Answer to Question #86011 in Molecular Physics | Thermodynamics for Beast

Question

Room temperature constant 20 ° C and relative humidity 20%. How many grams of water must be added to the room air to obtain relative humidity to rise to 60%?

Answer 1

Solution:

Relative humidity is the ratio of vapor density to saturation vapor density.

In the first case:

$$\phi_1=\frac{\rho_1} {\rho_0} 100\%$$

Vapor density equals to mass of vapor divided by the volume of the room:

$$\rho_1=\frac{m_1} V$$

Then:

$$\phi_1=\frac{m_1} {\rho_0 V} 100\%$$

Therefore:

$$m_1=\frac{\phi_1\rho_0 V} {100\%}$$

Similarly we can find mass of vapor after adding the water:

$$\phi_2=\frac{\rho_2} {\rho_0} 100\%=\frac{m_2} {\rho_0 V} 100\%$$

$$m_2=\frac{\phi_2\rho_0 V} {100\%}$$

Mass of water that must be added to the room air:

$$\Delta{m}=\frac{\rho_0 V} {100\%} (\phi_2-\phi_1)$$

Saturation vapor density at $20^\circ\:C:$

$$\rho_0=17.3\:g/m^3$$

Then:

$$\Delta{m}=\frac{17.3 V} {100\%} (60\%-20\%)=6.92V\:(g)$$

Answer:

$$\Delta{m}=6.92V\:g$$