Answer to Question #86285 in Molecular Physics | Thermodynamics for Mal

Question

Room temperature is 10 ° C and relative humidity 75%. Temperature raised to 20 ° C. What is the relative humidity of the air now? How much water should be added to the air to get 50% relative humidity? The temperature remains constant.

Answer 1

Solution:

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

Relative humidity of the air at $10^\circ C:$

$$\phi_1=\frac{\rho} {\rho_{01}}100\%\:(1)$$

From previous equation we can find vapor density:

$$\rho=\frac{\phi_1 \rho_{01}} {100\%}\:(2)$$

Saturation vapor density at $10^\circ C\:\:\rho_{01}=9.4\:g/m^3.$

Relative humidity of the air at $20^\circ C:$

$$\phi_2=\frac{\rho} {\rho_{02}}100\%\:(3)$$

Saturation vapor density at $20^\circ C\:\:\rho_{02}=17.3\:g/m^3.$

Let's substitute (2) into (3):

$$\phi_2=\frac{\phi_1 \rho_{01}100\%} {\rho_{02}100\%}=\frac{\phi_1 \rho_{01}} {\rho_{02}}=\frac{75\%\cdot 9.4} {17.3}=41\%$$

Let's find how much water should be added to the air to get 50% relative humidity. Relative humidity of the air in this case can be found using equation:

$$\phi_3=\frac{\rho'} {\rho_{03}}100\%\:(4)$$

The temperature remains constant, thus $\rho_{03}=\rho_{02}=17.3\:g/m^3.$

Using equation (3), we can find vapor density before adding the water:

$$\rho=\frac{\phi_2 \rho_{02}} {100\%}\:(5)$$

Mass of vapor before adding the water:

$$m_1=\rho V=\frac{\phi_2\rho_{02} V} {100\%}\:(6)$$

Using equation (4), we can find vapor density after adding the water:

$$\rho'=\frac{\phi_3 \rho_{03}} {100\%}\:(7)$$

Mass of vapor after adding the water:

$$m_2=\rho' V=\frac{\phi_3\rho_{03} V} {100\%}=\frac{\phi_3\rho_{02} V} {100\%}\:(8)$$

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

$$\Delta{m}=m_2-m_1=\frac{\rho_{02} V} {100\%} (\phi_3-\phi_2)\:(9)$$

Finally:

$$\Delta{m}=\frac{17.3 V} {100\%} (50\%-41\%)=1.6V\:(g)$$

Answer: $\phi_2=41\%,\:\Delta{m}=1.6V\:g.$