To calculate the partial pressure of O2, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume = 2.97 L
n = number of moles
R = gas constant = 0.08314 L bar K^-1 mol^-1
T = temperature = 298 K
We can start by calculating the total number of moles of gas in the bottle:
n_total = PV/RT
n_total = (5.68 bar)(2.97 L)/(0.08314 L bar K^-1 mol^-1)(298 K)
n_total = 0.725 mol
We know that the mixture contains 0.225 mol N2, so we can calculate the number of moles of the other gases:
n_other = n_total - n_N2
n_other = 0.725 mol - 0.225 mol
n_other = 0.500 mol
We also know that the partial pressure of CO2 is 0.309 bar, so we can calculate the number of moles of CO2:
n_CO2 = P_CO2 V/RT
n_CO2 = (0.309 bar)(2.97 L)/(0.08314 L bar K^-1 mol^-1)(298 K)
n_CO2 = 0.0112 mol
Now we can use the mole fractions of O2 and N2 to calculate the partial pressure of O2:
X_O2 = n_O2/n_other
X_N2 = n_N2/n_other
We know that the mole fraction of N2 is 0.225/0.500 = 0.450, so:
X_N2 = 0.450
Therefore:
X_O2 = 1 - X_N2
X_O2 = 1 - 0.450
X_O2 = 0.550
Now we can use the ideal gas law to calculate the partial pressure of O2:
P_O2 = n_O2 RT/V
P_O2 = X_O2 n_other RT/V
P_O2 = (0.550)(0.500 mol)(0.08314 L bar K^-1 mol^-1)(298 K)/(2.97 L)
P_O2 = 0.876 bar
Therefore, the partial pressure of O2 in the mixture is 0.876 bar.