91Ó°ÊÓ

In the realm of thermodynamics, work plays a pivotal role when studying how systems like gases interact with their surroundings. Understanding work done is key to grasping how energy is transferred. Work is essentially a measure of energy transfer that occurs when an object moves because of force acting upon it. In thermodynamics, especially with gases, we often see this work done during a change in volume at a given pressure. The primary formula to calculate this is:\[ W = P \times \Delta V \]Where:

• \( W \) represents the work done.
• \( P \) stands for the constant pressure exerted on or by the gas.
• \( \Delta V \) is the change in volume.

It's important to remember that work done by a gas system is typically negative because the system uses its internal energy to perform that work. The term "against constant pressure" indicates that the external pressure remains unchanged throughout the process. This uniformity simplifies the calculation as it allows us to directly apply the formula without adjustment over varying pressures.In the context of our exercise, the work done was calculated as \(-900 \; \mathrm{J}\) due to the negative sign indicating the system's interaction with its environment.

In many thermodynamic processes, especially those involving gases, we may encounter scenarios where the pressure remains constant. This type of process is known as an isobaric process. An isobaric process is significant because it helps us understand how gases behave under steady pressure conditions, which are common in many practical applications such as engines and natural gas cycles.During a constant pressure process, the gas might undergo changes in volume and temperature, but the pressure continues to stay the same, simplifying certain calculations. This constancy often allows engineers and scientists to predict outcomes without needing to factor in variations in pressure.When a gas expands at a constant pressure, as seen in our exercise, it indicates an exchange of energy as the gas pushes against its surroundings. This is why the calculated work done has a precise value: the pressure of \(1 \times 10^5\; \mathrm{Nm}^{-2}\) remains unchanged as the gas expands from an initial to a final volume. The significance of a constant pressure process cannot be understated, as it allows us to utilize straightforward mathematical principles to deduce various thermodynamic variables.

Volume expansion is a crucial concept in understanding how gases behave when subjected to changes in energy and pressure. It describes how the volume of a gas increases when factors such as pressure decrease or the gas is heated, according to the ideal gas law.The ideal gas law \( PV = nRT \) gives an insightful perspective on the relationship between pressure \( P \), volume \( V \), and temperature \( T \) for an ideal gas. If the pressure is kept constant and the gas is allowed to expand, the volume change can be observed as the temperature or the amount of gas (number of moles \( n \)) increases.In our exercise, the gas expanded in volume from \(1 \times 10^{-3} \mathrm{~m}^3\) to \(1 \times 10^{-2} \mathrm{~m}^3\) under constant pressure. This volume change \(\Delta V\) is directly used to calculate the work done. The formula for calculating the volume expansion here became straightforward, as it only required a simple subtraction of initial volume from final volume: \[ \Delta V = V_f - V_i = 9 \times 10^{-3} \; \mathrm{m}^3 \]Such calculations not only highlight the quantitative change but also illustrate the energetic implications of gas behavior. Understanding volume expansion in this structured way is vital for comprehending broader topics in thermodynamics, especially when viewing how gases adapt to changes in their environment.