First Law and PV Work Quiz: Constant Pressure + PV Area

Concept: Work sign and path dependence. The sign of work follows the sign of Δv for positive pressure. Work depends on the path because the pressure during the volume change can vary between processes.

Concept: Free expansion + adiabatic condition. Insulated means q=0, and expansion into a vacuum means no work is done (w=0). Therefore Δu=0, so internal energy does not change.

Concept: Work requires opposing pressure. Mechanical work on the surroundings requires a force (pressure) opposing expansion. If external pressure is zero, there’s nothing to push against, so w=0.

Concept: PV area and work. Work is the area under the PV curve between the same volumes. Higher pressure at each volume means a larger area, so process A produces greater work.

Concept: Cycles and net work. A system can return to its original state (including temperature) and still do net work if it also has net heat transfer. That’s the basic idea behind heat engines.

Concept: First law for cycles. Over a cycle Δu=0, so 0=q_net−w_net. Therefore q_net = w_net, meaning net heat in equals net work out (using this convention).

Concept: State function over a cycle. A cycle returns the system to its initial state, so any state function has zero net change. Since u is a state function, Δu=0 for a complete cycle.

Concept: Rearranging the first law. From Δu=q−w, solve for q: q=Δu+w. q=80+50=130 J, meaning net heat enters the system.

Concept: Meaning of Δu sign. Δu<0 means the final internal energy is lower than the initial internal energy. This typically occurs when the system does more work than the heat it absorbs, or when it releases heat.

Concept: Using signs for q and w. Releasing heat means q=-40, and doing work means w=+150. Δu=q−w=-40-150=-190 J, so internal energy decreases.

Concept: PV work at constant pressure. For constant pressure expansion/compression, work equals pressure times volume change. This is a special case of w=∫p dv where p is constant.

Concept: Subtracting a negative. If w is negative, then -w is positive. This means compression tends to increase Δu for a given q because work is effectively adding energy to the system.

Concept: pΔv with compression. Δv=-3.0 L=-3.0×10^(-3) m³ and p=200,000 Pa. w=200,000(-0.003)=-600 J, negative as expected for compression.

Concept: Work sign for compression. Compression means Δv<0, so w=pΔv is negative (for positive pressure). Negative w indicates work is done on the gas, not by it.

Concept: First law calculation. With q=+600 J and w=+300 J, Δu=q−w=600−300=300 J. Internal energy increases because heat input exceeds work output.

Concept: PV area as work. Work is w=∫p dv, which geometrically equals the area under the p–v curve. The sign depends on whether volume increases (positive) or decreases (negative).

Concept: Work from pΔv with SI units. Convert pressure: 100 kPa = 100,000 Pa, and Δv=3.0×10^(-3) m³. Then w=(100,000)(3.0×10^(-3))=300 J.

Concept: Unit conversion liters to m³. 1 L equals 10^(-3) m³. So 3.0 L becomes 3.0×10^(-3) m³.

Concept: Calculating Δv. Volume change is final minus initial: 5.0−2.0=3.0 L. A positive result matches expansion.

Concept: Volume change sign. Expansion means final volume is larger than initial volume. Therefore Δv=v_f−v_i is positive.