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Ideal gases are a simplified model used to understand the behavior of gases under various conditions. They follow the ideal gas law, which is given by the formula \[ PV = nRT \] where:

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

An ideal gas assumes no intermolecular forces and that the volume occupied by the gas molecules themselves is negligible. This model helps us predict how gases will react to changes in pressure, volume, and temperature. Notice that in an isothermal process, where temperature remains constant, the internal energy of an ideal gas does not change.

Internal energy refers to the total energy contained within a system due to the kinetic energy of its particles. In the case of an ideal gas, the internal energy is entirely dependent on the temperature of the gas. This is because, for an ideal gas, internal energy is a function of the translational motion of particles, which is directly related to temperature.

During an isothermal process, where the temperature stays constant, the internal energy of an ideal gas does not change. Therefore, the change in internal energy, denoted as \( \Delta U \), is zero. Understanding this concept is crucial when analyzing processes involving heat and work in thermodynamic systems.

The first law of thermodynamics is all about the conservation of energy. It states that the change in internal energy \( \Delta U \) of a system is equal to the heat added to the system \( q \) plus the work done on the system or by the system \( w \). This can be expressed with the equation:\[ \Delta U = q + w \]
Since, in an isothermal process for ideal gases, the internal energy change \( \Delta U \) is zero, this equation simplifies to:\[ 0 = q + w \]With this simplification, it can be derived that the work done \( w \) is equal in magnitude but opposite in sign to the heat exchanged \( q \), specifically \( w = -q \). In the given problem where \( q = -76 \mathrm{~J} \), the work done would be \( w = 76 \mathrm{~J} \), as the system does work equal to the heat lost.