Heat capacity integration — Air has Cp ≈ 26.693 + 7.365×10^-3·T (J/mol·K), with T in K. How much heat (kJ) does 1 mol of air give off when cooled at constant pressure from 500°C to −100°C?

Introduction / Context:
When specific heat depends on temperature, accurate enthalpy changes require integrating Cp over the temperature range. This is common in combustion products, gas cooling, and cryogenic operations. Here, you must perform a straightforward definite integral for Cp(T) over the cooling path to find the heat released by 1 mol of air at constant pressure.

Given Data / Assumptions:

• Cp(T) = 26.693 + 7.36510^-3 * T (J/mol·K).
• Initial temperature T1 = 500°C = 773.15 K.
• Final temperature T2 = −100°C = 173.15 K.
• Constant pressure process; ideal-gas approximation for air.

Concept / Approach:

For a constant-pressure process, the molar enthalpy change is ΔH = ∫(from T2 to T1) Cp(T) dT if you want the magnitude of heat released (positive). The sign convention for cooling is that the system releases heat: q_p = −∫(from T2 to T1) Cp dT if proceeding from hot to cold. We will compute the magnitude of heat given off.

Step-by-Step Solution:

Verification / Alternative check:

A linear Cp(T) gives a quadratic antiderivative. A quick midpoint estimate using Cp at ~473 K also gives a close value, reinforcing the integral result.

Why Other Options Are Wrong:

10.73 and 16.15 kJ underestimate the integrated Cp; 18.33 kJ slightly overestimates; 20.50 kJ is too high for the specified range and Cp.

Common Pitfalls:

Using constant Cp instead of integrating; forgetting unit conversions from J to kJ; mixing Celsius with Kelvin in Cp expressions.

Final Answer:

18.11 kJ