1. Concept:The First Law of Thermodynamics dictates that the change in a system's internal energy is the difference between the heat added to it and the mechanical work it performs on its surroundings.
2. Formula:$$W = P \Delta V \quad \text{and} \quad \Delta U = Q - W$$
3. Solution:- First, calculate the mechanical work done by the gas during expansion: \(W = P \times \Delta V\).
- Substitute the given values: \(W = (1 \text{ N/m}^2) \times (0.1 \text{ m}^3) = 0.1 \text{ Joules}\).
- Next, apply the First Law of Thermodynamics. The heat provided to the gas is \(Q = +0.1 \text{ J}\).
- Calculate the change in internal energy: \(\Delta U = 0.1 \text{ J} - 0.1 \text{ J} = 0 \text{ J}\).
- Since the net change in internal energy is exactly zero, the internal energy strictly remains the same.
4. Why other options are incorrect:If a student incorrectly added the work instead of subtracting it (assuming work was done ON the gas instead of BY the gas), they would get an increase of 0.2 J. If they ignored the work entirely, they would assume it increases by 0.1 J. Proper sign convention is required to see they cancel out perfectly.
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