91Ó°ÊÓ

The first law of thermodynamics, also known as the law of energy conservation, is a fundamental principle that governs energy interactions. It states that the total change in a system's internal energy is equal to the heat added to the system minus the work done by the system. Mathematically, this is expressed as:\[ \Delta U = Q - W \]Where:

• \( \Delta U \) is the change in internal energy.
• \( Q \) is the heat added to the system.
• \( W \) is the work done by the system.

This equation reveals the delicate balance between heat added and work done. In our problem, we know \( Q = 1500 \, \text{J} \) and \( \Delta U = 4500 \, \text{J} \). By applying the first law, we are able to determine the work done during the process.

Internal energy is a measure of the total energy stored within a system. It includes all forms of energy such as kinetic and potential energy at the microscopic level. When a system undergoes a change, its internal energy might increase or decrease depending on the heat exchange and work interaction.
In the problem presented, the system experiences an increase in internal energy by \( 4500 \, \text{J} \). This suggests that the system has absorbed more energy through heat or work than it has lost. This change reflects the energy balance required in accordance to the first law of thermodynamics. The gain in internal energy can result in changes such as a temperature increase or phase transitions within the system.

The concept of work in thermodynamics refers to the transfer of energy that happens due to a force acting over a distance. When a system performs work on its surroundings, energy is transferred from the system to the surroundings. In the context of our problem, work done by the system is calculated through the relationship:\[ W = p \Delta V \]This formula tells us that when pressure \( p \) is constant, the work done equals the pressure multiplied by the change in volume \( \Delta V \).
In our example, the work done is \( -3000 \, \text{J} \), indicating that work is done on the system, not by it. A negative change in volume \( \Delta V = -0.010 \, \text{m}^3 \) reflects this transfer of energy into the system, thereby increasing its internal energy.

Constant pressure processes are common in thermodynamic interactions, particularly in those involving gases. When pressure remains fixed over a process, it simplifies calculations related to work and energy. Under constant pressure, the volume change directly contributes to work calculations, as described by the equation:\[ W = p \Delta V \]Here, the assumption of constant pressure enables us to compute the exact work done by or on the system.
In this problem, the pressure is constant, which allows us to calculate the pressure by rearranging the equation derived from the first law of thermodynamics. By substituting \( W = -3000 \, \text{J} \) and \( \Delta V = -0.010 \, \text{m}^3 \), we find the pressure is \( 300,000 \, \text{Pa} \) or \( 300 \, \text{kPa} \). This pressure value is a critical parameter that determines how the system interacts with its surroundings during the energy exchange.