1. Since no heat is supplied or extracted, the process is adiabatic, so $Q=0.$

2. By the first law of thermodynamics, $\Delta U = Q - W,$ where $W$ is the work done by the gas.

Thus, $\Delta U = -W.$

3. From the $P$-$V$ graph, pressure changes linearly from point $A$ to $B$. The work done is the area under the straight line between the two volumes.

For a linear $P$-$V$ variation, $W = \text{average pressure} \times \Delta V.$

From the figure, the pressure decreases from $3\,\text{N/m}^2$ to $1\,\text{N/m}^2$ while the volume increases from $1\,\text{m}^3$ to $4\,\text{m}^3$.

So, $W = \frac{3+1}{2}\times (4-1) = 2\times 3 = 6\,\text{J}.$

4. Therefore, $\Delta U = -6\,\text{J}.$

So the internal energy decreases by $6\,$J.

5. Comparing with the given options:

• A: $6\,$J
• B: $4.5\,$J
• C: $0$
• D: $-4.5\,$J

None matches $-6\,$J.

Hence the stored correct answer $B$ is inconsistent with the thermodynamic calculation based on the graph values described.