Question

four gases were combined in a gas cylinder with these partial pressures: 3.5 atm n2, 2.8 atm o2, 0.25 atm ar, and 0.15 atm he.
what is the total pressure inside the cylinder?
6.7 atm
complete
what is the mole fraction of n2 in the mixture?
0.52 atm
complete
what is the mole fraction of o2 in the mixture?
atm
what is the mole fraction of ar in the mixture?
atm
done

Explanation:

Step1: Recall Mole Fraction Formula

The formula for mole fraction ($X_a$) of a gas \(a\) is \(X_a=\frac{P_a}{P_T}\), where \(P_a\) is the partial pressure of gas \(a\) and \(P_T\) is the total pressure. We know \(P_T = 6.7\) atm (from the first part), \(P_{O_2}=2.8\) atm, \(P_{Ar}=0.25\) atm.

Step2: Calculate Mole Fraction of \(O_2\)

Using \(X_{O_2}=\frac{P_{O_2}}{P_T}\), substitute \(P_{O_2} = 2.8\) atm and \(P_T=6.7\) atm. So \(X_{O_2}=\frac{2.8}{6.7}\approx0.418\) (or round as needed, but let's do exact division).

Step3: Calculate Mole Fraction of \(Ar\)

Using \(X_{Ar}=\frac{P_{Ar}}{P_T}\), substitute \(P_{Ar} = 0.25\) atm and \(P_T = 6.7\) atm. So \(X_{Ar}=\frac{0.25}{6.7}\approx0.037\).

Answer:

Mole fraction of \(O_2\): \(\frac{2.8}{6.7}\approx0.42\) (or more precisely \(\approx0.4179\))
Mole fraction of \(Ar\): \(\frac{0.25}{6.7}\approx0.037\)

(If we follow the previous precision where \(N_2\) was \(0.52\) (since \(\frac{3.5}{6.7}\approx0.522\approx0.52\)), then for \(O_2\): \(\frac{2.8}{6.7}\approx0.42\) (because \(2.8\div6.7\approx0.4179\approx0.42\)) and \(Ar\): \(\frac{0.25}{6.7}\approx0.037\) (or \(0.04\) if rounding to two decimal places, but more accurately \(\approx0.037\)).)