Question

Air in human lungs has a temperature of 37.0°C and a saturation vapor density of 44.0 g/m³.

(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 × 10³ N/m².

Answer 1

(a). The maximum loss of water vapor by the person is
`8.8*10^(-2)\ g`

(b). The partial pressure of water vapor is
`6.29*10^(3)\ N/m^2`

Step-by-step explanation:

Given that,

Temperature = 37.0°C

Volume of air = 2.00 L

Density of vapor = 44.0 g/m³

We need to calculate the maximum loss of water vapor by the person

Using formula of density

`m=\rho_(air)* V_(air)`

Put the value into the formula

`m=44.0*2.00*10^(-3)`

`m=0.088\ g`

`m=8.8*10^(-2)\ g`

(b). We need to calculate the partial pressure of water vapor

Using formula of pressure

`PV=nRT`

`P=(nRT)/(V)`

`P=(\rho* R* T)/(M)`

Put the value into the formula

`P=(44*8.314*(37+273))/(18.01528)`

`P=6294.82\ N/m^2`

`P=6.29*10^(3)\ N/m^2`

Hence, (a). The maximum loss of water vapor by the person is
`8.8*10^(-2)\ g`

(b). The partial pressure of water vapor is
`6.29*10^(3)\ N/m^2`