Problem 32 - Chemistry: The Central Science

Question

How much work (in J) is involved in a chemical reaction if the volume decreases from \(5.00\) to \(1.26 \mathrm{~L}\) against a constant pressure of \(0.857 \mathrm{~atm}\) ?

Step-by-Step Solution

Answer

The work involved in the chemical reaction is approximately 324,722 J (Joules).

Step 1: 1. Identify the given parameters

We are given: Initial volume, V1 = 5.00 L Final volume, V2 = 1.26 L Constant pressure, P = 0.857 atm

Step 2: 2. Convert the units to SI units

We need to convert the volumes from liters to cubic meters (m^3) and the pressure from atm to Pascals (Pa). 1 L = 0.001 m^3 and 1 atm = 101325 Pa So, V1 = 5.00 L × 0.001 m^3/L = 0.005 m^3 V2 = 1.26 L × 0.001 m^3/L = 0.00126 m^3 P = 0.857 atm × 101325 Pa/atm = 86814.025 Pa

Step 3: 3. Use the formula for work done in an isobaric process

The work done (W) in an isobaric process can be calculated using the formula: W = -P × (V2 - V1) Plug in the values: W = -86814.025 Pa × (0.00126 m^3 - 0.005 m^3)

Step 4: 4. Calculate the work done

Now perform the calculations: W = -86814.025 Pa × (-0.00374 m^3) W = 324722.0135 J

Step 5: 5. Write down the final answer

The work involved in the chemical reaction is approximately 324,722 J (Joules).

Key Concepts

Work in chemical reactions

In thermodynamics, work is a crucial concept, especially when considering chemical reactions. Whenever a chemical reaction involves a change in volume, work is done. For example, in reactions that produce gases, the system might expand against an external pressure, performing work on its surroundings. When the system's volume decreases, as in the given problem, it is the surroundings that can do work on the system.
This work can be quantified and is usually expressed in joules in the International System of Units (SI). Understanding the work involved in chemical reactions sheds light on how energy is transferred between a system and its surroundings. This helps in explaining various spontaneous processes and maintaining energy conservation during chemical reactions.

Isobaric process

An isobaric process is a process in which the pressure remains constant. In our example, the reaction takes place at a constant pressure of 0.857 atm. This means that the change in volume happens without altering the pressure value.
Isobaric processes are significant because they simplify calculations due to the constant pressure condition. In systems where such processes occur, they are often accompanied by heat transfer, leading to work being done when volume changes. Since pressure is constant, the energy calculations involve comparing initial and final states. Such a condition allows understanding of work and heat flow in systems, particularly in engines and atmospheric processes.
Isobaric processes are described mathematically with the equation for work:
\( W = -P \times \Delta V \)
where \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.

Pressure-volume work

Pressure-volume work occurs during a volume change in systems, especially during expansion or compression. It represents the work done by or on the system through changes in volume under a given pressure. It is a central concept in thermodynamics and plays an important role in reactions occurring in open systems.
When a system expands, it does work on its surroundings, while compression means work is done on the system. The formula for calculating pressure-volume work in an isobaric scenario is:
\[ W = -P \times (V_{f} - V_{i}) \]
Here, \( P \) represents the pressure, \( V_{f} \) the final volume, and \( V_{i} \) the initial volume.
In the provided exercise, this concept helps determine the work involved when the volume decreases, implying compression at constant pressure. That allows the precise calculation of energy changes associated with the process. Understanding pressure-volume work helps connect observable phenomena to quantitative thermodynamic principles.