Question

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(1.00 \mathrm{~atm}\) 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 change in internal energy \(\Delta E\) for the condensation of 1 mole of water at 1 atm and 100°C is -37.555 kJ.

Step-by-step solution

Convert the given information to the appropriate units.

We are given the volume of H₂O(g) as 30.6 L and the density of H₂O(l) as 0.996 g/cm³. We need to convert these to the appropriate units: 1. Convert the volume of H₂O(g) from L to cm³: \(30.6 \mathrm{~L} \times 1000 \mathrm{~cm^{3}/L} = 30600 \mathrm{~cm^{3}}\) 2. Convert the density of H₂O(l) to kg/m³: \(0.996 \mathrm{~g/cm^{3}} \times 1000 \mathrm{~kg/m^{3}/(g/cm^{3})} = 996 \mathrm{~kg/m^{3}}\)

Find the volume of H₂O(l).

For 1 mole of H₂O(l), the mass is \(1 \times 18.0153 \mathrm{~g/mol} = 18.0153 \mathrm{~g}\). Thus, we can find the volume of H₂O(l) at the given conditions using the density: $$V_{H_{2}O(l)} = \frac{m}{\rho} = \frac{18.0153 \mathrm{~g}}{0.996 \mathrm{~g/cm^{3}}} = 18.073 \mathrm{~cm^{3}}$$

Calculate the change in volume ΔV.

Now we will find the change in volume during the condensation process: $$\Delta V = V_{H_{2}O(l)} - V_{H_{2}O(g)} = 18.073 \mathrm{~cm^{3}} - 30600 \mathrm{~cm^{3}} = -30581.927 \mathrm{~cm^{3}}$$

Calculate the work done on the system w.

We can find the work done on the system by using the constant pressure: $$w = -P \Delta V$$ We are given the pressure P as 1 atm, and we should convert it to the SI unit of pressure (Pa): $$P = 1 \mathrm{~atm} \times 101325 \mathrm{~Pa/atm} = 101325 \mathrm{~Pa}$$ Now we can calculate the work done: $$w = -101325 \mathrm{~Pa} \times (-30581.927 \mathrm{~cm^{3}}) \times \frac{1 \mathrm{~J}}{100 \mathrm{~Pa \cdot cm^{3}}} = 3105.06 \mathrm{~J} = 3.105 \mathrm{~kJ}$$

Calculate the change in internal energy ΔE.

Finally, we can find ΔE using the First Law of Thermodynamics: $$\Delta E = q + w$$ We are given the heat released as q = -40.66 kJ. Therefore, $$\Delta E = (-40.66 \mathrm{~kJ}) + (3.105 \mathrm{~kJ}) = -37.555 \mathrm{~kJ}$$ Thus, the change in internal energy for the condensation of 1 mole of water at 1 atm and 100°C is -37.555 kJ.

Key concepts

Internal Energy Change

When studying thermodynamics, a fundamental aspect to understand is the concept of internal energy change. Internal energy, often denoted as \(E\), represents the total energy stored within a system. When a process such as condensation occurs, changes in internal energy are critical to assess. The change in internal energy is identified as \(\Delta E\) and can be calculated using the first law of thermodynamics. This law allows us to see how energy flows into or out of a system. During the condensation of water vapor to liquid, some energy is released as heat and reduces the internal energy of the system.
In essence, \(\Delta E\) reflects the net balance between the heat exchange \(q\) and the work \(w\) done by or on the system, expressed as \(\Delta E = q + w\). For the condensation of one mole of water, when the heat release is \(-40.66 \, \text{kJ}\) and the work done on the system is \(3.105 \, \text{kJ}\), the internal energy change is calculated to be \(-37.555 \, \text{kJ}\). This tells us that the water molecules release energy during condensation, which manifests as both work and heat.

First Law of Thermodynamics

The first law of thermodynamics is central to understanding processes within closed systems, such as the condensation process. It is a law of conservation stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of thermodynamics, this law can be expressed simply: \(\Delta E = q + w\).
This equation tells us that the change in internal energy \(\Delta E\) of a system equals the sum of the heat \(q\) added to the system and the work \(w\) done on it. This principle is crucial when analyzing energy changes during a phase transition, as it provides the framework for calculating how much energy is involved.

• \(q\) represents heat exchange. Heat flowing in is positive, while heat exiting is negative.
• \(w\) signifies work done. Work done on the system is positive; work done by the system is negative.

The calculated \(\Delta E = -37.555 \, \text{kJ}\) for our condensation exercise was derived using this principle, highlighting how thermodynamic processes balance energy transformations.

Condensation Process

The condensation process represents a physical transition where a gas changes into a liquid. It is integral to many natural and industrial processes, serving as a prime example of a phase change that involves energy transfer. When water vapor at \(1.00 \, \text{atm}\) and \(100^{\circ}\text{C}\) condenses into liquid water, energy in the form of heat is released, indicating an exothermic process. This is reflected in the heat \(-40.66 \, \text{kJ}\) released per mole of \(\text{H}_2\text{O}\) during the exercise.
Understanding condensation involves recognizing how volume and density changes impact the process. The drastic reduction in volume from gas to liquid (30600 \(\text{cm}^3\) to 18.073 \(\text{cm}^3\)) signifies how molecules compact closer together in a liquid form. Furthermore, the density of \(0.996 \, \text{g/cm}^3\) further indicates the compact nature of the liquid phase. Such measurements allow for accurate calculations in thermodynamic problems, ensuring the energy balances computed are correct according to the principles of thermodynamics.