Problem 26 - Chemistry The Central Science In Si Units

Question

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is cooled by removing \(0.655 \mathrm{~kJ}\) of heat. It shrinks on cooling, and the atmosphere does \(382 \mathrm{~J}\) of work on the balloon. (b) A \(100.0-\mathrm{g}\) bar of gold is heated from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) during which it absorbs \(322 \mathrm{~J}\) of heat. Assume the volume of the gold bar remains constant.

Step-by-Step Solution

Answer

(a) Change: -273 J, Exothermic; (b) Change: 322 J, Endothermic.

Step 1: Understanding the First Law of Thermodynamics

The First Law of Thermodynamics can be expressed as \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is the heat exchanged, and \( w \) is the work done on the system.

Step 2: Solving Part (a) - Compute the Change in Internal Energy for the Balloon

Given: \( q = -0.655 \text{ kJ} = -655 \text{ J} \) (since heat is removed), and work done on the balloon \( w = 382 \text{ J} \).Apply the formula: \[ \Delta U = q + w = (-655 \text{ J}) + 382 \text{ J} = -273 \text{ J}. \] The change in internal energy \( \Delta U \) is \(-273 \text{ J} \), meaning the process is exothermic as the internal energy decreases.

Step 3: Solving Part (b) - Compute the Change in Internal Energy for the Gold Bar

Given: \( q = 322 \text{ J} \) (the gold bar absorbs heat), and since the volume remains constant \( w = 0 \text{ J} \) (no work is done).Apply the formula: \[ \Delta U = q + w = 322 \text{ J} + 0 \text{ J} = 322 \text{ J}. \] The change in internal energy \( \Delta U \) is \( 322 \text{ J} \), meaning the process is endothermic as the internal energy increases.

Key Concepts

Change in Internal Energy

In the context of thermodynamics, the change in internal energy is a crucial concept that connects heat exchanges and work done with the energy state of a system. The First Law of Thermodynamics provides the framework for understanding this by stating that \( \Delta U = q + w \). This equation means:

• \( \Delta U \) is the change in internal energy.
• \( q \) is the heat added to the system (positive if added, negative if removed).
• \( w \) is the work done on the system (positive if work is done on the system, negative if done by the system).

Let's dive into a couple of practical examples to see how this works in real-life scenarios:- In the first example with a balloon, the equation \( \Delta U = (-655 \text{ J}) + (382 \text{ J}) \) led us to a negative change in internal energy of \(-273 \text{ J}\). This indicates energy has left the system, decreasing its internal energy.- In the second scenario with the gold bar, the energy change \( \Delta U = 322 \text{ J} \) shows that energy was absorbed by the system, increasing its internal energy. Both examples illustrate how the internal energy change depends on the balance between heat exchange and work done.

Endothermic and Exothermic Processes

A process can either absorb or release energy, and this is where understanding endothermic and exothermic reactions comes into play. The distinctions between these processes are vital because they determine how a system interacts with its surroundings temporarily:

• Endothermic Process: This occurs when a system absorbs heat from its surroundings, resulting in an increase in internal energy. The example of the gold bar heating up is endothermic, as it absorbed \( 322 \text{ J} \) of heat.
• Exothermic Process: In this scenario, the system releases heat to its surroundings, leading to a decrease in internal energy. For instance, when the balloon loses \( 0.655 \text{ kJ} \) and undergoes \( 382 \text{ J} \) of work, the net result is a loss of \( 273 \text{ J} \), characterizing an exothermic process.

These processes are central to chemical reactions and physical changes. Recognizing whether a process is endothermic or exothermic not only predicts energy changes but also gives insights into reaction mechanisms.

Heat Exchange Calculations

Heat exchange calculations help determine the heat absorbed or released during a process. Using the formula \( \Delta U = q + w \), we can decipher the heat flow characteristics in various situations:

• For calculations, always express heat \( q \) in joules. Convert kilojoules to joules when necessary (e.g., \( -0.655 \text{ kJ} \) to \( -655 \text{ J} \)).
• Consider the work term \( w \) based on whether work is done on or by the system. Recall that no work is involved if there's constant volume, as seen in the gold bar example (\( w = 0 \text{ J} \)).

Calculating the heat exchange requires a careful understanding of these conversions and contexts:- Use signs carefully; a negative \( q \) indicates heat leaving the system, while a positive \( q \) denotes heat absorption.- Follow the conventions for \( w \) strictly, as they affect the outcome's sign and magnitude.By being meticulous with heat exchanges in your calculations, you will be better equipped to understand a system's overall energy dynamics.