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To determine the partial pressure of xenon (Xe) in the gas mixture, you should follow these steps:
1. Understand the Relationship Between Total and Partial Pressures:
The total pressure of a gas mixture is the sum of the partial pressures of the individual gases in the mixture. This can be expressed as:
[tex]\[ P_{\text{total}} = P_{\text{He}} + P_{\text{Ar}} + P_{\text{Xe}} \][/tex]
2. Identify the Given Values:
- Total pressure ([tex]\(P_{\text{total}}\)[/tex]): 2.10 atm
- Partial pressure of helium ([tex]\(P_{\text{He}}\)[/tex]): 0.300 atm
- Partial pressure of argon ([tex]\(P_{\text{Ar}}\)[/tex]): 0.450 atm
3. Set Up the Equation:
Using the given values, substitute them into the equation:
[tex]\[ 2.10 \, \text{atm} = 0.300 \, \text{atm} + 0.450 \, \text{atm} + P_{\text{Xe}} \][/tex]
4. Solve for the Partial Pressure of Xenon ([tex]\(P_{\text{Xe}}\)[/tex]):
Rearrange the equation to isolate [tex]\(P_{\text{Xe}}\)[/tex]:
[tex]\[ P_{\text{Xe}} = 2.10 \, \text{atm} - (0.300 \, \text{atm} + 0.450 \, \text{atm}) \][/tex]
[tex]\[ P_{\text{Xe}} = 2.10 \, \text{atm} - 0.750 \, \text{atm} \][/tex]
[tex]\[ P_{\text{Xe}} = 1.35 \, \text{atm} \][/tex]
Thus, the partial pressure of Xe in the mixture is [tex]\(1.35\)[/tex] atm.
1. Understand the Relationship Between Total and Partial Pressures:
The total pressure of a gas mixture is the sum of the partial pressures of the individual gases in the mixture. This can be expressed as:
[tex]\[ P_{\text{total}} = P_{\text{He}} + P_{\text{Ar}} + P_{\text{Xe}} \][/tex]
2. Identify the Given Values:
- Total pressure ([tex]\(P_{\text{total}}\)[/tex]): 2.10 atm
- Partial pressure of helium ([tex]\(P_{\text{He}}\)[/tex]): 0.300 atm
- Partial pressure of argon ([tex]\(P_{\text{Ar}}\)[/tex]): 0.450 atm
3. Set Up the Equation:
Using the given values, substitute them into the equation:
[tex]\[ 2.10 \, \text{atm} = 0.300 \, \text{atm} + 0.450 \, \text{atm} + P_{\text{Xe}} \][/tex]
4. Solve for the Partial Pressure of Xenon ([tex]\(P_{\text{Xe}}\)[/tex]):
Rearrange the equation to isolate [tex]\(P_{\text{Xe}}\)[/tex]:
[tex]\[ P_{\text{Xe}} = 2.10 \, \text{atm} - (0.300 \, \text{atm} + 0.450 \, \text{atm}) \][/tex]
[tex]\[ P_{\text{Xe}} = 2.10 \, \text{atm} - 0.750 \, \text{atm} \][/tex]
[tex]\[ P_{\text{Xe}} = 1.35 \, \text{atm} \][/tex]
Thus, the partial pressure of Xe in the mixture is [tex]\(1.35\)[/tex] atm.
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