Suggested languages for you:

Americas

Europe

Q57PE

Expert-verifiedFound in: Page 472

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 1.01×10 ^{5} N/m^{2}. (a) What is the partial pressure of oxygen at sea level? (b) If the diver breathes a gas mixture at a pressure of 2.00×10^{6} N/m^{2}, what percent oxygen should it be to have the same oxygen partial pressure as at sea level?**

**Answer**

The partial pressure of oxygen at sea level is 2.111×10^{4} N/m^{2} and the percentage of oxygen at a pressure of 2.00×10^{6} N/m^{2 }is^{ }1.06.

**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;**

**${{\mathit{P}}}_{{\mathbf{O}}_{\mathbf{2}}}{\mathbf{=}}{\mathbf{\u200a}}{\mathit{f}}{\mathit{r}}{\mathit{a}}{\mathit{c}}{\mathit{t}}{\mathit{i}}{\mathit{o}}{\mathit{n}}{\mathbf{\u200a}}{\mathbf{\u200a}}{\mathit{o}}{\mathit{f}}{\mathbf{\u200a}}{\mathbf{\u200a}}{{\mathit{O}}}_{{\mathbf{2}}}{\mathbf{\times}}{{\mathit{P}}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}$ ………………………….(1)**

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

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

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

Substitute these values in equation (1)

${}_{{\text{O}}_{2}}=1.01\times {10}^{5}N/{m}^{2}\times 0.209\phantom{\rule{0ex}{0ex}}=2.111\times {10}^{4}N/{m}^{2}$

In this scenario, the total pressure has changed to 2.00×10^{6} N/m^{2}. We have to calculate the percentage of oxygen with the calculated measurement of partial pressure at sea level. Substitute the values in equation (1).

$2.111\times {10}^{4}N/{m}^{2}\u200a=fraction\u200a\u200aof\u200a\u200a{O}_{2}\times 2.00\times {10}^{6}N/{m}^{2}\u200a\phantom{\rule{0ex}{0ex}}fraction\u200a\u200aof\u200a\u200a{O}_{2}=\frac{2.111\times {10}^{4}N/{m}^{2}\u200a}{2.00\times {10}^{6}N/{m}^{2}\u200a}\phantom{\rule{0ex}{0ex}}=0.0106\phantom{\rule{0ex}{0ex}}=1.06\%\phantom{\rule{0ex}{0ex}}$

94% of StudySmarter users get better grades.

Sign up for free