Questions: A mixture of He, N2, and Ar has a pressure of 29.8 atm at 28.0°C. If the partial pressure of He is 2371 torr and that of Ar is 967 mm Hg, what is the partial pressure of N2? PN2= atm

A mixture of He, N2, and Ar has a pressure of 29.8 atm at 28.0°C. If the partial pressure of He is 2371 torr and that of Ar is 967 mm Hg, what is the partial pressure of N2? PN2= atm

Transcript text: A mixture of $\mathrm{He}, \mathrm{N}_{2}$, and Ar has a pressure of 29.8 atm at $28.0^{\circ} \mathrm{C}$. If the partial pressure of He is 2371 torr and that of Ar is 967 mm Hg , what is the partial pressure of $\mathrm{N}_{2}$ ? $P_{\mathrm{N}_{2}}=$ $\square$ atm

Solution

Solution Steps

Step 1: Convert Partial Pressures to Consistent Units

First, we need to convert the partial pressures of helium (He) and argon (Ar) from torr and mm Hg to atm, since the total pressure is given in atm.

Convert He from torr to atm:

\[ P_{\text{He}} = \frac{2371 \, \text{torr}}{760 \, \text{torr/atm}} = 3.1197 \, \text{atm} \]
Convert Ar from mm Hg to atm:

\[ P_{\text{Ar}} = \frac{967 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 1.2724 \, \text{atm} \]

Step 2: Use Dalton's Law of Partial Pressures

According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of each component:

\[ P_{\text{total}} = P_{\text{He}} + P_{\text{N}_2} + P_{\text{Ar}} \]

Given that $P_{\text{total}} = 29.8 \, \text{atm}$, we can solve for the partial pressure of nitrogen ($P_{\text{N}_2}$):

\[ P_{\text{N}_2} = P_{\text{total}} - P_{\text{He}} - P_{\text{Ar}} \]

Step 3: Calculate the Partial Pressure of $ \mathrm{N}_2 $

Substitute the known values into the equation:

\[ P_{\text{N}_2} = 29.8 \, \text{atm} - 3.1197 \, \text{atm} - 1.2724 \, \text{atm} \]

\[ P_{\text{N}_2} = 25.4079 \, \text{atm} \]