Thermodynamics — Set 3

Correct Answer: D. Open system

• **Open system** = An open system allows both mass and energy to cross its boundary; examples include turbines, compressors, and living organisms. • **Mass flow** — the presence of mass exchange distinguishes an open system from a closed system, which allows energy but not mass to cross the boundary. • 💡 Wrong-option analysis: Closed system: a closed system allows energy exchange but not mass exchange; Isolated system: an isolated system allows neither mass nor energy exchange with surroundings; Adiabatic system: adiabatic means no heat exchange, but mass exchange may still be possible.

Correct Answer: A. Closed system

• **Closed system** = A closed system allows heat and work to cross its boundary but does not permit any mass to enter or leave. • **Sealed piston-cylinder** — a gas enclosed in a sealed cylinder with a movable piston is the classic example of a closed system allowing energy but not mass transfer. • 💡 Wrong-option analysis: Isolated system: an isolated system allows neither energy nor mass exchange, which is more restrictive than closed; Perfectly rigid system: rigid means no work is done (no volume change), but the system can still exchange heat; Open system: an open system allows both mass and energy exchange, unlike a closed system.

Correct Answer: A. Isolated system

• **Isolated system** = An isolated system has no interaction with its surroundings: neither mass nor energy in any form can cross its boundary. • **Second law application** — the isolated system concept is central to the second law; entropy of a truly isolated system can never decrease over time. • 💡 Wrong-option analysis: Open system: an open system allows both mass and energy exchange, which is the opposite of isolated; Closed system: a closed system allows energy exchange but not mass exchange; Isothermal system: isothermal describes constant temperature, not the absence of mass and energy exchange.

Correct Answer: A. Pressure

• **Pressure** = Pressure does not change when the system is subdivided; it is the same in each part as in the whole, making it an intensive property. • **Independent of size** — intensive properties such as temperature, density, and pressure are independent of the amount of substance, so they are the same for any sample of a homogeneous system. • 💡 Wrong-option analysis: Internal energy: internal energy scales with the amount of substance, so doubling the mass doubles U, making it extensive; Mass: mass increases proportionally with the size of the system, so it is extensive; Volume: volume scales with system size and doubles when mass doubles, so it is extensive.

Correct Answer: C. Volume

• **Volume** = Volume depends on the amount of matter in the system; doubling the mass at fixed density doubles the volume, so it is an extensive property. • **Additive** — extensive properties are additive: the total volume of two combined subsystems equals the sum of their individual volumes. • 💡 Wrong-option analysis: Density: density is mass per unit volume and is the same for each part of a homogeneous system, making it intensive; Temperature: temperature is the same throughout a system at thermal equilibrium and does not add up when systems are combined; Pressure: pressure is uniform in a system at mechanical equilibrium and does not change with system size.

Correct Answer: C. Heat

• **Heat** = Heat is not a property stored by the system; it depends on the specific path taken between states, so it is a path function denoted δQ, not dQ. • **Path-dependent** — the same initial and final states can be connected by an isothermal path, an adiabatic path, or others, each transferring a different amount of heat. • 💡 Wrong-option analysis: Enthalpy: enthalpy H = U + PV is a state function whose value is fixed once the state is fixed; Internal energy: internal energy depends only on the state of an ideal gas, not on the path; Entropy: entropy is a state function defined by the integral of δQrev/T, independent of path.

Correct Answer: B. It is zero

• **It is zero** = Work done by a gas is W = ∫ P dV; in an isochoric process dV = 0 throughout, so the integral is zero regardless of how pressure changes. • **ΔU = Q** — because W = 0, all heat supplied to the gas goes entirely into increasing its internal energy, making Q = ΔU = nCvΔT. • 💡 Wrong-option analysis: It equals nRT: nRT is the product PV from the ideal gas equation, not the work done in an isochoric process; It is negative: negative work requires the gas to be compressed (dV < 0), but volume is fixed in an isochoric process; It is maximum: maximum work is done in a reversible isothermal expansion, not in a constant-volume process.

Correct Answer: C. The area under the curve

• **The area under the curve** = Since W = ∫ P dV, the work done by the gas equals the area enclosed between the process curve and the volume axis on a P-V diagram. • **Sign rule** — the area is positive when volume increases (expansion, gas does work) and negative when volume decreases (compression, work done on gas). • 💡 Wrong-option analysis: The intercept on the P-axis: the P-intercept is a single point representing a pressure value at zero volume, with no area and no thermodynamic work meaning; The slope of the curve: the slope dP/dV is related to the compressibility of the gas, not to the work done; The area above the curve: the region above the curve on a P-V diagram has no standard thermodynamic significance for work calculations.

Correct Answer: C. A reversible engine

• **A reversible engine** = Carnot's theorem states that no engine operating between the same two temperature reservoirs can be more efficient than a reversible engine. • **Upper bound** — the Carnot efficiency η = 1 − Tc/Th is the absolute maximum; all real irreversible engines are less efficient due to friction and heat leaks. • 💡 Wrong-option analysis: An engine with the largest cylinder: cylinder size has no direct bearing on thermodynamic efficiency; An engine with maximum friction: friction is a dissipative irreversibility that reduces efficiency below the Carnot limit; Any irreversible engine: irreversible engines always have efficiency strictly less than the Carnot (reversible) limit.

Correct Answer: D. Second law

• **Second law** = A perpetual motion machine of the second kind would convert heat from a single reservoir entirely into work in a cycle, which the Kelvin-Planck statement of the second law declares impossible. • **100% efficiency** — such a machine would have η = 1, requiring Tc = 0 K (absolute zero) which is physically unattainable, further confirming the violation. • 💡 Wrong-option analysis: First law: the first law forbids creating energy from nothing (perpetual motion of the first kind), not converting heat completely into work; Zeroth law: the zeroth law defines thermal equilibrium and temperature; it says nothing about engine efficiency; Third law: the third law states that absolute zero cannot be reached in a finite number of steps, but it does not directly address engine efficiency.