Thermodynamics — Set 4

Correct Answer: C. J/K

• **J/K** = From the definition dS = δQrev/T, entropy has units of energy per unit temperature, which is joules per kelvin. • **Boltzmann constant** — the Boltzmann constant k_B = 1.38 × 10⁻²³ J/K connects microscopic disorder to macroscopic entropy via S = k_B ln W. • 💡 Wrong-option analysis: K/J: this is the inverse of the correct unit and would make entropy decrease when heat is added; J: joules are units of energy or heat, not of entropy which requires dividing by temperature; N·m/s: this equals watts, the unit of power, which is unrelated to entropy.

Correct Answer: A. Temperature

• **Temperature** = For an ideal gas, internal energy is purely kinetic energy of molecules; since kinetic energy depends only on temperature, U depends only on T (and on the number of moles). • **U = nCvT** — for a fixed amount of ideal gas, internal energy changes only when temperature changes, independent of how pressure or volume vary. • 💡 Wrong-option analysis: Volume: for a real gas, intermolecular forces make U depend on volume, but for an ideal gas with no interactions, U is volume-independent; Number of moles only: mole number alone sets the amount, but temperature determines the energy per mole; Pressure: changing pressure at constant temperature (isothermal) causes no change in U for an ideal gas.

Correct Answer: A. It remains unchanged

• **It remains unchanged** = In free expansion into vacuum, W = 0 (no opposing pressure) and Q = 0 (insulated walls); by the first law ΔU = 0, and since U depends only on T for an ideal gas, temperature does not change. • **Joule's experiment** — James Joule confirmed this result experimentally; temperature of an ideal gas is unchanged by free expansion, validating ΔU = 0 for ideal gases. • 💡 Wrong-option analysis: It decreases: a temperature drop would require either work done against an external force or heat leaving the system, neither of which occurs in free expansion into vacuum; It increases: an increase would need either heat input or work done on the gas, but Q = 0 and W = 0 here; It becomes absolute zero: this would require removing all internal energy, which violates conservation since ΔU = Q − W = 0.

Correct Answer: D. Heat absorbed equals work done

• **Heat absorbed equals work done** = In an isothermal process for an ideal gas, temperature is constant so ΔU = 0; the first law then gives Q = ΔU + W = W, meaning all absorbed heat converts to work. • **Q = W = nRT ln(V2/V1)** — in isothermal expansion, the gas absorbs exactly this amount of heat from the reservoir to maintain constant temperature while doing work. • 💡 Wrong-option analysis: Heat absorbed is zero: if Q = 0 with ΔU = 0, then W = 0 too, but isothermal expansion clearly produces work by pushing the piston; Internal energy increases: ΔU = nCvΔT = 0 because temperature is constant, so internal energy cannot increase; Work done is zero: W = 0 only in an isochoric (constant volume) process; an isothermal expansion involves a volume change and therefore nonzero work.

Correct Answer: D. A device that converts heat into work while operating in a cycle

• **A device that converts heat into work while operating in a cycle** = A heat engine absorbs heat from a high-temperature source, converts part of it into useful work, and rejects the remainder to a low-temperature sink in a repeating cycle. • **Efficiency η = W/Qh** — the fraction of absorbed heat converted to work; the second law limits this below 1, so some heat must always be rejected. • 💡 Wrong-option analysis: A device that creates energy from nothing: this describes a perpetual motion machine of the first kind, which violates energy conservation; A device that keeps temperature always constant: an isothermal device does not necessarily do cyclical work; a heat engine operates between two temperatures; A device that converts work entirely into heat: this describes a dissipative device like a brake; a heat engine operates in the opposite direction.

Correct Answer: B. COPheat pump = COPrefrigerator + 1

• **COP_heat pump = COP_refrigerator + 1** = A heat pump delivers Qh = Qc + W to the hot space; dividing by W gives COP_hp = Qc/W + 1 = COP_ref + 1. • **Energy balance** — since Qh = Qc + W by energy conservation, the heat pump always delivers more energy to the hot space than the work consumed, making its COP always greater than 1. • 💡 Wrong-option analysis: COP_heat pump = 1/COP_refrigerator: the reciprocal relation has no thermodynamic basis and would give wrong values; COP_heat pump = COP_refrigerator: they would be equal only if W = 0, which would violate the second law; COP_heat pump = COP_refrigerator − 1: this would give a smaller value for the heat pump, contradicting the energy balance Qh = Qc + W.

Correct Answer: D. TV^(γ-1) = constant

• **TV^(γ−1) = constant** = Combining PV^γ = constant with the ideal gas law PV = nRT eliminates P to give TV^(γ−1) = constant for a reversible adiabatic process. • **Cooling on expansion** — TV^(γ−1) = constant shows that as volume increases, temperature decreases, explaining why an adiabatically expanding gas cools. • 💡 Wrong-option analysis: T^γV = constant: placing γ on T instead of (γ−1) on V is incorrect and contradicts the derivation from PV^γ = constant; TV = constant: this would require γ − 1 = 1, i.e., γ = 2, which is not physical for common gases; T/V = constant: this would imply temperature increases with volume in an adiabatic process, which contradicts the cooling effect.

Correct Answer: D. 5/3

• **5/3** = For a monoatomic ideal gas, Cv = 3R/2 and Cp = 5R/2, giving γ = Cp/Cv = (5R/2)/(3R/2) = 5/3 ≈ 1.67. • **Noble gases** — helium, neon, and argon are monoatomic and experimentally confirm γ ≈ 5/3; this value governs adiabatic processes in these gases. • 💡 Wrong-option analysis: 7/5: this is γ for a diatomic gas at ordinary temperatures with five active degrees of freedom; 1.4: this is the numerical equivalent of 7/5, also for diatomic gases; 1.2: this would correspond to a gas with ten degrees of freedom, not seen in common monoatomic or diatomic gases.

Correct Answer: B. 7/5

• **7/5** = A diatomic ideal gas at ordinary temperatures has five active degrees of freedom (three translational, two rotational); equipartition gives Cv = 5R/2 and Cp = 7R/2, so γ = 7/5 = 1.4. • **N₂ and O₂** — nitrogen and oxygen, the main components of air, are diatomic with γ ≈ 1.4, which is why this value is used in calculations involving sound speed and adiabatic processes in air. • 💡 Wrong-option analysis: 5/3: this is γ for monoatomic gases with three translational degrees of freedom; 1.0: γ = 1 would require Cp = Cv, meaning R = 0, which is impossible for a real gas; 2.0: this would require Cv = Cp/2, i.e., Cv = R, corresponding to only two degrees of freedom, not physical for a diatomic gas.

Correct Answer: C. 4.186 J

• **4.186 J** = One calorie is defined as the heat required to raise 1 gram of water by 1°C at 15°C; this equals approximately 4.186 joules. • **Joule's mechanical equivalent** — James Joule determined this conversion factor experimentally using a paddle-wheel apparatus, establishing the equivalence of heat and mechanical energy. • 💡 Wrong-option analysis: 0.4186 J: this is the reciprocal relationship — 1 joule equals approximately 0.2390 calories, not 0.4186; 1.000 J: equating one calorie to one joule would make calorie and joule identical, but they differ by a factor of about 4.186; 9.81 J: 9.81 is the acceleration due to gravity in m/s², which is unrelated to the calorie-joule conversion.