Problem 40 - Chemistry An Atoms First Approach

Question

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}\) occupies a volume of 30.6 L. When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}, 40.66 \mathrm{kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{g} / \mathrm{cm}^{3},\) calculate \(\Delta E\) for the condensation of 1 mole of water at 1.00 atm and \(100 .^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The change in internal energy, \(\Delta E\), for the condensation of 1 mole of water at 1.00 atm and \(100^{\circ}\)C is -37.562 kJ.

Step 1: 1. Calculate the number of moles of water vapor

We are given that 1 mole of water vapor is being condensed. So, n = 1 moles of \(\mathrm{H}_{2} \mathrm{O}(g)\).

Step 2: 2. Determine the heat, q, released during condensation

We are given that 40.66 kJ of heat is released during the condensation process. Since it is released, it will be a negative value. So, q = -40.66 kJ.

Step 3: 3. Calculate the initial and final volume of the system

The initial volume of the system is given as 30.6 L, which is the volume of 1 mole of water vapor. To find the final volume, we can use the given density of liquid water at this temperature and pressure, which is 0.996 g/cm³. First, find the mass of 1 mole of water: mass = (1 mole) × (18.015 g/mol) = 18.015 g Now, we can calculate the volume of liquid water using the density formula: volume = mass/density Convert the density to g/L: density = 0.996 g/cm³ × \(\frac{1000 \mathrm{cm}^{3}}{\mathrm{L}}\) = 996 g/L The final volume, V₂, can be calculated as: V₂ = \(\frac{18.015 \mathrm{g}}{996 \mathrm{g}/\mathrm{L}}\) = 0.0181 L

Step 4: 4. Calculate the work, w, done by the system

The work done by the system can be calculated using the formula for work done by a gas at constant pressure: \[w = -P(V_{2}-V_{1})\] We are given that the pressure, P, is 1.00 atm. First, let's convert pressure to J/L units by using the conversion factor: 1 atm = 101.325 J/L P = 1.00 atm × \(\frac{101.325 \mathrm{J}}{\mathrm{L}}\) = 101.325 J/L Now, we can substitute the given values into the formula for work: w = -101.325 J/L × (0.0181 L - 30.6 L) = -101.325 J/L × (-30.582 L) = 3098.00 J

Step 5: 5. Calculate the change in internal energy, \(\Delta E\)

Now we apply the first law of thermodynamics, which states that the change in internal energy, \(\Delta E\), can be calculated as a sum of heat, q, and work, w: \(\Delta E = q + w\) Plugging in the values obtained in Steps 2 and 4: \(\Delta E = -40.66 \mathrm{kJ} + 3098 \mathrm{J}\) Since the energy units are different, we must convert one of them to be consistent. Let's convert 3098 J to kJ: 3098 J = 3.098 kJ Now we find \(\Delta E\): \(\Delta E = -40.66 \mathrm{kJ} + 3.098 \mathrm{kJ} = -37.562 \mathrm{kJ}\) Hence, the change in internal energy, \(\Delta E\), for the condensation of 1 mole of water at 1.00 atm and \(100^{\circ}\)C is -37.562 kJ.

Key Concepts

Change in Internal Energy

The change in internal energy, often denoted as \( \Delta E \), is a vital concept in thermodynamics that describes the total change in an energy system. According to the first law of thermodynamics, the change in internal energy is the sum of the heat exchanged \( q \) with its surroundings and the work \( w \) done by the system:

• \( \Delta E = q + w \)
• \( q \) represents heat, positive when absorbed and negative when released.
• \( w \) represents work, positive when done on the system and negative when the system does work.

In the condensation of water, the system releases energy as heat is transferred from the gas to the liquid state. This transfer is expressed as a negative \( q \), showing a release of 40.66 kJ. The work done due to volume change from gas to liquid at a constant atmospheric pressure also affects \( \Delta E \). Converting the work into consistent units, the total change sums up to \(-37.562\, \text{kJ}\) during the water condensation process.

Condensation Process

The condensation process is where a gas transforms into a liquid. This phase change involves the release of energy in the form of heat. For water,

• The condensation occurs at constant temperature and pressure, specifically at \( 100^{\circ} \text{C} \) and \( 1.00 \text{atm} \).
• During the transition from steam to liquid water, intermolecular forces bring molecules closer together, releasing thermal energy.
• In our scenario, 40.66 kJ of heat energy is released when 1 mole of water vapor condenses.

This release of energy is significant as it aids in understanding how energy interactions occur in natural processes like weather systems or even in industrial applications. Understanding this can help explain phenomena like cloud formation where water vapor condenses, releasing energy that can affect atmospheric temperatures.

Mole Concept

The mole concept is fundamental in chemistry as it is a basic unit for measuring particles of a substance. A mole represents \( 6.022 \times 10^{23} \) entities, whether they are atoms, molecules, or ions. In thermodynamics, the mole concept helps calculate physical properties by connecting them to measurable quantities like mass and volume:

• In our example, 1 mole of water vapor is condensed, meaning exactly one Avogadro's number of water molecules undergo the phase change.
• The molar mass of water is 18.015 g/mol, which also relates to the water's density in finding the final liquid volume.

This concept allows converting microscopic molecular changes to macroscopic phenomena, essential for laboratory settings and industrial processes where large quantities are involved.

Heat Release

Heat release is a key aspect of many thermodynamic processes, especially phase changes such as condensation. It refers to the energy released when a system releases heat into its surroundings:

• In this exercise, the heat release is quantified as \(-40.66 \text{ kJ}\), indicating exothermic reaction where heat is released as water transitions from gas to liquid.
• It is achieved because energy is given off when steam's gaseous molecules lose kinetic energy, slowing down, and allowing intermolecular forces to attract molecules into a liquid form.

This released heat is not lost but transferred to the surroundings, influencing the system's internal energy change. Heat release is a common concept in various natural phenomena and industrial operations, from weather changes to power generation.

Work Calculation

Work calculation in thermodynamics often involves changes in volume under pressure. When gas condenses to a liquid, the system's volume decreases, allowing us to calculate work:

• The work \( w \) is given by \( w = -P(V_{2} - V_{1}) \), where \( P \) is the pressure and \( V_{2} \) and \( V_{1} \) are final and initial volumes, respectively.
• In this problem, water vapor condenses from 30.6 L to 0.0181 L, crucial for calculating work done by the system. With \( P = 101.325 \text{ J/L} \) (converted from atm), the work done is calculated as \( 3098 \text{ J} \) or \( 3.098 \text{ kJ} \).

This work calculation is vital in determining total energy changes in systems, helping provide insights into processes such as engine cycles and atmospheric mechanics.