex]
and,
[tex]\text{Moles of }CH_4=\frac{\text{Given mass }CH_4}{\text{Molar mass }CH_4}=\frac{9.00g}{16g/mol}=0.562mol[/tex]
Now we have to calculate the total moles of gas.
Total moles of gas = Moles of oxygen + Moles of methane
Total moles of gas = 0.188 + 0.562
Total moles of gas = 0.75 mol
Now we have to calculate the total pressure of gas by using ideal gas equation.
[tex]PV=nRT[/tex]
where,
P = total pressure of gas = ?
V = total volume of gas = 15.0 L
n = total number of moles gas = 0.75 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of gas = [tex]0^oC=273+0=273K[/tex]
Putting values in above equation, we get:
[tex]P\times 15.0L=0.75mole\times (0.0821L.atm/mol.K)\times 273K[/tex]
[tex]P=1.12atm[/tex]
Now we have to calculate the mole fraction of oxygen and methane.
[tex]\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }O_2+\text{Moles of }CH_4}=\frac{0.188}{0.188+0.562}=0.251[/tex]
and,
[tex]\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }O_2+\text{Moles of }CH_4}=\frac{0.562}{0.188+0.562}=0.749[/tex]
Now we have to calculate the partial pressure of oxygen and methane.
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 = 1.12 atm
[tex]X_i[/tex] = mole fraction of gas
[tex]p_{O_2}=X_{O_2}\times p_T[/tex]
[tex]p_{O_2}=0.251\times 1.12atm=0.281atm[/tex]
and,
[tex]p_{CH_4}=X_{CH_4}\times p_T[/tex]
[tex]p_{CH_4}=0.749\times 1.12atm=0.839atm[/tex]
Thus, the partial pressure of [tex]O_2\text{ and }CH_4[/tex] is, 0.281 atm and 0.839 atm respectively and the total pressure in vessel is, 1.12 atm.
