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Q59PE

Expert-verifiedFound in: Page 472

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Question:**** ****Air in human lungs has a temperature of 37.0ºC and a saturation vapor density of 44.0 g/m ^{3}. (a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person? (b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of 6.31×10^{3} N/m^{2}.**

**Answer**

The maximum water vapor exhaled by the person is 0.088 g and the partial pressure of water vapor is equal to the vapor pressure.

**Density is the ratio of mass to volume. Therefore, mass is the product of density and volume.**

The loss of water vapor from the person’s body is equal to the mass of oxygen exhaled by the person. Here, volume and density are given.

$density\u200a=44.0g/{m}^{3}\phantom{\rule{0ex}{0ex}}volume\u200a=200\u200aL\phantom{\rule{0ex}{0ex}}=2.00\u200aL\left(\frac{1\u200a{m}^{3}}{1000L}\right)\phantom{\rule{0ex}{0ex}}=0.002{m}^{3}\phantom{\rule{0ex}{0ex}}$

$Mass\u200a=density\times volume\phantom{\rule{0ex}{0ex}}=44.0g/{m}^{3}\times 0.002{m}^{3}\phantom{\rule{0ex}{0ex}}=0.088g\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, 0.088 g of water vapor was lost from the person.

**According to the gas law,**

${\mathbf{\text{PV = nRT}}}\phantom{\rule{0ex}{0ex}}{\mathbf{\text{PV =}}}\frac{\mathbf{\text{mass}}}{\mathbf{\text{molar}}\mathbf{\u200a}\mathbf{\text{mass}}}{\mathbf{\text{RT}}}\phantom{\rule{0ex}{0ex}}{\mathbf{\text{P =}}}\frac{\text{mass}}{\text{molar}\u200a\text{mass\xd7v}}{\mathbf{\text{\xd7RT}}}\phantom{\rule{0ex}{0ex}}{\mathbf{\text{=}}}\frac{density}{\text{molar}\u200a\text{mass}}{\mathbf{\text{RT}}}$

**Here, is the pressure, is the volume, is the number of moles, is the universal gas constant and is the temperature.**

The value of density and temperature are given.

$Density\u200a=\u200a44.0g/{m}^{3}\text{}\phantom{\rule{0ex}{0ex}}T=\u200a37.0C\phantom{\rule{0ex}{0ex}}=(37.0+273.15)K\phantom{\rule{0ex}{0ex}}=310.15\u200aK\phantom{\rule{0ex}{0ex}}$

The value of the universal gas constant is 8.314 JK^{-1}mol^{-1} and the molar mass of water is 18g. The partial pressure of water vapor can be calculated by substituting these values in the expression of pressure.

$P=\frac{44.0g/{m}^{3}}{18g}\times 8.314\text{}J{K}^{-1}mo{l}^{-1}\times 310.15\phantom{\rule{0ex}{0ex}}=6.303\times {10}^{3}N/{m}^{3}$

This value is almost equal to the vapor pressure which is equal to 6.31×10^{3} N/m^{2}. So, partial pressure of water vapor and vapor pressure are equal.

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