Isochoric heating (constant volume) – energy and work effects When a gas is heated at constant volume, which statement best describes where the supplied heat goes and what happens to the system properties?

Introduction / Context:
Isochoric (constant-volume) processes occur in rigid vessels and also model heat rejection/addition legs in ideal cycles. Properly accounting for energy and work clarifies why cp and cv differ for gases.

Given Data / Assumptions:

• Closed system gas in a rigid container (dv = 0).
• Heat is supplied to the gas.
• Ideal-gas relation used for intuition; qualitative result holds broadly.

Concept / Approach:

The First law for a closed system is ΔU = Q − W. With constant volume, boundary work W_b = ∫ p dv = 0. Therefore, any heat added (Q > 0) increases internal energy: ΔU = Q. For an ideal gas, u = u(T), so increasing internal energy directly increases temperature and pressure (since p = R * T / v and v is fixed). No external boundary work is done because volume does not change.

Step-by-Step Solution:

Verification / Alternative check:

Calorimetry in rigid bombs demonstrates temperature rise with heat input and no boundary work; measured cv relates directly to the temperature change at constant volume.

Why Other Options Are Wrong:

External work during expansion does not occur at constant volume.Statements implying internal energy reduction contradict the First law for Q > 0 with W_b = 0.

Common Pitfalls:

Confusing pressure rise with work: pressure can rise without boundary work if volume is fixed.

Final Answer:

increases the internal energy of the gas and increases the temperature of the gas