Answer to Question #115478 in Molecular Physics | Thermodynamics for sathu

Question

You are breathing at a rate of 20 breaths per minute in a warm, dry environment.
Each time you take a breath, you breathe in 0.5×10−3 kg of air at a temperature of 24°C and with a relative humidity of 20%. You breathe out the same mass of air but at a temperature of 36°C and a relative humidity of 100%.
How much water vapour do you exhale in one hour under these circumstances? (to 2 s.f and in g)

Answer 1

Solution.

$n=$ 20 breaths per minute;

$m_a=0.5×10^{−3} kg;$

$t_1=24°C;$

$\phi_1=20\%;$

$t_2=36°C;$

$\phi_2=100\%;$

$\tau=1 hour;$

You take 20×60=1200 breaths per hour

In an hour you inhale 1200×0.5×10^−3 kg=600×10^−3 kg;

$\phi_1=\dfrac{\rho_{1a}}{\rho_{1s}}\sdot100\%;\implies \rho_{1a}=\dfrac{\phi_1\sdot\rho_{1s}}{100\%};$

$\rho_{1s}=21.8g/m^3$ - saturated vapor density at a temperature of 24^oC (table);

$\rho_{1a}=\dfrac{20\%\sdot21.8g/m^3}{100\%}=4.36g/m^3;$

$\phi_2=\dfrac{\rho_{2a}}{\rho_{2s}}\sdot100\%;\implies \rho_{2a}=\dfrac{\phi_2\sdot\rho_{2s}}{100\%};$

$\rho_{2s}=42.1g/m^3;$

$\rho_{2a}=42.1g/m^3;$

The density of dry air is equal to 1.189kg/m³;

$m_a=\rho V\implies V=\dfrac{m_a}{\rho}$;

$V=\dfrac{600\sdot 10^{-3}kg}{1.189kg/m^3}=504.63\sdot10^{-3}m^3;$

$m_v=\rho_{2a}V;$

$m_v=42.1g/m^3\sdot504.63\sdot10^{-3}m^3=21.24g;$

Answer: $m_v=21.24g.$