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 during the vaporization process is \(w = -2937.7 \, \text{J}\) and the change in internal energy is \(\Delta E = 27762.3 \, \text{J/mol}\).

Step-by-step solution

Calculate work done by the system

We are considering an isothermal process, so the temperature is constant. The formula for work done during constant temperature and pressure process on 1 mole of liquid is: \(w = - P\Delta V\) where: - P is the external pressure, and - ΔV is the change in volume during the vaporization process. The negative sign indicates that the work is done by the system, which means it's an expansion process against the surrounding pressure. However, we don't know the ΔV to determine the work. So, we need to find the change in volume during the vaporization process.

Calculate the change in volume during vaporization

We know that the boiling point of the liquid is 80°C, and it is vaporized under atmospheric pressure (1.00 atm). We are assuming that the vapor acts as an ideal gas. From the ideal gas equation, we can find the molar volume of the vapor: \(PV = nRT\) where: - n is the number of moles (in this case 1 mole), - R is the gas constant (0.0821 L atm/mol K, since the pressure is given in atm), and - T is the temperature in Kelvin (353.15 K). Solving for V: \(V = \frac{nRT}{P}\) Since we know that the molar volume of condensed liquid is negligible compared to molar volume of vapor, we can approximate the change in volume (ΔV) with the molar volume of the vapor (V): \(\Delta V \approx V\)

Plug in the values and calculate work done by the system

Now, we can plug in the values to calculate the work done during the vaporization process: \(w = - P\Delta V = -P\left(\frac{nRT}{P}\right)\) \(w = -nRT\) With n = 1, R = 0.0821 L atm/mol K, and T = 353.15 K: \(w = -(1)(0.0821)(353.15) = -29.0 \, \text{L atm}\) To convert the work to Joules multiply with 101.3 J/L atm: \(w = -29.0 \, \text{L atm} \times 101.3 \, \frac{\text{J}}{\text{L atm}} = -2937.7 \, \text{J}\) Now, we will calculate the change in internal energy (ΔE) for the vaporization process.

Use the first law of thermodynamics to calculate the change in internal energy

The first law of thermodynamics relates the heat absorbed by the system (Q), the work done by the system (w), and the change in internal energy (ΔE): \(\Delta E = Q + w\) We know that the enthalpy of vaporization (ΔH_vap) is 30.7 kJ/mol, and since the process is occurring at constant pressure, we have: \(Q = \Delta H_{\text { vap }}\) Convert ΔH_vap to Joules: ΔH_vap = (30.7 kJ/mol) × (1000 J/1 kJ) = 30700 J/mol Now plug in the values for Q and w to find ΔE: \(\Delta E = Q + w = 30700 - 2937.7 = 27762.3 \, \text{J/mol}\) Our final results are: \(w = -2937.7 \, \text{J}\) \(\Delta E = 27762.3 \, \text{J/mol}\)

Key concepts

Vaporization

Vaporization is the process in which a liquid is converted into a gas. It occurs at a specific temperature known as the boiling point. It's fascinating because during vaporization, molecules in the liquid gain enough energy to overcome intermolecular forces and enter the gaseous state.
This phase change takes energy, commonly called the heat of vaporization or enthalpy of vaporization.

• Vaporization requires energy to break the bonds between molecules in a liquid.
• This energy moves molecules far apart, transitioning them to a gaseous state.

In our exercise, 1 mole of a liquid vaporizes at 80°C. Here, the enthalpy of vaporization given is 30.7 kJ/mol, meaning that this amount of energy is needed to vaporize one mole of liquid at this temperature.

Enthalpy

Enthalpy, denoted by \(H\), is a measure of the total energy of a thermodynamic system. It includes internal energy plus the energy required to make room for it by displacing its environment.
It's particularly useful in chemical reactions occurring at constant pressure, where heat transfer is involved.

• Enthalpy change, \(\Delta H\), represents the heat absorbed or released during a process at constant pressure.
• Positive \(\Delta H\) indicates endothermic processes, where heat is absorbed, like in vaporization.

In the given exercise, the enthalpy of vaporization (\(\Delta H_{\text{vap}}\)) is used to determine the heat required for the phase change at constant atmospheric pressure, helping us understand the energy dynamics during vaporization.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in physical chemistry describing the behavior of an ideal gas. It's expressed as \(PV = nRT\), where:

• \(P\) is pressure.
• \(V\) is volume.
• \(n\) is the number of moles.
• \(R\) is the universal gas constant.
• \(T\) is temperature in Kelvin.

This law helps us comprehend how different macroscopic properties of gases are interrelated. In our problem, assuming the vapor behaves as an ideal gas, the Ideal Gas Law helps find the change in volume when 1 mole of liquid is vaporized.
This conversion is crucial for calculating work done during the process, as it assists in determining the molar volume of the vapor, demonstrating the integral role of gas laws in thermodynamics.

First Law of Thermodynamics

The First Law of Thermodynamics, often referred to as the law of energy conservation, states that energy cannot be created or destroyed; it can only change forms. The formula \(\Delta E = Q + w\) signifies that the change in internal energy (\(\Delta E\)) of a system is the sum of the heat added (\(Q\)) and the work done (\(w\)) on the system.

• \(Q\) represents heat absorbed or released, with heat absorbed being positive.
• \(w\) indicates work done by or on the system. Work done on the system is positive, while work done by the system is negative.

In this exercise, the First Law is used to calculate the change in internal energy (\