Q57PE - College Physics (Urone)

Question

Question: A deep-sea diver should breathe a gas mixture that has the same oxygen partial pressure as at sea level, where dry air contains 20.9% oxygen and has a total pressure of 1.01×10⁵ N/m². (a) What is the partial pressure of oxygen at sea level? (b) If the diver breathes a gas mixture at a pressure of 2.00×10⁶ N/m², what percent oxygen should it be to have the same oxygen partial pressure as at sea level?

Step-by-Step Solution

Answer

Answer

The partial pressure of oxygen at sea level is 2.111×10⁴ N/m² and the percentage of oxygen at a pressure of 2.00×10⁶ N/m² is 1.06.

Step 1: Definition of partial pressure

Partial pressure is the pressure exerted by individual gases in a mixture. Partial pressure is the product of the fraction of oxygen present and total pressure. The partial pressure of oxygen is expressed as follows;

${\mathit{P}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{=}\mathbf{ }\mathit{f}\mathit{r}\mathit{a}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathbf{ }{\mathit{O}}_{\mathbf{2}}\mathbf{\times }{\mathit{P}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}$ ………………………….(1)

Where is the partial pressure of oxygen and is the total pressure

Step 2: Solution of part a

The fraction of oxygen and the total pressure is given as;

$$\begin{gathered} {\mathbf{fractionofO}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{9}\mathbf{\%} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{209} \\ {\mathbf{P}}_{\mathbf{total}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}} \end{gathered}$$

Substitute these values in equation (1)

$$\begin{gathered} {}_{{\text{O}}_2}=1.01\times {10}^5 N/m^2\times 0.209 \\ =2.111\times {10}^{4}N/m^2 \end{gathered}$$

Step 3: Solution of part b

In this scenario, the total pressure has changed to 2.00×10⁶ N/m². We have to calculate the percentage of oxygen with the calculated measurement of partial pressure at sea level. Substitute the values in equation (1).

 $$\begin{gathered} 2.111\times {10}^{4}N/m^2 =fraction of O_2\times 2.00\times {10}^{6}N/m^2 \\ fraction of O_2=\frac{{\displaystyle 2.111\times {10}^{4}N/m^2 }}{{\displaystyle 2.00\times {10}^{6}N/m^2 }} \\ =0.0106 \\ =1.06\% \end{gathered}$$