Heating a gas at constant pressure – distribution of the supplied heat When a gas is heated at constant pressure (isobaric process), the supplied heat is used to accomplish which of the following?

Introduction / Context:Specific heats at constant pressure (cp) and constant volume (cv) differ because, at constant pressure, a portion of the heat input must push back the surroundings as the gas expands. This explains why cp > cv for ideal gases.

Given Data / Assumptions:

• Ideal-gas model for clarity.
• Isobaric (constant-pressure) heating with a movable boundary (e.g., piston).
• No phase change occurs.

Concept / Approach:First law for a quasi-static process: δQ = dU + δW. At constant pressure, δW = p dV. For an ideal gas, dU = m * cv * dT, and δQ = m * cp * dT, with cp = cv + R (per unit mass basis), showing that part of the heat goes into internal energy rise and part into boundary work.

Step-by-Step Solution:Write δQ = dU + p dV.Use ideal-gas relation at constant p: p dV = m R dT.Relate internal energy change: dU = m cv dT.Thus δQ = m (cv dT + R dT) = m cp dT, where cp = cv + R.Interpretation: heat input raises internal energy and performs external work simultaneously.

Verification / Alternative check:Calorimetry experiments consistently measure cp > cv, directly confirming that additional heat beyond dU is required to provide boundary work at constant pressure.

Why Other Options Are Wrong:

• “Internal energy only” corresponds to constant-volume heating.
• “External work only” would imply dU = 0, which is false for temperature rise in a gas.
• “None” and “condensing” do not describe an isobaric heating of a gas.

Common Pitfalls:Confusing heat (a path function) with internal energy (a state property). The split cp = cv + R captures the dual role of the heat input at constant pressure.

Final Answer:increasing the internal energy of gas and also for doing some external work