91Ó°ÊÓ

A monatomic ideal gas consists of single-atom molecules, like helium or neon. Ideal gases follow the ideal gas law, which assumes that:

• Particles are in constant random motion, colliding elastically with each other and the walls of the container.
• The volume of gas particles themselves is negligible compared to the volume of their container.
• No forces act upon the gas particles except during collisions.

These properties make ideal gases simple to analyze using fundamental thermodynamic equations. When dealing with a monatomic ideal gas, we often engage with concepts such as pressure, volume, temperature, and internal energy. Its behavior is governed by statistical mechanics and can be described accurately by mathematical models like the one used in the exercise.

Internal energy in thermodynamics refers to the total energy contained within a system, largely due to the random motion of its particles. For a monatomic ideal gas, the internal energy depends only on its temperature and is given by:\[ U = \frac{3}{2} nRT \]Here, \( n \) is the number of moles and \( R \) is the ideal gas constant. Internal energy changes when heat is added or removed or when work is done by or on the system. According to the first law of thermodynamics, \[ \Delta U = Q - W \]where \( Q \) is the heat transferred to the system, and \( W \) is the work done by the system. In the given problem, the internal energy decreases by 1300 J as the gas performs work while absorbing heat, illustrating these dynamic exchanges.

The temperature of a gas often changes when heat is absorbed or when work is performed. In the context of a monatomic ideal gas, the change in temperature is directly linked to a change in its internal energy.For this exercise, we apply the equation relating change in internal energy to temperature change:\[ \Delta U = \frac{3}{2} nR \Delta T \]Given the change in internal energy \( \Delta U = -1300 \text{ J} \) and known values for \( n \) and \( R \), we can solve for the temperature change \( \Delta T \). Substituting the given values shows that \( \Delta T \approx -208.5 \text{ K} \), indicating a decrease in temperature due to a net energy reduction.

Work done by a gas involves energy being transferred as the gas expands or contracts against an external pressure. In this problem, 2500 J of work is done by the monatomic ideal gas, which expands while doing so. This work contributes to the decrease in the system’s internal energy as part of the energy is used to perform work rather than increasing the internal energy. This principle highlights the trade-off in energy distribution according to the first law of thermodynamics:\[ \Delta U = Q - W \]As the gas does work, its tendency is to lose internal energy, emphasizing that energy within a system can take different forms, such as kinetic (increasing motion) or being utilized in external work.

Heat absorbed by a system is one form of energy transfer that can lead to an increase in the system’s internal energy. It involves the introduction of external energy into the gas, which could result in increased particle motion, increasing the temperature.In this scenario, the monatomic ideal gas absorbs 1200 J of heat. However, because the gas also does 2500 J of work, the overall internal energy decreases. This net effect on internal energy is quantitatively expressed as:\[ \Delta U = Q - W = 1200 \text{ J} - 2500 \text{ J} = -1300 \text{ J} \]Hence, even as it absorbs heat, more energy is expended in doing work, resulting in a decrease in temperature, demonstrating how heat and work are interconnected in thermodynamic processes.