Question

When 1 mole of benzene is vaporized at a constant pressure of \(1.00\) atm and at its boiling point of \(353.0 \mathrm{~K}, 30.79 \mathrm{~kJ}\) of energy (heat) is absorbed and the volume change is \(+28.90 \mathrm{~L}\). What are \(\Delta E\) and \(\Delta H\) for this process?

Short answer

For this process, the change in internal energy (ΔE) is \(1.508 \mathrm{~kJ}\), and the change in enthalpy (ΔH) is \(30.79 \mathrm{~kJ}\).

Step-by-step solution

Calculate the work done on the system

We can use the formula for work, which is w = -PΔV. Since we are given the pressure P = 1.00 atm and the volume change ΔV = +28.90 L, we can plug these values into the formula: w = -PΔV = -(1.00 atm)(28.90 L) To continue, we need to convert pressure and volume from atm and L, respectively, to Joules (J). We can use the conversion factor: 1 L·atm = 101.325 J. So, we have: w = -(1.00 × 28.90 × 101.325 J) = -29282.025 J ≈ -29.282 kJ

Determine the change in internal energy ΔE

According to the first law of thermodynamics, ΔE = q + w. We are given the heat absorbed (q) as 30.79 kJ, and we calculated work (w) as -29.282 kJ. So, we can now calculate ΔE: ΔE = q + w = 30.79 kJ - 29.282 kJ = 1.508 kJ

Determine the change in enthalpy ΔH

To calculate the change in enthalpy ΔH, we use the formula ΔH = ΔE + PΔV, where we already have ΔE = 1.508 kJ, P = 1.00 atm, and ΔV = 28.90 L: ΔH = ΔE + PΔV = 1.508 kJ + (1.00 atm × 28.90 L) Now, we need to convert (1.00 atm × 28.90 L) to Joules using the conversion factor 1 L·atm = 101.325 J: ΔH = 1.508 kJ + (1.00 × 28.90 × 101.325 J) = 1.508 kJ + 29282.025 J = 1.508 kJ + 29.282 kJ = 30.79 kJ So, for this process: - ΔE = 1.508 kJ - ΔH = 30.79 kJ

Key concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that describes the conservation of energy in a system. It states that energy cannot be created or destroyed but can only be transformed from one form to another. This law is essentially a statement about the conservation of total energy.
In the context of the given problem, when benzene evaporates, energy transitions from heat to other forms within the system. The First Law is expressed mathematically as \[ \Delta E = q + w \] where:

• \( \Delta E \) is the change in internal energy.
• \( q \) represents the heat exchanged with the surroundings. If heat is absorbed by the system, \( q \) is positive.
• \( w \) is the work done on or by the system. This value is negative if work is done by the system, as in the expansion during vaporization.

Understanding this law is vital for exploring how energy changes manifest in physical and chemical processes, and it's crucial for performing energy calculations in thermodynamic problems.

Enthalpy Change

Enthalpy is another important concept in the study of thermodynamics. It is a measure of the total heat content of a system and is defined as the internal energy plus the product of pressure and volume, mathematically given as \[ H = E + PV \].
Change in enthalpy, \( \Delta H \), is especially useful in situations where processes are carried out at constant pressure, which is common in many chemical reactions.
In the given exercise, the change in enthalpy (\( \Delta H \)) is calculated as the sum of the change in internal energy (\( \Delta E \)) and the work done by pressure-volume changes. Specifically:

• \( \Delta H = \Delta E + P \Delta V \)
• Here, \( \Delta E \) is the internal energy change calculated earlier as \( 1.508 \) kJ, and \( P \Delta V \) is calculated as \( 29.282 \) kJ.
• This results in a total \( \Delta H \) of \( 30.79 \) kJ.

This measurement provides insight into the energy exchange as heat during a phase change at boiling point.

Internal Energy

Internal energy, denoted by \( E \), is the total energy contained within a system due to the random motions and interactions of its molecules. It includes energy associated with the translational, rotational, and vibrational motion of molecules, as well as potential energy between molecules.
The change in internal energy \( \Delta E \) can be influenced by heat transfer and work done on or by the system. Calculating \( \Delta E \) involves using the First Law of Thermodynamics:

• \( \Delta E = q + w \)
• Where \( q \) is the heat absorbed, which is \( 30.79 \) kJ from the exercise, and \( w \) is the work done, calculated as \( -29.282 \) kJ.
• The resulting \( \Delta E \) is therefore \( 1.508 \) kJ.

This value indicates the net energy change within the benzene during evaporation, accounting for both absorbed heat and mechanical work.

Work and Energy Conversion

Work and energy conversion play significant roles in thermodynamic processes and transformations. Work in thermodynamics often involves pressure-volume changes, as substances like gases expand or compress within a system.
In the context of the given problem, work is calculated using the formula:

• \( w = -P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the volume change.
• For the vaporization of benzene, this results in \( w = -29.282 \) kJ.

The negative sign denotes work done by the system during expansion. This work reflects in energy conversion processes where the system's internal energy decreases, transferring energy in the form of work to the surroundings.
Grasping how work interacts with energy in thermodynamic systems is critical. It aids in understanding how different energy types interconvert, crucial for both theoretical studies and practical applications in energy management.