Respuesta :
Answer: The partial pressure of oxygen gas is 2.76 bar
Explanation:
To calculate the number of moles, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 5.57 bar
V = Volume of the gas = 2.19 L
T = Temperature of the gas = 298 K
R = Gas constant = [tex]0.0831\text{ L bar }mol^{-1}K^{-1}[/tex]
n = Total number of moles = ?
Putting values in above equation, we get:
[tex]5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol[/tex]
To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex] ........(1)
where,
[tex]p_A[/tex] = partial pressure of carbon dioxide = 0.318 bar
[tex]p_T[/tex] = total pressure = 5.57 bar
[tex]\chi_A[/tex] = mole fraction of carbon dioxide = ?
Putting values in above equation, we get:
[tex]0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571[/tex]
- Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
We are given:
Moles of nitrogen gas = 0.221 moles
Mole fraction of nitrogen gas, [tex]\chi_{N_2}=\frac{0.221}{0.493}=0.448[/tex]
Calculating the partial pressure of oxygen gas by using equation 1, we get:
Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949
Total pressure of the system = 5.57 bar
Putting values in equation 1, we get:
[tex]p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar[/tex]
Hence, the partial pressure of oxygen gas is 2.76 bar
