Answer to Question #91168 in Molecular Physics | Thermodynamics for Deepak patwa

Question

A mixture of air and water vapour at 1.013 bar and 16 degree C has a dew point of 5degree C.calculate the relative and specific humidity's.

Answer 1

Using table for saturated vapor

We have:

Partial pressure for saturated vapor for 16 degree C

$$p''_s=0.01818 (bar)$$

Specific volume

$$\frac{1}{\rho''}=147.2 (m^3/kg)$$

From this equation

$$p''=\frac{\rho''RT}{M}$$

molar mass of water

$$M=0.018 (kg/mol)$$

For relative humidity

$$\phi=\frac{p''}{p''_s}\cdot 100$$ $$\phi=\frac{RT}{\frac{1}{\rho''}Mp''_s}\cdot 100$$ $$\phi=\frac{8.31\cdot 289}{147.2 \cdot 0.018 \cdot 1818}\cdot 100=50 \%$$ $$E=mc^2$$

Partial pressure of air we can receive from Dalton's law

$$p=p'+p''$$ $$p'=p-p''$$

$$p'=101300-906=1.00394 (bar)$$

Further, for air density from gas equation

$$\rho'=\frac{p'M}{RT}$$

$$\rho'=\frac{101394\cdot 0.029}{8.31\cdot 289}=1.212 (kg/m^3)$$

Density of moist air

$$\rho=\rho'+\rho''$$

Specific humidity

$$SH=\frac{\rho''}{\rho'+\rho''}$$

$$SH=\frac{\frac{1}{147.2}}{1.212+\frac{1}{147.2}}=0.0056$$