First law at constant pressure (closed system): A gas mixture expands from 0.03 m^3 to 0.06 m^3 at a constant pressure of 1 MPa while absorbing 84 kJ of heat. Determine the change in internal energy ΔU.

Introduction / Context:
For a closed system undergoing a quasi-equilibrium process at constant pressure, the first law of thermodynamics connects heat transfer, boundary work, and the change in internal energy. This problem reinforces correct work calculation and sign convention to obtain ΔU.

Given Data / Assumptions:

• Constant pressure p = 1 MPa throughout the expansion.
• Initial volume V1 = 0.03 m^3, final volume V2 = 0.06 m^3.
• Heat added to the system Q = +84 kJ.
• Closed system; kinetic and potential energy changes are negligible.

Concept / Approach:
First law for a closed system: Q − W = ΔU, where W is boundary work. For a constant-pressure process, W = p * ΔV. Ensure consistent units (MPa·m^3 = kJ) and correct signs (work done by the system is positive, thus subtracted from Q to get ΔU).

Step-by-Step Solution:

Verification / Alternative check:
Unit check: 1 MPa·m^3 = 1,000 kPa·m^3 = 1,000 kN/m^2 * m^3 = 1,000 kN·m = 1,000 kJ/m^3 * m^3 → kJ, consistent. The positive ΔU indicates that not all added heat went into boundary work; internal energy increased.

Why Other Options Are Wrong:

• 30 kJ: This is the boundary work, not ΔU.
• 84 kJ: Ignores work; would be correct only if ΔV = 0.
• 114 kJ: Sum of Q and W; violates first law sign convention.

Common Pitfalls:
Mixing sign conventions; forgetting that constant pressure work equals area under the p–V rectangle; unit mismatch between MPa and kJ.

Final Answer:
54 kJ