Question

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .{ }^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H_{\text {vap }}\) for the liquid is \(30.7 \mathrm{~kJ} / \mathrm{mol}\) at \(80 .{ }^{\circ} \mathrm{C}\).

Short answer

The work done (w) during vaporization is \(-2.94 \, kJ/mol\), and the change in internal energy (ΔE) is \(27.76 \, kJ/mol\).

Step-by-step solution

Identify the relevant equations

We will use the first law of thermodynamics to relate the change in internal energy (ΔE), heat (q), and work (w) for the process: ΔE = q + w Since the process occurs at constant pressure, we can also use the equation relating the change in enthalpy (ΔH_vap) to the heat (q): ΔH_vap = q (at constant pressure) Finally, we can calculate the work (w) done by the system on the surroundings during vaporization using the equation: w = -PΔV

Calculate work done (w) during vaporization

To calculate the work done during vaporization, we need to determine the change in volume (ΔV) of the system. Since 1 mole of liquid is vaporized, we can use the ideal gas equation for 1 mole of vapor at the final state: PV = nRT Where: P = pressure (1.00 atm) V = volume of vapor in L n = moles (1 mole) R = ideal gas constant (0.0821 L atm / K mol) T = temperature in Kelvin (80°C + 273.15 = 353.15 K) V = nRT / P = (1 mol)(0.0821 L atm / K mol)(353.15 K) / 1 atm = 29.02 L Since the volume of the liquid can be considered negligible compared to the volume of the vapor, we can assume that ΔV = V (volume of the vapor) - 0 (volume of the liquid). Now, we can calculate the work done (w) during vaporization: w = -PΔV = -1 atm × 29.02 L = -29.02 L atm

Convert work to the same unit as ΔH_vap

To calculate ΔE, we will need the work (w) to be in the same unit as ΔH_vap (kJ/mol). We can convert L atm to J using the conversion factor 101.3 J = 1 L atm: w = -29.02 L atm × (101.3 J / 1 L atm) = -2937.66 J And then convert Joules to kJ: w = -2937.66 J × (1 kJ / 1000 J) = -2.94 kJ/mol

Calculate the change in internal energy (ΔE)

Now, we can use the relationship between ΔE, ΔH_vap, and w to find the change in internal energy during vaporization: ΔE = q + w ΔH_vap = q (as the process is at constant pressure) ΔE = ΔH_vap + w Substitute the known values: ΔE = 30.7 kJ/mol + (-2.94 kJ/mol) = 27.76 kJ/mol

Write down the work done and the change in internal energy

The work done (w) during vaporization is -2.94 kJ/mol, and the change in internal energy (ΔE) is 27.76 kJ/mol.

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a crucial concept in thermodynamics that refers to the heat content change of a system when it undergoes a process at constant pressure. It's a measure of the total heat flow into or out of a system, embracing both the internal energy change and the work done by the system as it expands or contracts. In the context of our exercise, \( \Delta H_{\text{vap}} \) represents the enthalpy change during vaporization, which is the heat required to convert 1 mole of liquid into its vapor without a change in pressure.

The enthalpy of vaporization, quite specifically, is a key quantity in describing phase changes, as it provides insight into the energy needed for a substance to overcome intermolecular forces and transition from the liquid to the gaseous state. The equation \( \Delta H_{\text{vap}} = q \) (at constant pressure) highlighted in our exercise allows us to directly use the given enthalpy change value as the heat input for the vaporization of one mole of liquid, which is crucial for further calculations of internal energy.

Vaporization

Vaporization is a phase transition that occurs when a substance changes from its liquid form to a gas. The process requires energy, as it involves the breaking of intermolecular forces holding the liquid molecules together. The amount of energy required for vaporization is particular to each substance and is quantified by the enthalpy of vaporization \( \Delta H_{\text{vap}} \). In our exercise, this energy is provided as 30.7 kJ/mol for the transition at the given boiling point.

Understanding the Energy Involved

As a liquid boils and vaporizes into a gas, it not only absorbs heat energy corresponding to its enthalpy of vaporization but also does work on its surroundings by expanding. This work is directly associated with the change in volume \( (\Delta V) \) of the system. It's important to note that the energy absorbed during vaporization is primarily used to overcome attractive intermolecular forces, and not all of this energy results in a temperature change. Rather, it triggers the transformative action from liquid to gas, which is signified by the occurrence of boiling.

Internal Energy

Internal energy, usually symbolized as \( E \) or \( U \), represents the total energy contained within a system. This energy comprises the kinetic energy of particles moving and vibrating, alongside potential energy associated with the positions and interactions between particles. Internal energy is pivotal because it embodies the total energy changes that occur within a closed system.

The First Law of Thermodynamics is essential here; it states that the change in internal energy \( (\Delta E) \) of a closed system is equal to the heat added to the system (q) plus the work done on the system (w). Mathematically, it is articulated as \( \Delta E = q + w \). In the textbook problem, we use the notion of internal energy to understand the energy dynamics when a liquid is vaporized at its boiling point. We have calculated the work done by the system and have used the provided enthalpy of vaporization to assess the heat added to the system during vaporization. By combining these with the first law, we have deduced the total internal energy change during the process.

It is fascinating that while enthalpy change provides a macroscopic view by including work done at constant pressure, the concept of internal energy dwells on a microcosmic level, reflecting changes in particle motion and interactions without concern for the pressure-volume work.