Chapter U3.66, Problem 11E

(a)

Interpretation Introduction

Interpretation:

The humidity is $65\%$ during daytime when the temperature is $30°\text{C}$, if the temperature drops to $25°\text{C}$ at night then the expectation of fog needs to be explained.

Concept Introduction:

Water vapor concentration in air is termed as humidity. There is a limit on the number density of molecules of water which may be in the air for a given temperature. The air saturates at a certain point due to which no net evaporation takes place further.

Answer to Problem 11E

The actual vapor density is $1.105\text{mol}\text{per}\text{1000}\text{mL}$ which is smaller than the maximum vapor density at $25°\text{C}$ $1.25$ mol per $\text{1}000\text{ L}$. Hence, there is no chance of precipitation.

Explanation of Solution

The percent of the maximum humidity for a specified temperature is termed as relative humidity. When the air holds the maximum amount of water vapor for the temperature of the air, it is at $\text{1}00\%$ relative humidity.

The maximum vapor density at $\text{3}0°\text{C}$ is $\text{1}.\text{7}$ mol per $\text{1}000\text{ L}$, so, if this value is $0.\text{85}$, at $\text{3}0°\text{C}$ in a region, then the relative humidity is,

$\left(\frac{0.\text{85}}{\text{1}.\text{7}}\right)\times \text{ 1}00\%=0.5\times 100\%$

       $=50\%$

The graph showing variation in vapor density with temperature represents the maximum number density, that is, for air maximum vapor density is in temperatures range between $-\text{1}°\text{C}$ and $\text{4}0°\text{C}$. The representation of maximum vapor density is at Y-axis while the representation of the temperature in $°\text{C}$ is at x-axis.

The value of y, for $\text{x }=\text{ 30}$, is $1.7$ mol per $\text{1}000\text{ L}$. Therefore, water vapor density at $\text{1}00\%$ relative humidity for $\text{3}0°\text{C}$ is $1.7$ mol per $\text{1}000\text{ L}$ ....... (1)

The value of y, for $\text{x }=\text{ 25}$, is $1.25$ mol per $\text{1}000\text{ L}$. Therefore, water vapor density at $\text{1}00\%$ relative humidity for $25°\text{C}$ is $1.25$ mol per $\text{1}000\text{ L}$ ....... (2)

Given,

The relative humidity is $65\%$.

Calculate vapor density at $65\%$ humidity at $\text{3}0°\text{C}$: Put the vapor density value from equation (1).

$\text{Vapor density at}\text{65\%}\text{humidity at}\text{30}°\text{C = 65\%}\text{of}\text{maximum}\text{vapor}\text{density}\text{at}\text{30}°\text{C}$

                  $\text{= 65\%}\times \text{1}\text{.7}\text{mol}\text{per}\text{1000}\text{mL}$

                  $\text{= 0}\text{.65}\times \text{1}\text{.7}\text{mol}\text{per}\text{1000}\text{mL}$

                  $=1.105\text{mol}\text{per}\text{1000}\text{mL}$

At night when the temperature is $25°\text{C}$ the vapor density will be the same. Therefore, maximum vapor density is $1.25$ mol per $\text{1}000\text{ L}$ from equation (2).

The actual vapor density $1.105\text{mol}\text{per}\text{1000}\text{mL}$ is smaller than the maximum vapor density at $25°\text{C}$ $1.25$ mol per $\text{1}000\text{ L}$. Hence, there is no chance of precipitation.

(b)

Interpretation Introduction

Interpretation:

The humidity is $65\%$ during daytime when the temperature is $30°\text{C}$, if the temperature drops to $20°\text{C}$ at night then the expectation of fog needs to be explained.

Concept Introduction:

Water vapor concentration in air is termed as humidity. There is a limit on the number density of molecules of water which may be in the air for a given temperature. The air saturates at a certain point due to which no net evaporation takes place further.

Answer to Problem 11E

The actual vapor density $1.105\text{mol}\text{per}\text{1000}\text{mL}$ is greater than the maximum vapor density at $20°\text{C}$ $0.90$ mol per $\text{1}000\text{ L}$. Hence, there is a chance of precipitation.

Explanation of Solution

The percent of the maximum humidity for a specified temperature is termed as relative humidity. When air holds the maximum concentration of water vapor for the temperature of the air, it is at $\text{1}00\%$ relative humidity.

The maximum vapor density at $\text{3}0°\text{C}$ is $\text{1}.\text{7}$ mol per $\text{1}000\text{ L}$. So, if this value is $0.\text{85}$, at $\text{3}0°\text{C}$ in a region, then the relative humidity is,

$\left(\frac{0.\text{85}}{\text{1}.\text{7}}\right)\times \text{ 1}00\%=0.5\times 100\%$

       $=50\%$

The graph showing variation in vapor density with temperature represents the maximum number density, that is, for air maximum vapor density is in temperatures range between $-\text{1}°\text{C}$ and $\text{4}0°\text{C}$. The representation of maximum vapor density is at Y-axis while the representation of the temperature in $°\text{C}$ is at x-axis.

The value of y, for $\text{x }=\text{ 30}$, is $1.7$ mol per $\text{1}000\text{ L}$. Therefore, water vapor density at $\text{1}00\%$ relative humidity for $\text{3}0°\text{C}$ is $1.7$ mol per $\text{1}000\text{ L}$ ....... (1)

The value of y, for $\text{x }=\text{ 20}$, is $0.90$ mol per $\text{1}000\text{ L}$. Therefore, water vapor density at $\text{1}00\%$ relative humidity for $20°\text{C}$ is $0.90$ mol per $\text{1}000\text{ L}$ ....... (2)

Given,

The relative humidity is $65\%$.

Calculate vapor density at $65\%$ humidity at $\text{3}0°\text{C}$: Put the vapor density value from equation (1).

$\text{Vapor density at}\text{65\%}\text{humidity at}\text{30}°\text{C = 65\%}\text{of}\text{maximum}\text{vapor}\text{density}\text{at}\text{30}°\text{C}$

                  $\text{= 65\%}\times \text{1}\text{.7}\text{mol}\text{per}\text{1000}\text{mL}$

                  $\text{= 0}\text{.65}\times \text{1}\text{.7}\text{mol}\text{per}\text{1000}\text{mL}$

                  $=1.105\text{mol}\text{per}\text{1000}\text{mL}$

At night when the temperature is $20°\text{C}$ the vapor density will be the same. Therefore, maximum vapor density is $0.90$ mol per $\text{1}000\text{ L}$ from equation (2).

The actual vapor density $1.105\text{mol}\text{per}\text{1000}\text{mL}$ is greater than the maximum vapor density at $20°\text{C}$ $0.90$ mol per $\text{1}000\text{ L}$. Hence, there is a chance of precipitation.