Question #115478
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)
1
Expert's answer
2020-05-12T10:23:32-0400

Solution.

n=n= 20 breaths per minute;

ma=0.5×103kg;m_a=0.5×10^{−3} kg;

t1=24°C;t_1=24°C;

ϕ1=20%;\phi_1=20\%;

t2=36°C;t_2=36°C;

ϕ2=100%;\phi_2=100\%;

τ=1hour;\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;

ϕ1=ρ1aρ1s100%;    ρ1a=ϕ1ρ1s100%;\phi_1=\dfrac{\rho_{1a}}{\rho_{1s}}\sdot100\%;\implies \rho_{1a}=\dfrac{\phi_1\sdot\rho_{1s}}{100\%};

ρ1s=21.8g/m3\rho_{1s}=21.8g/m^3 - saturated vapor density at a temperature of 24oC (table);

ρ1a=20%21.8g/m3100%=4.36g/m3;\rho_{1a}=\dfrac{20\%\sdot21.8g/m^3}{100\%}=4.36g/m^3;

ϕ2=ρ2aρ2s100%;    ρ2a=ϕ2ρ2s100%;\phi_2=\dfrac{\rho_{2a}}{\rho_{2s}}\sdot100\%;\implies \rho_{2a}=\dfrac{\phi_2\sdot\rho_{2s}}{100\%};

ρ2s=42.1g/m3;\rho_{2s}=42.1g/m^3;

ρ2a=42.1g/m3;\rho_{2a}=42.1g/m^3;

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

ma=ρV    V=maρm_a=\rho V\implies V=\dfrac{m_a}{\rho};

V=600103kg1.189kg/m3=504.63103m3;V=\dfrac{600\sdot 10^{-3}kg}{1.189kg/m^3}=504.63\sdot10^{-3}m^3;

mv=ρ2aV;m_v=\rho_{2a}V;

mv=42.1g/m3504.63103m3=21.24g;m_v=42.1g/m^3\sdot504.63\sdot10^{-3}m^3=21.24g;

Answer: mv=21.24g.m_v=21.24g.






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