Close the first-law energy account

Write the system boundary and signs first

The system is the retained gas. Heat Q is positive when it enters the gas. Work W is positive when the gas does work on its surroundings. With no changes in bulk kinetic or gravitational potential energy, conservation gives ΔU = Q − W, where ΔU is the change in internal energy.

An equivalent convention uses work on the gas, Won = −W, giving ΔU = Q + Won. Both conventions are valid if used consistently. Read the words “by” and “on” before substituting numbers. Neither heat nor work is stored as a separate substance inside the gas.

The OpenStax first-law reference explains this state-versus-transfer distinction. The first law is an energy balance, not a complete model of allowed process direction or efficiency; those require further thermodynamic reasoning.

Supply an internal-energy model

This investigation specifies one mole of a classical monatomic ideal gas with only translational thermal energy. Its energy changes follow ΔU = ³⁄₂nRΔT. Since PV = nRT, the same change is ΔU = ³⁄₂(PfVf − PiVi). Pressure-volume products in L kPa are joules.

The displayed state energy uses U = ³⁄₂PV with one common reference. Only differences enter the balance. This model is not the total internal-energy formula for every molecular gas, near a phase change, or when additional internal modes are important. No species-specific real-gas validity range is established by the controls.

Initially PiVi = 200 × 20 = 4000 J, giving modeled Uinitial = 6000 J and Tinitial approximately 481.116 K. At the default final state 300 kPa, 30 L, PfVf = 9000 J, Ufinal = 13500 J, and Tfinal approximately 1082.511 K. The internal-energy increase is therefore 7500 J for either path.

Reconcile the two heat inputs

Rearrange the first law to Q = ΔU + W. Path A does 3000 J of work and needs Q = 7500 + 3000 = 10500 J of net heat input. Path B does 2000 J and needs 9500 J. The extra 1000 J of heat along A supplies its extra 1000 J of work; the final internal energy is the same.

Check the legs as well as the totals. Path A first raises pressure at 20 L: ΔU = 3000 J, W = 0, Q = 3000 J. Its expansion at 300 kPa adds ΔU = 4500 J, W = 3000 J and Q = 7500 J. The two heat inputs sum to 10500 J.

Path B first expands at 200 kPa: ΔU = 3000 J, W = 2000 J and Q = 5000 J. Raising pressure at 30 L adds ΔU = 4500 J, W = 0 and Q = 4500 J. Each row independently satisfies Q − W = ΔU. Do not confuse net path heat with a general engine’s gross heat intake: a different path can have positive and negative heat legs that must be tracked separately.

Expansion need not imply a temperature rise

Set the final state to 100 kPa and 30 L. Final PV is 3000 J, so ΔU = −1500 J: the gas ends colder than it began despite expanding. Path A does 1000 J of work, giving net Q = −500 J. Path B does 2000 J, giving net Q = +500 J.

Both paths have the same cooling, but one exports net heat while the other receives net heat. Heat input alone does not determine the sign of temperature change: outgoing work also matters. Inspect the individual legs to see how cooling and heating stages combine.

At fixed volume, W = 0 and heat equals ΔU under the stated restrictions. At fixed pressure, some heat can accompany boundary work as well as changing internal energy. For this monatomic ideal gas, those relations lead to molar capacities Cv = ³⁄₂R and Cp = ⁵⁄₂R. These constants apply to this model and to heat capacity under the named constraint, not to arbitrary paths or materials.

Identify what is still missing

The ledger supplies the heat needed to realize each prescribed leg; it does not calculate a heater design, reservoir temperature schedule, duration or entropy production. A path is quasistatic by assumption, but slow motion alone does not prove every real process is reversible. The first law also does not say that all internal energy can be converted into useful work in a cycle.

Transfer using a different work convention

A gas rejects 400 J of heat while its surroundings do 900 J of work on it. Find ΔU using both work conventions. A second path between the same two states has the gas doing 200 J of work. Find its net heat input.

Check the first-law transfer

Q = −400 J and work by the gas is W = −900 J, so ΔU = −400 − (−900) = 500 J. Using work on it instead gives −400 + 900 = 500 J. The second path must have the same ΔU. With W = +200 J, Q = ΔU + W = 700 J into the gas. The different heat transfers are consistent with the same endpoints.