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 \(\mathrm{L}\) . 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 \(\mathrm{atm}\) and \(100 .^{\circ} \mathrm{C}\)

Short answer

The internal energy change for the condensation of 1 mole of water at 1.00 atm and 100°C is approximately -43.8 kJ/mol.

Step-by-step solution

Determine the initial and final states of water

Initially, water is in the gaseous state and occupies a volume of 30.6 L at 1.00 atm pressure and 100°C temperature. Upon condensation, water will be in the liquid state at the same pressure and temperature.

Calculate the mass and volume of liquid water

We are given the density of liquid water at 100°C and 1.00 atm pressure as 0.996 g/cm³. Since there is 1 mole of water, we can compute the mass using the molar mass of water (\(18.015 \thinspace g/mol\)): mass = (1 mol) × (18.015 g/mol) = 18.015 g Then, we can calculate the volume of liquid water using the density: volume = (mass) / (density) = (18.015 g) / (0.996 g/cm³) = 18.086 cm³ Note that this volume will need to be converted from cm³ to L.

Convert the volume of liquid water from cm³ to L

To convert from cm³ to L, we can use the following conversion factor: 1 L = 1000 cm³. So, our volume of liquid water in L will be: volume = (18.086 cm³) × (1 L / 1000 cm³) = 0.018086 L

Calculate the work done during condensation

As the volume of water decreases during condensation (from 30.6 L to 0.018086 L), we can calculate the work done by the system on the surroundings. We will use the formula for the work done during an isothermal process at constant pressure, which is \(w = -P \Delta V\). Here, \(P\) is the pressure (1.00 atm) and \(\Delta V\) is the change in volume (\(0.018086 - 30.6 \thinspace L\)). We must convert the pressure to a consistent unit, the International System of Units (SI) unit for pressure is pascal (1 atm = 101325 Pa): Pressure in Pa = \(1 \thinspace atm \times \frac{101325 \thinspace Pa}{1 \thinspace atm} = 101325 \thinspace Pa\) Now, we can calculate the work done: \(w = -P \Delta V = -(101325 \thinspace Pa)(0.018086 - 30.6 \thinspace L)\) We need to convert the unit of volume from L to m³, using the conversion factor 1 L = 0.001 m³: \(\Delta V = (0.018086 - 30.6) \times 0.001 \thinspace m^3 = -0.030582 \thinspace m^3\) Now we can calculate the work done in Joules (1 J = 1 Pa × 1 m³): \(w = -(101325 \thinspace Pa)(-0.030582 \thinspace m^3) = 3101 \thinspace J\)

Calculate the change in internal energy

Using the first law of thermodynamics, \(\Delta E = q - w\), we can calculate the change in internal energy. The heat released during condensation was given as \(40.66 \thinspace kJ\) (keep in mind that this is an exothermic process so the heat value is negative), which is equal to \(-40660 \thinspace J\). Then we can plug the values of \(q\) and \(w\) to find the change in internal energy: \(\Delta E = (-40660 \thinspace J) - (3101 \thinspace J)\) \(\Delta E = -43761 \thinspace J\) So, the internal energy change for the condensation of 1 mole of water at 1.00 atm and 100°C is approximately -43.8 kJ/mol.

Key concepts

Internal Energy Change

When discussing internal energy change, we're essentially exploring how the energy within a system varies. Internal energy encompasses kinetic and potential energy of molecules.
During a condensation process, such as when water vapor becomes liquid water, energy shifts occur. Released energy manifests as heat because the gaseous state has higher energy than the liquid state.
The internal energy change can be calculated using the formula related to the first law of thermodynamics:

• \( \Delta E = q - w \)
• Where \( q \) is the heat exchange, and \( w \) is the work done

In our exercise, the effective work is determined by the pressure and the change in volume, essential components of reshaping this energy balance.
By plugging in values for heat and work, we discover the internal energy change, crucial for understanding energy movements within and out of a chemical system.
It showcases how manipulations within physical processes reverberate in energy flow.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the entire process. In thermodynamics, this implies that temperature does not fluctuate while other physical changes take place in the substance.
Experiments or calculations involving isothermal processes heavily rely on this constancy for predicting other variables accurately.
In a situation where water transitions from steam to liquid at 100°C, maintaining an unvarying temperature simplifies the thermodynamic analysis.
The simplicity surfaces as the temperature acts as an anchored baseline, preventing additional variables from skewing the energy equation between states.
During such processes, any heat absorbed or released by the system precisely balances the work done by or on the system. This balance makes it theoretically smoother to calculate parameters like work and heat transfer. Within our calculation:

• Work done is derived from volume change and pressure
• The consistent temperature simplifies these calculations

By recognizing the constancy, we apply the principles of isothermal processes to effectively evaluate the energy exchanges that occur.

Chemical Thermodynamics

Chemical thermodynamics explores energy changes associated with chemical reactions and phase transformations. It provides pivotal insight into why reactions occur, how they proceed, and the energy dynamics involved.
This field contemplates the laws of thermodynamics, notably focusing on how these laws influence chemical reactions, including the concepts of enthalpy, entropy, and Gibbs free energy. In our condensation exercise, our grasp of chemical thermodynamics enabled us to illuminate the heat released when gaseous water condenses into liquid.
The laws of thermodynamics guide us in analyzing how heat transition influences energized change during physical chemical processes.
In this scenario:

• Enthalpy reflects the heat exchange at constant pressure
• Understanding the change in internal energy \( \Delta E \) showcases the remnants after accounting for work

Chemical thermodynamics serves as a blueprint for unraveling energy transformations, allowing for a clear visualization of how reactions translate to observable energy exchanges.