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\begin{tabular}{|l|l|}
\hline Ideal gas law & [tex]$PV=nRT$[/tex] \\
\hline \multirow{3}{*}{Ideal gas constant} & [tex]$R=8.314 \frac{L \cdot kPa}{mol \cdot K}$[/tex] \\
& or & \\
& [tex]$R=0.0821 \frac{L \cdot atm}{mol \cdot K}$[/tex] \\
\hline Standard atmospheric pressure & [tex]$1 \, atm = 101.3 \, kPa$[/tex] \\
\hline Conversion to Kelvin & [tex]$K = \text{°C} + 273.15$[/tex] \\
\hline
\end{tabular}

A scuba diver's air tank contains oxygen, helium, and nitrogen at a total pressure of 205 atmospheres. The partial pressure of nitrogen is 143 atmospheres, and the partial pressure of helium is 41 atmospheres. What is the partial pressure of oxygen in the tank?

A. 21 atm
B. 103 atm
C. 307 atm
D. 389 atm



Answer :

GinnyAnswer
To find the partial pressure of oxygen in the scuba diver's air tank, we will use the concept of total pressure and partial pressures in a mixture of gases. According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.

Given:
- The total pressure in the tank is 205 atmospheres.
- The partial pressure of nitrogen (P[tex]\(_{\text{N}_2}\)[/tex]) is 143 atmospheres.
- The partial pressure of helium (P[tex]\(_{\text{He}}\)[/tex]) is 41 atmospheres.

To find the partial pressure of oxygen (P[tex]\(_{\text{O}_2}\)[/tex]), we use the equation:

[tex]\[ P_{\text{total}} = P_{\text{N}_2} + P_{\text{He}} + P_{\text{O}_2} \][/tex]

Rearranging to isolate P[tex]\(_{\text{O}_2}\)[/tex], we get:

[tex]\[ P_{\text{O}_2} = P_{\text{total}} - P_{\text{N}_2} - P_{\text{He}} \][/tex]

Substitute the given values into the equation:

[tex]\[ P_{\text{O}_2} = 205 \text{ atm} - 143 \text{ atm} - 41 \text{ atm} \][/tex]

Perform the subtraction:

[tex]\[ P_{\text{O}_2} = 205 \text{ atm} - 184 \text{ atm} = 21 \text{ atm} \][/tex]

Therefore, the partial pressure of oxygen in the tank is 21 atmospheres.

The correct answer is:

A. [tex]\( 21 \)[/tex] atm

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