Heating a Gas at Constant Pressure – What Does the Supplied Heat Do? For a fixed mass of ideal gas heated at constant pressure (isobaric process), identify which effects always occur under the ideal-gas model.

Introduction / Context:
Processes at constant pressure are common in heating ducts, boilers, and combustors. Under the ideal-gas assumption, understanding how heat input splits between raising temperature and doing boundary work clarifies energy accounting and helps prevent misconceptions in first-law applications.

Given Data / Assumptions:

• Fixed mass of an ideal gas.
• Isobaric heating (p = constant).
• Quasi-equilibrium path with negligible kinetic/potential energy changes.

Concept / Approach:

For an ideal gas, enthalpy h depends only on temperature, so dh = Cp * dT. During isobaric heating, the gas expands as temperature rises (since p is fixed), performing boundary work W = ∫ p dV. The first law for a closed system gives δQ = dU + δW. Because temperature increases, both internal energy U (via Cv * dT) and enthalpy h rise; simultaneously, boundary work is positive. Among the options provided, the correct guaranteed outcomes are the temperature rise and the performance of external work, collectively listed in option (D).

Step-by-Step Solution:

Verification / Alternative check:

Using enthalpy: δQ = dH for isobaric processes in closed systems with only p–V work, and dH = m * Cp * dT, which is positive; the accompanying expansion implies positive work.

Why Other Options Are Wrong:

(A) alone is incomplete; although U increases, the option does not acknowledge the simultaneous work. (E) claims no work, contradicting expansion at constant pressure. (B) or (C) alone omit the other inevitable effect.

Common Pitfalls:

Assuming δQ equals only temperature rise (forgetting work), or using Cv instead of Cp when analyzing isobaric heating. Also, confusing constant-pressure (p constant) with constant-volume (V constant) heating.

Final Answer:

Both (B) and (C)