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An ideal gas is a theoretical construct that helps us simplify and understand how gases behave under various conditions. It is composed of a large number of molecules that are in constant motion, following the basic principles of classical mechanics. The defining properties of an ideal gas include:

• There are no forces of attraction or repulsion between the gas molecules.
• The molecules occupy no volume themselves; they are considered point particles.
• All collisions between molecules or with container walls are perfectly elastic, meaning no energy is lost.

These assumptions allow us to use the ideal gas law, which relates pressure (\(P\), volume (\(V\), temperature (\(T\)), and number of moles (\(n\), through the equation: \(PV = nRT\), where \(R\) is the universal gas constant. While real gases do not perfectly follow these assumptions, especially under high pressure or low temperature, the ideal gas model provides a close approximation for many gases under typical conditions.

In an isochoric process, the volume of the gas remains constant. Because the container is rigid, like in the exercise mentioned, the gas cannot expand or compress, making it an isochoric or "constant-volume" process.
This means:

• The work done by the gas, which is calculated using \(W = P\Delta V\), where \(\Delta V\) is the change in volume, equals zero, as \(\Delta V = 0\).

During an isochoric process, any heat added to the system translates entirely to an increase in the internal energy of the gas. The key takeaway is that in an isochoric process:

• Volume remains constant.
• No work is done by the gas.
• Changes in heat directly affect the internal energy.

This type of process is often analyzed using the first law of thermodynamics, as it conveniently highlights how heat input modifies the energy of the system.

The first law of thermodynamics is fundamental as it represents the principle of conservation of energy within a thermodynamic system. It succinctly states:
\(\Delta U = Q - W,\) 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.

In the context of an isochoric process, as we have discussed:

• There is no change in volume, which means \(W = 0\).
• Therefore, the equation simplifies to \(\Delta U = Q\).

This means all the heat energy transferred into the gas is used to increase its internal energy, without performing any work on the surroundings.
In our exercise, since the heat added is \(2.0 \times 10^4\, J\), and no work is done, this entire amount results in an equal increase of the internal energy of the gas. Understanding this law helps us analyze and predict energy changes in systems, providing a foundational tool extensively used in various engineering and physical sciences.