Question

a gas mixture contains an equal number of moles of he and ne. the total pressure of the mixture is 3.2 atm
part a
what are the partial pressures of he and ne?
∘ ( p_{\text{he}} = 2.1 , \text{atm}; , p_{\text{ne}} = 1.1 , \text{atm} )
∘ ( p_{\text{he}} = 1.1 , \text{atm}; , p_{\text{ne}} = 2.1 , \text{atm} )
∘ ( p_{\text{he}} = 1.6 , \text{atm}; , p_{\text{ne}} = 1.6 , \text{atm} )
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Explanation:

Step1: Recall Dalton's Law and Mole Fraction

Dalton's law states that partial pressure of a gas ($P_i$) is $P_i = X_i \times P_{total}$, where $X_i$ is the mole fraction. Since moles of He ($n_{He}$) and Ne ($n_{Ne}$) are equal, $n_{He}=n_{Ne}=n$. Mole fraction of He ($X_{He}$) is $\frac{n}{n + n}=\frac{1}{2}$, same for Ne ($X_{Ne}=\frac{1}{2}$).

Step2: Calculate Partial Pressures

Given $P_{total}=3.2$ atm. For He: $P_{He}=X_{He} \times P_{total}=\frac{1}{2} \times 3.2 = 1.6$ atm. For Ne: $P_{Ne}=X_{Ne} \times P_{total}=\frac{1}{2} \times 3.2 = 1.6$ atm. So the option with $P_{He}=1.6$ atm and $P_{Ne}=1.6$ atm is correct.

Answer:

The option with $P_{\text{He}} = 1.6$ atm; $P_{\text{Ne}} = 1.6$ atm (the third option in the list of choices).