Question

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

Answer 1

Answer:

(a). The maximum loss of water vapor by the person is $8.8\times10^{-2}\ g$

(b). The partial pressure of water vapor is $6.29\times10^{3}\ N/m^2$

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

$m=\rho_{air}\times V_{air}$

Put the value into the formula

$m=44.0\times2.00\times10^{-3}$

$m=0.088\ g$

$m=8.8\times10^{-2}\ g$

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

Using formula of pressure

$PV=nRT$

$P=\dfrac{nRT}{V}$

$P=\dfrac{\rho\times R\times T}{M}$

Put the value into the formula

$P=\dfrac{44\times8.314\times(37+273)}{18.01528}$

$P=6294.82\ N/m^2$

$P=6.29\times10^{3}\ N/m^2$

Hence, (a). The maximum loss of water vapor by the person is $8.8\times10^{-2}\ g$

(b). The partial pressure of water vapor is $6.29\times10^{3}\ N/m^2$