Question

Air in human lungs has a temperature of 37/5 °C and a saturation vapor density of 0.044/5 kg/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?
a) 0.055/5 g
b) 0.042/5 g
c) 0.033/5 g
d) 0.027/5 g

Answer 1

The maximum loss of water vapor by the person is 0.088 g.

Step-by-step explanation:

The maximum loss of water vapor by the person can be calculated by multiplying the saturation vapor density of air by the change in volume:

Mass of water vapor lost = saturation vapor density x change in volume

Given that the saturation vapor density of air is 44.0 g/m³ and the change in volume is 2.00 L, we can calculate the maximum loss of water vapor:

Maximum loss of water vapor = 44.0 g/m³ x (2.00 L / 1000 L) = 0.088 g

Therefore, the maximum loss of water vapor by the person is 0.088 g.