Closed, rigid vessel heating: Three moles of N2 at 30°C are heated at constant volume to 250°C. With Cv = 20.8 J mol^-1 °C^-1, what total heat is required (neglect vessel heat capacity)?

Introduction / Context:
This problem is a straightforward application of constant-volume heating of an ideal gas. In a rigid vessel, the volume does not change, so boundary work is zero. The heat added equals the change in internal energy, which is n * Cv * ΔT for ideal gases. Mastering such calculations is essential for quick energy balances in reactors and closed vessel heating steps.

Given Data / Assumptions:

• Gas: nitrogen, treated as ideal with temperature-independent heat capacities over the range.
• n = 3 mol, T1 = 30°C, T2 = 250°C.
• Cv = 20.8 J mol^-1 °C^-1, Cp = 29.1 J mol^-1 °C^-1 (Cp is not used here).
• Rigid vessel implies constant volume, so W = 0.

Concept / Approach:
For an ideal gas at constant volume, ΔU = n * Cv * ΔT and Q = ΔU because there is no boundary work. Temperature change is computed on the Celsius scale because only differences matter (ΔT is the same in K and °C). Substitute the given values to find Q.

Step-by-Step Solution:
Compute temperature rise: ΔT = 250 - 30 = 220 °C.Write the relation at constant volume: Q = n * Cv * ΔT.Insert numbers: Q = 3 * 20.8 * 220 J.Evaluate product: 3 * 20.8 = 62.4; 62.4 * 220 = 13728 J.Select the option that matches 13728 J.

Verification / Alternative check:
As a check, if Cp had been used accidentally, the heat would be higher (3 * 29.1 * 220 ≈ 19194 J), which does not match the constant-volume constraint and confirms that Cv is the correct property here.

Why Other Options Are Wrong:
19206 J: close to using Cp instead of Cv.4576 J and 12712 J: result from arithmetic or unit mistakes, or using wrong ΔT.

Common Pitfalls:
Using Cp instead of Cv; converting temperatures to Kelvin for initial and final values but forgetting that ΔT is unchanged; including boundary work in a rigid vessel where it is zero.

Final Answer:
13728 J