Question

the USEPA assumes that each person breathes 20 cubic meters of air per day. If the air we breathe in has 50% relative humidity at 20 °C, and we breathe it out with 100% relative humidity at 37 °C, how many milliliters of water would we need to drink each day just to replace the water lost by breathing? The vapor pressure of water at 20 °C is 17.5 mmHg and at 37 °C it is 47.1 mmHg. Assume the volume of air
breathed in is the same as that breathed out.

Answer 1

Answer on Question #47306 – Chemistry – Inorganic Chemistry

Question

The USEPA assumes that each person breathes 20 cubic meters of air per day. If the air we breathe in has 50% relative humidity at 20 °C, and we breathe it out with 100% relative humidity at 37 °C, how many milliliters of water would we need to drink each day just to replace the water lost by breathing? The vapor pressure of water at 20 °C is 17.5 mmHg and at 37 °C it is 47.1 mmHg. Assume the volume of air breathed in is the same as that breathed out.

Given

V = 20 m³/day

φ_in = 50%

φ_out = 100%

T_in = 20 °C = 293 K

T_out = 37 °C = 310 K

e_*w(20°C) = 17.5 mmHg

e'_w (37°C) = 47.1 mmHg

V_w = ?

Solution

Relative humidity by definition is the ratio of the partial pressure of water vapor (p_w) in the mixture to the equilibrium vapor pressure of water (p'_w) at a given temperature.

$$\varphi_{in} = \frac{p_w^{in}}{p_w^* (20^\circ C)} 100\%$$ $$\varphi_{out} = \frac{p_w^{out}}{p_w^* (37^\circ C)} 100\%$$

Hence

$$p_w^{in} = \frac{\varphi_{in} \cdot p_w^* (20^\circ C)}{100\%} = \frac{50\% \cdot 17.5\ \text{mmHg}}{100\%} = 8.75\ \text{mmHg} = 1167\ \text{Pa}$$ $$p_w^{out} = \frac{\varphi_{out} \cdot p_w^* (37^\circ C)}{100\%} = \frac{100\% \cdot 47.1\ \text{mmHg}}{100\%} = 47.1\ \text{mmHg} = 6279\ \text{Pa}$$

According to ideal gas equation

$$pV = \frac{m}{M} RT$$

where M – molar weight (for water M = 18 g/mol), p – pressure (which should be replaced by partial pressure of water in our case), V – volume, R – ideal gas constant (R = 8.314 J/mol·K), T – absolute temperature.

Mass of water breathed in and out can be calculated from the ideal gas equation. When substituting volume in m³/day, and molar weight in g/mol, the result is obtained in g/day.

$$m_{in} = \frac{p_w^{in} \cdot V \cdot M}{R \cdot T_{in}} = \frac{1167 \cdot 20 \cdot 18}{8.314 \cdot 293} = 172\ \text{g/day}$$ $$m_{out} = \frac{p_w^{out} \cdot V \cdot M}{R \cdot T_{out}} = \frac{6279 \cdot 20 \cdot 18}{8.314 \cdot 310} = 877\ \text{g/day}$$

Mass of water lost by breathing each day is

$$m = m_{out} - m_{in} = 877 - 172 = 705\ \text{g/day}$$

Volume of water needed to be drunk a day to replace the water lost by breathing

$$V_w = \frac{m}{\rho} = \frac{705}{0.9982} = 706\ \text{ml/day}$$

Answer: to replace the water lost by breathing one need to drink 706 ml of water each day.

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