Chapter 22, Problem 22.27E

Interpretation Introduction

(a)

Interpretation:

The reason corresponding to the fact that the vapor pressure of a liquid increases as the radius of the droplet decreases is to be stated. The implications for condensation process and for atmospheric processes like raindrop formation are to be stated.

Concept introduction:

The Kelvin equation is given by an expression shown below.

$\ln \frac{p_{\text{vapor}}}{p{°}_{\text{vapor}}}=\frac{2\gamma \cdot \overline{V}}{r\cdot RT}$

Where,

• $p_{\text{vapor}}$ is the vapor pressure of the droplet.

• $p{°}_{\text{vapor}}$ is the vapor pressure of the bulk liquid.

• $\overline{V}$ is the molar volume of the liquid.

• $R$ is the Gas constant.

• $T$ is the temperature.

• $\gamma$ is the surface tension.

• $r$ is the radius of droplet.

Answer to Problem 22.27E

The vapor pressure of a liquid increases as the radius of the droplet decreases because value of natural log ($p_{\text{vapor}}$) will increase by decreasing the radius of droplet. The formation of droplets will be very slow during condensation and the formation of raindrop because vapour pressure of the initial droplet is very large and results in the high rate of evaporation.

Explanation of Solution

The Kelvin equation is given by an expression shown below.

$\ln \frac{p_{\text{vapor}}}{p{°}_{\text{vapor}}}=\frac{2\gamma \cdot \overline{V}}{r\cdot RT}$

Where,

• $p_{\text{vapor}}$ is the vapor pressure of the droplet.

• $p{°}_{\text{vapor}}$ is the vapor pressure of the bulk liquid.

• $\overline{V}$ is the molar volume of the liquid.

• $R$ is the Gas constant.

• $T$ is the temperature.

• $\gamma$ is the surface tension.

• $r$ is the radius of droplet.

In the above equation, when the radius (in the denominator) of droplet decreases, the overall right hand side fraction will increase. Therefore, the value of natural log ($p_{\text{vapor}}$) will also increase.

The vapour pressure of the initial droplet is very large during the condensation and the formation of raindrop. This leads to the increase in the rate of evaporation and makes the formation of droplet difficult. Therefore, the formation of droplets will be very slow.

Conclusion

The vapor pressure of a liquid increases as the radius of the droplet decreases because value of natural log ($p_{\text{vapor}}$) will increase by decreasing the radius of droplet. The formation of droplets will be very slow during condensation and the formation of raindrop because vapour pressure of the initial droplet is very large and results in the high rate of evaporation.

Interpretation Introduction

(b)

Interpretation:

The vapor pressure of a droplet of water is to be predicted.

Concept introduction:

The Kelvin equation is given by an expression shown below.

$\ln \frac{p_{\text{vapor}}}{p{°}_{\text{vapor}}}=\frac{2\gamma \cdot \overline{V}}{r\cdot RT}$

Where,

• $p_{\text{vapor}}$ is the vapor pressure of the droplet.

• $p{°}_{\text{vapor}}$ is the vapor pressure of the bulk liquid.

• $\overline{V}$ is the molar volume of the liquid.

• $R$ is the Gas constant.

• $T$ is the temperature.

• $\gamma$ is the surface tension.

• $r$ is the radius of droplet.

Answer to Problem 22.27E

The vapor pressure of a droplet of water is $25.06\text{mmHg}$.

Explanation of Solution

The Kelvin equation is given by an expression shown below.

$\ln \frac{p_{\text{vapor}}}{p{°}_{\text{vapor}}}=\frac{2\gamma \cdot \overline{V}}{r\cdot RT}$ …(1)

Where,

• $p_{\text{vapor}}$ is the vapor pressure of the droplet.

• $p{°}_{\text{vapor}}$ is the vapor pressure of the bulk liquid.

• $\overline{V}$ is the molar volume of the liquid.

• $R$ is the Gas constant.

• $T$ is the temperature.

• $\gamma$ is the surface tension.

• $r$ is the radius of droplet.

The molar volume of water is calculated by the density formula shown below.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ …(2)

It is known that the density of water is $\text{0}\text{.997}\text{g}/\text{mL}$. The molar mass of the water is $18.016{\text{g mol}}^{-1}$.

Substitute the density of water in equation (2).

$\begin{array}{rll}\text{0}\text{.997}\text{g}/\text{mL}& =& \frac{18.016{\text{g mol}}^{-1}}{\text{Volume}}\\ \text{Volume}& =& \frac{18.016{\text{g mol}}^{-1}}{\text{0}\text{.997}\text{g}/\text{mL}\times \left(\frac{{10}^3\text{mL}}{1\text{L}}\right)}\\ & =& 1.807\times {10}^{-2}{\text{Lmol}}^{-1}\end{array}$

Therefore, the molar volume of water is $1.807\times {10}^{-2}{\text{Lmol}}^{-1}$.

The radius of droplet, temperature and vapor pressure of bulk water are given as $\text{20}\text{.0}\text{nm}$, $\text{298}\text{K}$ and $\text{23}\text{.77}\text{mmHg}$ respectively.

Substitute the value of $r$, $T$ and $p{°}_{\text{vapor}}$ in equation (1).

$\begin{array}{rll}\ln \frac{p_{\text{vapor}}}{\text{23}\text{.77}\text{mmHg}}& =& \frac{2\left(72.75\text{erg}{\text{cm}}^{-2}\right)\cdot \left(1.807\times {10}^{-2}{\text{Lmol}}^{-1}\right)}{\left(20.0\text{nm}\right)\cdot \left(8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\\ & =& 5.30\times {10}^{-5}\text{erg}{\text{cm}}^{-2}{\text{nm}}^{-1}\text{L}{\text{J}}^{-1}\end{array}$

The conversion of joule into erg is shown below.

$\begin{array}{rll}\ln \frac{p_{\text{vapor}}}{\text{23}\text{.77}\text{mmHg}}& =& \left[5.30\times {10}^{-5}\text{erg}{\text{cm}}^{-2}{\text{nm}}^{-1}\text{L}{\text{J}}^{-1}\times \frac{\text{1J}}{{10}^7\text{erg}}\right]\\ & =& 5.30\times {10}^{-2}{\text{cm}}^{-2}{\text{nm}}^{-1}\text{L}\end{array}$

The conversion of nm into cm is shown below.

$\begin{array}{rll}\ln \frac{p_{\text{vapor}}}{\text{23}\text{.77}\text{mmHg}}& =& \left[5.30\times {10}^{-5}{\text{cm}}^{-2}{\text{nm}}^{-1}\text{L}\times \frac{\text{1}\text{cm}}{{\text{10}}^{-7}\text{nm}}\right]\\ & =& 5.30\times {10}^{-2}\text{L}{\text{cm}}^{-1}\end{array}$

The conversion of L into cm is shown below.

$\begin{array}{rll}\ln \frac{p_{\text{vapor}}}{\text{23}\text{.77}\text{mmHg}}& =& \left[5.30\times {10}^{-5}{\text{cm}}^{-1}\text{L}\times \left(\frac{1{\text{dm}}^3}{1\text{L}}\right)\times {\left(\frac{10\text{cm}}{1\text{dm}}\right)}^3\right]\\ & =& 5.30\times {10}^{-2}\end{array}$

The above expression is solved as follow.

$\begin{array}{rll}\frac{p_{\text{vapor}}}{\text{23}\text{.77}\text{mmHg}}& =& e^{5.30\times {10}^{-2}}\\ p_{\text{vapor}}& =& e^{5.30\times {10}^{-2}}\times \text{23}\text{.77}\text{mmHg}\\ & =& 1.0544\times \text{23}\text{.77}\text{mmHg}\\ & =& 25.06\text{mmHg}\end{array}$

Therefore, the vapor pressure of a droplet of water is $25.06\text{mmHg}$.

Conclusion

The vapor pressure of a droplet of water is $25.06\text{mmHg}$.