91Ó°ÊÓ

An isobaric process in thermodynamics is one where the pressure remains constant throughout the process. This means that even though the volume and temperature of the gas might change, the pressure holding the gas constant is unchanging. In the context of the Ideal Gas Law, this can be visualized by the equation \( PV = nRT \), where if pressure \( P \) is constant, any increase in volume \( V \) would require a proportional increase in temperature \( T \).
Understanding isobaric processes is crucial because it simplifies many calculations in thermodynamics. We can isolate the effects of volume and temperature changes, making predictions about energy transfer and other phenomena more straightforward.For the given problem, the process was isobaric, meaning the pressure remained constant at \(7.60 \times 10^{5} \, \text{Pa}\). Despite volume increase, this pressure value didn't change, allowing us to effectively calculate the work done by the gas and the heat added during this process.

When gas expands under constant pressure, work is done by the gas. The work done \( W \) during an isobaric process can be calculated using the formula:

• \( W = P \Delta V \)

Here \( P \) is the pressure and \( \Delta V \) is the change in volume. This relationship is useful because it directly ties the macroscopic properties of pressure and volume to the energy considerations in a system.
In our example, the work done by the gas is determined given the constant pressure \(7.60 \times 10^{5} \, \text{Pa}\) and the change in volume. With \( \Delta V = 0.35 \times 10^{-3} \, \text{m}^3\), the work done is calculated as \( 266 \, \text{J} \). This energy is quite literally the effort exerted by the gas as it expands against the atmospheric pressure, which translates into mechanical energy output.

A monatomic ideal gas consists of single-atom molecules, like helium or neon. These gases are "ideal" under high-temperature and low-pressure conditions, meaning they perfectly follow the Ideal Gas Law: \( PV = nRT \). For monatomic gases, internal energy is solely a function of temperature, making many calculations simpler.In an isobaric process, the change in internal energy \( \Delta U \) for monatomic gases is given by:

• \( \Delta U = \frac{3}{2} nR \Delta T \)

However, in our scenario, since these changes occur under constant pressure, we use work \( W \) to express changes: \( \Delta U = \frac{3}{2} W \). This reflects the energy associated with movements of molecules resulting from volume changes.
This concept helps us understand that for monatomic gases, the relationship between the work, heat, and internal energy is straightforward, as they completely utilize energy from heat in changing their internal temperature without complex interactions.