Question

Calculate the work done (in joules) when \(1.0 \mathrm{~mole}\) of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\). The volumes of 1 mole of water and ice at \(0^{\circ} \mathrm{C}\) are 0.0180 and \(0.0196 \mathrm{~L},\) respectively. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)

Short answer

The work done is approximately 0.162 J.

Step-by-step solution

Recognizing the Formula for Work

To calculate the work done when water freezes, we will use the formula for work done by a system at constant pressure: \[ W = P \Delta V \] where \( W \) is work, \( P \) is pressure, and \( \Delta V \) is the change in volume of the substance.

Calculating Volume Change

We need to determine the change in volume \( \Delta V \) when water turns into ice. The volume of water is 0.0180 L, and the volume of ice is 0.0196 L, hence:\[ \Delta V = V_{\text{ice}} - V_{\text{water}} = 0.0196 \text{ L} - 0.0180 \text{ L} = 0.0016 \text{ L} \]

Applying the Pressure Value

The pressure given is 1.0 atm. Hence, the problem occurs at constant pressure of 1.0 atm, which will be applied in our work equation.

Calculating with the Conversion Factor

Now, substitute \( P = 1.0 \text{ atm} \) and \( \Delta V = 0.0016 \text{ L} \) into the work formula. Apply the conversion factor where 1 L·atm = 101.3 J:\[ W = (1.0 \text{ atm})(0.0016 \text{ L}) \times 101.3 \frac{J}{L\cdot atm} = 0.0016 \times 101.3 \text{ J} \]

Calculating the Final Work Value

By performing the multiplication:\[ W = 0.16208 \text{ J} \]Therefore, the work done when 1 mole of water freezes at the given conditions is approximately 0.162 J.

Key concepts

Work Done

Work done in thermodynamics often involves understanding how forces apply over a distance. In the context of this exercise, when water freezes, it undergoes a change in volume, which leads to the concept of work being done.

Let's break down the process:

• The formula for calculating work is given by: \[ W = P \Delta V \]- where \( W \) is the work done, \( P \) is the pressure, and \( \Delta V \) is the change in volume.
• Volume change is crucial here. Even a small adjustment in volume, like water turning to ice, results in work done based on the pressure applied.
• Pressure is usually constant in such problems, simplifying the calculation.
- Hence, in this example, the calculated work represents the energy exchanged due to the phase change from liquid to solid at given pressure conditions.
It is clear that temperature and pressure conditions will affect work when phases change. By freezing the water at atmospheric pressure, this small work done illustrates how energy transformations don't always involve large mechanical movements, even in seemingly simple phase changes.

Phase Change

Phase change involves transitioning between solid, liquid, and gaseous states. In this problem, water transforms from a liquid to a solid.

Here's how it works:

• When water freezes to become ice, it undergoes a phase change from liquid to solid. This transition involves changes in energy without altering the temperature.
• The volume expands slightly during this process; the important detail here is that the density of ice is less than that of water.
• In thermodynamics, such transitions involve latent heat, which is the heat absorbed or released during a phase change at constant temperature and pressure.

By understanding the behavior of substances during phase changes, one can predict and calculate the associated work or energy changes involved. These transformations play a critical role in many practical and industrial applications, such as refrigeration and material processing, showcasing the vast influence of phase changes on natural and engineered processes.

Constant Pressure Process

A constant pressure process is a common scenario in thermodynamics where changes happen while the pressure remains steady.

Let's explore its significance:

• In thermodynamics, processes like the one in this exercise are performed at atmospheric pressure (1 atm), which is often considered constant for calculations.
• The formula for work expressed as \( W = P \Delta V \) derives its simplicity and usefulness from the assumption of constant pressure.
• In this way, work calculations can be simplified significantly, since pressure is a known constant rather than a variable.

Constant pressure conditions are frequently encountered both in lab experiments and real-world applications; they ease the understanding of how energy is transferred during various physical and chemical changes. This concept is particularly practical in cases like phase changes, such as freezing or boiling, where maintaining a constant pressure helps predict outcomes and calculate energy changes effectively.