A mixture of gas expands from \(0.03 \text{ m}^3\) to \(0.06 \text{ m}^3\) at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is
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Keep unit consistency intact! Work done at constant pressure can be quickly found by multiplying pressure in MPa directly with volume in $\text{m}^3$, which gives work in MJ: $W = 1 \text{ MPa} \times 0.03 \text{ m}^3 = 0.03 \text{ MJ} = 30 \text{ kJ}$. Then, $\Delta U = 84 - 30 = 54 \text{ kJ}$.
Step 1: Use MPa and cubic metres together as a shortcut for kJ.
A useful fact worth remembering is that \(1 \text{ MPa} \times 1 \text{ m}^3 = 1000 \text{ kJ}\), since \(10^6 \text{ Pa} \times 1 \text{ m}^3 = 10^6 \text{ J} = 1000 \text{ kJ}\). This lets us skip converting to base SI units and back.
Step 2: Compute the boundary work directly in kJ.
The volume change is \(\Delta V = 0.06 - 0.03 = 0.03 \text{ m}^3\), so:
\[
W = P \, \Delta V = 1 \text{ MPa} \times 0.03 \text{ m}^3 = 0.03 \times 1000 \text{ kJ} = 30 \text{ kJ}
\]
Step 3: Apply the first law to isolate the internal energy change.
\[
\Delta U = Q - W = 84 - 30 = 54 \text{ kJ}
\]
\[
\boxed{\Delta U = 54 \text{ kJ}}
\]
This matches option 2.