Chapter 4, Problem 4A.1AST

Interpretation Introduction

Interpretation:

The work done by water when it freezes at $\text{0}°\text{C}$ has to be determined.

Concept Introduction:

The system can done two kind of works.  The one is the work of expansion which is the work that arises from the change in the volume of the system.  On the other hand, the second work done is the work of non-expansion which does not involve any change in the volume of the system.  The expression for calculating the work of expansion is shown below.

    $w=-P_{\text{ex}}\Delta V$

Explanation of Solution

As per the data given in the question, the densities of water and ice at $\text{0}°\text{C}$ is $\text{1}\text{.00}\text{g}\cdot {\text{cm}}^{-3}$ and $\text{0}\text{.92}\text{g}\cdot {\text{cm}}^{-3}$ respectively.  The applied opposing pressure is $\text{1070}\text{atm}$.

Therefore, the initial and final volume of $\text{100}\text{g}$ water is $92.0{\text{cm}}^3$ and $\text{100}{\text{cm}}^3$ when water is in the form of solid and liquid respectively.

The change in volume is calculated by using the relation given below.

  $\Delta V=V_f-V_i$        (1)

Where,

• $\Delta V$ is the change in volume.
• $V_i$ is the initial volume.
• $V_f$ is the final volume.

The value of $V_i$ is $92.0{\text{cm}}^3$.

The value of $V_f$ is $\text{100}{\text{cm}}^3$.

Substitute the value of $V_i$ and $V_f$ in equation (1).

    $\begin{array}{c}\Delta V=\text{100}{\text{cm}}^3-92.0{\text{cm}}^3\\ =8.00{\text{cm}}^3\times \frac{0.001\text{L}}{\text{1}{\text{cm}}^3}\\ =8\times {10}^{-3}\text{L}\end{array}$

Therefore, the change in volume of the given system is $8\times {10}^{-3}\text{L}$.

The relation for calculating the work of expansion by the automobile engine is shown below.

  $w=-P_{\text{ex}}\Delta V$        (2)

Where,

• $w$ is the work done.
• $P_{\text{ex}}$ is the external pressure.
• $\Delta V$ is the change in volume.

The value of $P_{\text{ex}}$ is $\text{1070}\text{atm}$.

The value of $\Delta V$ is $8\times {10}^{-3}\text{L}$.

Substitute the value of $P_{\text{ex}}$ and $\Delta V$ in equation (1).

    $\begin{array}{c}w=-\left(1070\text{atm}\right)\times \left(8\times {10}^{-3}\text{L}\right)\\ =-8.56\text{L}\cdot \text{atm}\times \frac{\text{101}\text{.325}\text{J}}{1\text{L}\cdot \text{atm}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =-0.86\text{kJ}\end{array}$

Thus, the work done by the water when it freezes at $\text{0}°\text{C}$ is $\underset{\_}{-0.86kJ}$.