Cooling a Sealed Rigid Tank
A rigid, insulated steel tank of fixed volume 0.020 m³ holds 0.50 mol of a gas modeled as ideal and monatomic. Its initial temperature is 600 K. A small electric heater inside the tank is switched on and delivers 900 J of energy while the tank's insulation keeps heat exchange with the surroundings at zero. A student claims that because the tank is rigid, the gas does no work and its temperature rise can be found from ΔU = Q alone, treating the heater energy as heat. A second student insists the heater energy is work, not heat, and that adding it as Q double-counts. The gas remains a single phase throughout.
1. Identify the thermodynamic process and decide whether the heater energy should be classified as heat Q or work W for the first law. Explain why the numerical result for ΔU is the same either way.
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The volume is fixed, so the gas does zero boundary work (∫P dV = 0). The first law is ΔU = Q − W (or Q + W depending on sign convention); either way the 900 J enters the system through a non-mechanical channel (electrical energy dissipated in the resistor). Classifying it as Q_in or as W_on changes only the bookkeeping, not ΔU: ΔU = 900 J. The rigid tank guarantees no PΔV work, so no work term is missing. The insulation guarantees no heat leak to the surroundings. The key judgment is that the energy transfer is 900 J into the gas regardless of label.
2. Compute the final temperature and the final pressure. Use R = 8.31 J/(mol·K).
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For a monatomic ideal gas, C_V = (3/2)R = 12.47 J/(mol·K). ΔU = n C_V ΔT → ΔT = 900 / (0.50 × 12.47) = 900 / 6.23 ≈ 144 K. T_f = 600 + 144 ≈ 744 K. Pressure from ideal gas law: P = nRT/V. P_i = (0.50)(8.31)(600)/0.020 = 2493/0.020 ≈ 1.25×10⁵ Pa. P_f = (0.50)(8.31)(744)/0.020 ≈ 3091/0.020 ≈ 1.55×10⁵ Pa. Check: P_f/P_i = T_f/T_i = 744/600 = 1.24, consistent with constant V and n.
3. A third student proposes the temperature rise could be estimated using C_P instead of C_V 'because pressure changes.' Evaluate this reasoning.
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It is incorrect. C_P applies to a constant-pressure process where the gas expands and does PΔV work; here V is fixed, so no expansion work occurs and all internal energy change shows up as ΔT. Using C_P = (5/2)R would give ΔT = 900/(0.50×20.8) ≈ 86.6 K, understating the rise by about 40%. The pressure rises as a consequence of the temperature rise at fixed volume, not because the process is isobaric. The correct heat capacity is selected by the constraint (constant V), not by which state variable happens to change.