91Ó°ÊÓ

In thermodynamics, when we talk about "work done by gas," we're referring to the energy transferred by the gas when it expands or is compressed against a pressure.
For a gas, the work done can be calculated using the formula:

• \( W = P \cdot (V_f - V_i) \)

Here, \( W \) is the work done, \( P \) is the constant pressure, and \( V_f - V_i \) is the change in volume.
In the given exercise, the gas is compressed, meaning the final volume \( V_f \) is less than the initial volume \( V_i \), which results in a negative work value \( (-1.15 \times 10^5 \, \text{J}) \).
This negative sign signifies that work is being done on the gas rather than by the gas. So, even though the gas itself isn't actively doing work (like expanding), outside forces are doing work on it to compress it.

Heat flow (\( Q \)) is a fundamental concept in thermodynamics that refers to the transfer of thermal energy in and out of a system.
In this problem, we need to determine if the gas is gaining or losing heat using the first law of thermodynamics.

• The first law of thermodynamics can be expressed as: \( \Delta U = Q - W \)

This formula tells us that the change in internal energy (\( \Delta U \) ) of a system is equal to the heat added to the system, minus the work done by the system.
By rearranging this equation, we find:

• \( Q = \Delta U + W \)

Substituting in the given values:

• \( \Delta U = -1.40 \times 10^5 \, \text{J} \)
• \( W = -1.15 \times 10^5 \, \text{J} \)
• \( Q = -2.55 \times 10^5 \, \text{J} \)

The negative sign indicates that heat is leaving the system, which means heat is flowing out of the gas.

Internal energy change (\( \Delta U \)) refers to the change in the total energy contained within the gas. This can happen through changes in temperature, phase, or volume.
In the context of the first law of thermodynamics, the change in internal energy is directly tied to both the work done and heat transfer.
The equation\( \Delta U = Q - W \) summarizes this relationship.
When the internal energy of the gas decreases, as in this exercise, it implies that either work is done on the gas, heat is lost, or both.
Here, with a decrease of \(-1.40 \times 10^5 \, \text{J}\), we know:

• The compression of the gas represents work done on it
• Negative heat flow (\(-2.55 \times 10^5 \, \text{J}\)) suggests heat was removed

Despite changes, this problem's context shows that the nature of the gas (ideal or real) doesn’t impact these calculations much, as these depend mainly on pressures and volumes involved.