Problem Analysis: Internal Energy Change

The question asks us to find the change in the internal energy ($\Delta U$) of a gas system. We are given two key pieces of information:

• The amount of heat absorbed by the system (Q).
• The amount of work done *on* the system (W).

We need to use the fundamental principles of thermodynamics to relate these quantities.

Thermodynamics: First Law Formula

The relationship between internal energy change, heat transfer, and work done is described by the First Law of Thermodynamics. The formula is:

$ \Delta U = Q + W $

Where:

• $ \Delta U $ represents the change in internal energy of the system.
• $ Q $ represents the heat added to the system. A positive value means heat is absorbed by the system.
• $ W $ represents the work done on the system. A positive value means work is done *on* the system.

It's crucial to get the signs right:

• Heat absorbed by the system: $ Q $ is positive.
• Heat released by the system: $ Q $ is negative.
• Work done *on* the system: $ W $ is positive.
• Work done *by* the system: $ W $ is negative.

Calculation: Internal Energy Change

Let's identify the values given in the problem:

• Heat absorbed, $ Q = +300 \, \text{J} $ (positive because the system is absorbing heat).
• Work done *on* the system, $ W = +60 \, \text{J} $ (positive because work is done on the system).

Now, we substitute these values into the First Law of Thermodynamics equation:

$ \Delta U = Q + W $

$ \Delta U = (300 \, \text{J}) + (60 \, \text{J}) $

$ \Delta U = 360 \, \text{J} $

Final Result

The change in the internal energy ($\Delta U$) of the gas system is 360 J.