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Q49PE

Expert-verifiedFound in: Page 471

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Question: Dry air is 78.1% nitrogen. What is the partial pressure of nitrogen when the atmospheric pressure is 1.01×10 ^{5} N/m^{2}?**

**Answer**

The partial pressure of nitrogen is $7.89\times {10}^{4}N/{m}^{2}$

**We apply the formula for partial pressure which is to find partial pressure by multiplying nitrogen percentage by total pressure.**

Total pressure = Atmospheric pressure =$\text{1.01}\times {\text{10}}^{\text{5}}{\text{N/m}}^{2}$

Percentage of nitrogen =78.1%

The formula for partial pressure ${\text{P}}_{\text{partial}}\text{=}\frac{\text{\% Nitrogen}}{\text{100}}\times {\text{P}}_{\text{total}}$

Here, P_{partial} partial pressure and P_{total } total pressure.

The partial pressure of nitrogen in the atmosphere

${\text{P}}_{\text{partial}}=\frac{\%Nitrogen}{100}\times {\text{P}}_{\text{total}}\phantom{\rule{0ex}{0ex}}{\text{P}}_{\text{partial}}=\frac{78.1(1.01\times {10}^{5}{\text{N/m}}^{2})}{100}\times {\text{P}}_{\text{total}}\phantom{\rule{0ex}{0ex}}{\text{=7.89\xd710}}^{4}{\text{N/m}}^{2}$

Therefore, partial pressure of nitrogen is \({\rm{7}}{\rm{.89 \times 1}}{{\rm{0}}^{\rm{4}}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\)

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