Chapter 22, Problem 22.5E

Interpretation Introduction

(a)

Interpretation:

The work done needed to increase the surface area of a pool of chloroform by $50.0{\text{cm}}^{\text{2}}$ is to be calculated.

Concept introduction:

The atoms or molecules at the surface of the liquid are considered as a film. The atoms or molecules in bulk imply a force on the atoms or molecules at the surface of the liquid and tends to minimize the surface area of the liquid. This phenomenon is known as the surface tension of the liquid.

Answer to Problem 22.5E

The work done required to increase the surface area of a pool of chloroform by $50.0{\text{cm}}^{\text{2}}$ is $\text{1}\text{.355}\times {10}^{-4}\text{J}$.

Explanation of Solution

The work done to increase the surface area is calculated by the formula shown below.

$w=\gamma \cdot \Delta A$ …(1)

Where,

• $\gamma$ is the surface tension.

• $\Delta A$ is the change in surface area.

The conversion of $\text{dyn}{\text{cm}}^{-1}$ into $\text{erg}{\text{cm}}^{-2}$ is shown below.

$1\text{dyn}{\text{cm}}^{-1}=1\text{erg}{\text{cm}}^{-2}$

Therefore, conversion of $27.1\text{dyn}{\text{cm}}^{-1}$ into $\text{erg}{\text{cm}}^{-2}$ is shown below.

$27.1\text{dyn}{\text{cm}}^{-1}=27.1\text{erg}{\text{cm}}^{-2}$

The value of $\gamma$ and $\Delta A$ are $27.1\text{erg}{\text{cm}}^{-2}$ and $50.0{\text{cm}}^{\text{2}}$.

Substitute the values of $\gamma$ and $\Delta A$ in equation (1).

$\begin{array}{rll}w& =& 27.1\text{erg}{\text{cm}}^{-2}\times 50.0{\text{cm}}^{\text{2}}\\ & =& 1355\text{erg}\end{array}$

The conversion of $\text{erg}$ into $\text{J}$ is shown below.

$1\text{erg}={10}^{-7}\text{J}$

Therefore,

$\begin{array}{rll}\text{1355}\text{erg}& =& \text{1355}\times {10}^{-7}\text{J}\\ & =& \text{1}\text{.355}\times {10}^{-4}\text{J}\end{array}$

Conclusion

The work done to increase the surface area of pool of chloroform by $50.0{\text{cm}}^{\text{2}}$ is $\text{1}\text{.355}\times {10}^{-4}\text{J}$.

Interpretation Introduction

(b)

Interpretation:

The work done to make the film of chloroform having area of $0.010{\text{m}}^{\text{2}}$ is to be calculated.

Concept introduction:

The atoms or molecules at the surface of the liquid are considered as a film. The atoms or molecules in bulk imply a force on the atoms or molecules at the surface of the liquid and tends to minimize the surface area of the liquid. This phenomenon is known as the surface tension of the liquid.

Answer to Problem 22.5E

The work done to make the film of chloroform having area of $0.010{\text{m}}^{\text{2}}$ is $2.71\times {10}^{-4}\text{J}$.

Explanation of Solution

The making of film is same as increasing the surface area of a liquid. Therefore, the work done to make the film of chloroform is calculated by the formula given below.

$w=\gamma \cdot \Delta A$ …(1)

Where,

• $\gamma$ is the surface tension.

• $\Delta A$ is the change in surface area.

The conversion of $\text{dyn}{\text{cm}}^{-1}$ into $\text{erg}{\text{cm}}^{-2}$ is shown below.

$1\text{dyn}{\text{cm}}^{-1}=1\text{erg}{\text{cm}}^{-2}$

Therefore, the conversion of $27.1\text{dyn}{\text{cm}}^{-1}$ into $\text{erg}{\text{cm}}^{-2}$ is shown below.

$27.1\text{dyn}{\text{cm}}^{-1}=27.1\text{erg}{\text{cm}}^{-2}$

The conversion of ${\text{m}}^{\text{2}}$ into ${\text{cm}}^{\text{2}}$ is done as shown below.

$1{\text{m}}^{\text{2}}={10}^4{\text{cm}}^{\text{2}}$

Therefore, the conversion of $0.010{\text{m}}^{\text{2}}$ into ${\text{cm}}^{\text{2}}$ is done as shown below.

$\begin{array}{rll}0.010{\text{m}}^{\text{2}}& =& 0.010\times {10}^4{\text{cm}}^{\text{2}}\\ & =& {10}^2{\text{cm}}^{\text{2}}\end{array}$

The value of $\gamma$ and $\Delta A$ are $27.1\text{erg}{\text{cm}}^{-2}$ and ${10}^2{\text{cm}}^{\text{2}}$.

Substitute the values of $\gamma$ and $\Delta A$ in equation (1).

$\begin{array}{rll}w& =& \gamma \cdot \Delta A\\ & =& 27.1\text{erg}{\text{cm}}^{-2}\times {10}^2{\text{cm}}^{\text{2}}\\ & =& 2710\text{erg}\end{array}$

The conversion of $\text{erg}$ into $\text{J}$ is shown below.

$1\text{erg}={10}^{-7}\text{J}$

Therefore,

$\begin{array}{rll}2710\text{erg}& =& 2710\times {10}^{-7}\text{J}\\ & =& 2.71\times {10}^{-4}\text{J}\end{array}$

Conclusion

The work done to make the film of chloroform having area of $0.010{\text{m}}^{\text{2}}$ is $2.71\times {10}^{-4}\text{J}$.