SSRAO

Explanation:
Intensive properties are independent of the mass or size of the system. Temperature, pressure, and density are intensive properties. Volume, mass, and internal energy are extensive properties because they depend on the amount of matter present.

Explanation:
The zeroth law states that if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law establishes the concept of temperature and forms the basis for thermometry (temperature measurement).

Explanation:
A closed system (also called a control mass) allows energy transfer (heat and work) across its boundary but does not allow mass transfer. An isolated system allows neither mass nor energy transfer. An open system allows both mass and energy transfer.

Explanation:
At constant pressure, when heat is added, the gas expands and does work (pdV work). This work must also be supplied in addition to raising the internal energy. Hence more heat is required at constant pressure than at constant volume for the same temperature rise. Mathematically, Cp - Cv = R (for ideal gas), so Cp > Cv always.

Explanation:
The triple point is the unique temperature and pressure at which all three phases of a pure substance (solid, liquid, and vapour) coexist in thermodynamic equilibrium. For water, the triple point is at 0.01°C and 611.73 Pa.

Explanation:
For a reversible isothermal process of an ideal gas, T = constant. The work done is W = ∫p dV = ∫(mRT/V) dV = mRT ln(V2/V1). Since pV = mRT, this can also be written as W = p1V1 ln(V2/V1) = mRT ln(p1/p2).

Explanation:
The second law of thermodynamics states that the entropy of an isolated system always increases for irreversible processes and remains constant for reversible processes. Entropy is a state function (not a path function) and is transferred by heat, not work. This is the principle of increase of entropy.

Explanation:
The Clausius inequality states that ∮(δQ/T) ≤ 0 for any thermodynamic cycle. The equality holds for reversible cycles (Carnot cycle), and the inequality (<0) holds for irreversible cycles. This is a mathematical statement of the second law of thermodynamics.

Explanation:
For a reversible adiabatic (isentropic) process of an ideal gas, pVγ = constant, where γ = Cp/Cv is the ratio of specific heats. This is derived from the first law with δQ = 0 and the ideal gas relations. For air, γ ≈ 1.4.

Explanation:
At the critical point, the properties of the liquid and vapour phases become identical. Above the critical temperature and pressure, the substance exists as a supercritical fluid and cannot be liquefied by pressure alone. For water, the critical point is at 374.14°C and 22.09 MPa.

Explanation:
The Joule-Thomson coefficient μJT = (∂T/∂p)H. For an ideal gas, the internal energy depends only on temperature, and enthalpy also depends only on temperature. Therefore, during throttling (isenthalpic process), the temperature of an ideal gas does not change, making μJT = 0.

Explanation:
By the first law of thermodynamics: ΔU = Q - W. Here Q = +50 kJ (heat added to system) and W = +30 kJ (work done by system). Therefore ΔU = 50 - 30 = 20 kJ. The internal energy increases by 20 kJ.

Explanation:
Heat (Q) and work (W) are not thermodynamic properties; they are energy interactions that occur at the boundary of a system during a process. They are path functions. Enthalpy (H), entropy (S), and internal energy (U) are thermodynamic properties (state functions) that depend only on the state of the system.

Explanation:
For an ideal gas, the internal energy U depends only on temperature (Joule's law). Since enthalpy H = U + pV = U + mRT (using ideal gas law pV = mRT), and both U and mRT depend only on temperature, enthalpy of an ideal gas is a function of temperature only.

Explanation:
The Carnot efficiency is the maximum possible efficiency for a heat engine operating between two thermal reservoirs. η_Carnot = 1 - TL/TH = (TH - TL)/TH, where TH and TL are the absolute temperatures (in Kelvin) of the hot source and cold sink respectively. This is derived from the second law of thermodynamics.

Explanation:
Specific volume (v) is defined as the volume per unit mass: v = V/m (m³/kg). It is the reciprocal of density (ρ = m/V = 1/v). Specific volume is an intensive property, while volume is an extensive property.

Explanation:
For a reversible process, the entropy change of the universe (system + surroundings) is zero: ΔS_universe = 0. For an irreversible process, ΔS_universe > 0. The entropy of the universe can never decrease (second law). A reversible process is an idealization where no entropy is generated.

Explanation:
In the two-phase region, pressure and temperature are not independent (they are related by the saturation curve). Therefore, the state is specified by pressure (or temperature) and one more independent property such as quality (dryness fraction x = mg/(mf + mg)), specific volume, specific enthalpy, or specific entropy.

Explanation:
Dryness fraction (quality) x = mg/(mf + mg), where mg is the mass of saturated vapour and mf is the mass of saturated liquid in the wet mixture. x = 0 for saturated liquid, x = 1 for saturated vapour, and 0 < x < 1 for wet steam. It is used to determine properties of wet steam by interpolation.

Explanation:
The first law of thermodynamics introduces the concept of internal energy (U). It states that energy is conserved: ΔU = Q - W. The zeroth law introduces temperature, the second law introduces entropy, and the third law states that entropy approaches zero as temperature approaches absolute zero.

Explanation:
In a throttling process (flow through a restriction like a valve or orifice), there is no heat transfer (Q = 0), no work done (W = 0), and negligible change in kinetic and potential energy. Applying the steady-flow energy equation: h1 = h2, so enthalpy remains constant. This is used in refrigeration expansion valves.

Explanation:
The Clausius statement: 'It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.' This means heat cannot flow spontaneously from cold to hot without external work input (as in a refrigerator).

Explanation:
For a diatomic ideal gas (N2, O2, air) at room temperature, the degrees of freedom f = 5 (3 translational + 2 rotational). Cv = (f/2)R = 2.5R, Cp = Cv + R = 3.5R, so γ = Cp/Cv = 3.5/2.5 = 1.4. For monatomic gases (He, Ar), γ = 1.67. For triatomic gases, γ ≈ 1.3.

Explanation:
Carnot efficiency η = 1 - TL/TH = 1 - 300/800 = 1 - 0.375 = 0.625 = 62.5%. Work output W = η × QH = 0.625 × 1000 = 625 kJ. Heat rejected QL = QH - W = 1000 - 625 = 375 kJ.

Explanation:
Joule's experiment demonstrated that the internal energy of an ideal gas depends only on temperature. This is because ideal gas molecules have no intermolecular forces, so potential energy is zero. Internal energy U = mCvT for an ideal gas. This is known as Joule's law.

Explanation:
For a polytropic process pVn = constant, the work done is W = ∫p dV = (p1V1 - p2V2)/(n-1) for n ≠ 1. This can also be written as W = mR(T1-T2)/(n-1). For n=1 (isothermal), W = p1V1 ln(V2/V1). For n=0 (isobaric), W = p(V2-V1). For n=γ (adiabatic), W = (p1V1-p2V2)/(γ-1).

Explanation:
For expansion between the same pressure limits, isothermal expansion produces maximum work because the temperature (and hence pressure) is maintained at the highest possible level throughout the process. The work done in isothermal expansion W = p1V1 ln(V2/V1) is greater than adiabatic expansion work for the same pressure ratio.

Explanation:
The Kelvin-Planck statement: 'It is impossible to construct a device that operates in a thermodynamic cycle and produces no effect other than the absorption of energy from a single thermal reservoir and the performance of an equivalent amount of work.' This means a perpetual motion machine of the second kind (PMM2) is impossible.

Explanation:
The Steady Flow Energy Equation (SFEE) per unit mass is: q - w = (h2 - h1) + (V2² - V1²)/2 + g(z2 - z1), where h is specific enthalpy, V is velocity, z is elevation, q is heat transfer per unit mass, and w is shaft work per unit mass. This applies to turbines, compressors, nozzles, and heat exchangers.

Explanation:
Exergy (availability) is a combined property of the system and its environment (dead state). It represents the maximum useful work extractable as the system reaches equilibrium with the environment at T0, p0. Exergy = (U - U0) + p0(V - V0) - T0(S - S0) + KE + PE. It depends on both the system state and the dead state.

Explanation:
An isobaric process is one in which pressure remains constant (p = constant). An isochoric process has constant volume (V = constant). An isothermal process has constant temperature (T = constant). An isentropic process has constant entropy (s = constant), which is a reversible adiabatic process.

Explanation:
In an isochoric (constant volume) process, dV = 0, so the boundary work W = ∫p dV = 0. All the heat added goes into increasing the internal energy: Q = ΔU = mCv(T2 - T1). No work is done by or on the system in a constant volume process.

Explanation:
An isentropic process is one in which entropy remains constant (ds = 0). It is a reversible adiabatic process. For an ideal gas, an isentropic process follows pVγ = constant. Isentropic processes are used as ideal models for turbines, compressors, and nozzles.

Explanation:
For an ideal gas, the Mayer's relation gives Cp - Cv = R, where R is the specific gas constant (J/kg·K). This is derived from the definitions of Cp and Cv and the ideal gas equation pV = mRT. For air, R = 287 J/kg·K, Cp = 1005 J/kg·K, Cv = 718 J/kg·K, and Cp - Cv = 287 J/kg·K.

Explanation:
Applying Gibbs phase rule: F = C - P + 2 = 1 - 2 + 2 = 1. This means in the two-phase region of a pure substance, only one independent variable (either temperature or pressure) is needed to specify the state. Temperature and pressure are not independent in the two-phase region (they are related by the saturation curve).

Explanation:
COP of refrigerator = Desired effect / Work input = QL / W = QL / (QH - QL), where QL is heat absorbed from the cold reservoir and QH is heat rejected to the hot reservoir. For a Carnot refrigerator, COP = TL / (TH - TL). COP can be greater than 1, unlike thermal efficiency.

Explanation:
The Helmholtz function A = U - TS gives dA = -SdT - pdV. From the exactness condition (∂²A/∂T∂V = ∂²A/∂V∂T), we get (∂S/∂V)T = (∂p/∂T)V. This Maxwell relation is useful for relating entropy changes to measurable p-V-T properties.

Explanation:
The inversion temperature is the temperature above which a gas heats up during throttling (Joule-Thomson effect) and below which it cools down. Above the critical temperature, a gas cannot be liquefied by pressure alone. However, the question specifically asks about the Joule-Thomson inversion temperature, which is the temperature at which μJT changes sign. For most gases, the inversion temperature is well above room temperature.

Explanation:
The boundary work done during a quasi-static process is W = ∫p dV, which is represented by the area under the process curve on a p-V diagram. The area under the T-s diagram represents heat transfer (Q = ∫T ds). The area enclosed by a cycle on a p-V diagram represents the net work output.

Explanation:
The specific enthalpy of wet steam is h = hf + x·hfg, where hf is the enthalpy of saturated liquid, hg is the enthalpy of saturated vapour, hfg = hg - hf is the latent heat of vaporisation, and x is the dryness fraction (0 ≤ x ≤ 1). Similarly, specific volume v = vf + x·vfg and specific entropy s = sf + x·sfg.

Explanation:
The Kelvin (thermodynamic) temperature scale is defined by assigning the value 273.16 K to the triple point of water. This is the internationally agreed reference point. The ice point (0°C) corresponds to 273.15 K, and the steam point (100°C) corresponds to 373.15 K. The absolute zero is 0 K = -273.15°C.

Explanation:
In a polytropic process pVn = constant: n=0 gives isobaric (constant pressure), n=1 gives isothermal (constant temperature), n=γ gives isentropic/adiabatic (reversible adiabatic), and n=∞ gives isochoric (constant volume). For air, γ = 1.4, so a reversible adiabatic process has n = 1.4.

Explanation:
The Helmholtz function (Helmholtz free energy) is defined as A = U - TS, where U is internal energy, T is temperature, and S is entropy. It represents the maximum work obtainable from a closed system at constant temperature and volume. The Gibbs function G = H - TS = U + pV - TS represents maximum work at constant T and p.

Explanation:
In a free expansion (expansion into a vacuum), the external pressure is zero (pext = 0). Therefore, the work done W = ∫pext dV = 0. For an ideal gas, since Q = 0 (adiabatic) and W = 0, the internal energy change ΔU = 0, which means temperature remains constant. Free expansion is an irreversible process.

Explanation:
The Gibbs function (Gibbs free energy) is G = H - TS = U + pV - TS. Note that options B and C are equivalent since H = U + pV. The Gibbs function is useful for phase equilibrium and chemical reactions at constant temperature and pressure. At equilibrium, G is minimised. dG = -SdT + Vdp.

Explanation:
An open system (control volume) allows both mass and energy to cross its boundary. A steam turbine has steam flowing in and out (mass transfer) and produces work (energy transfer). A gas in a fixed piston-cylinder and a bomb calorimeter are closed systems. A thermos flask approximates an isolated system.

Explanation:
For a steady-flow device, the shaft work is given by W = -∫V dp (flow work or technical work). This is different from boundary work W = ∫p dV for closed systems. For an ideal gas in an isentropic process, the compressor work = γ/(γ-1) × mRT1[(p2/p1)^((γ-1)/γ) - 1]. The negative sign indicates work input for compressors.

Explanation:
The Carnot cycle consists of four reversible processes: (1) Isothermal heat addition at TH, (2) Isentropic (reversible adiabatic) expansion, (3) Isothermal heat rejection at TL, and (4) Isentropic compression. Since isentropic = reversible adiabatic, the cycle has two isothermal and two isentropic processes.

Explanation:
Exergy (availability) is the maximum useful work that can be obtained from a system as it interacts with the environment and reaches the dead state (equilibrium with environment at T0, p0). It measures the quality or work potential of energy. Entropy measures disorder, enthalpy is useful for flow processes, and internal energy is total microscopic energy.

Explanation:
The Clapeyron equation is (dp/dT)sat = hfg / (T × vfg), where hfg is the latent heat of vaporisation, T is the saturation temperature, and vfg = vg - vf is the change in specific volume. This equation relates the slope of the saturation curve to measurable thermodynamic properties and is used to calculate latent heat from p-V-T data.

Explanation:
The second law of thermodynamics determines: (1) the direction of spontaneous processes (entropy increases), (2) the maximum possible efficiency of heat engines (Carnot efficiency), (3) the quality of energy, and (4) the degree of irreversibility. The first law determines energy quantities (heat, work, internal energy changes).

Explanation:
Temperature is a measure of the average translational kinetic energy of gas molecules: (3/2)kT = (1/2)m<v²>, where k is Boltzmann's constant and <v²> is the mean square velocity. This is the kinetic theory interpretation of temperature. Higher temperature means faster molecular motion on average.

Explanation:
The Joule-Thomson effect (cooling during throttling for real gases below the inversion temperature) is used in the liquefaction of gases (Linde-Hampson process). When a gas at high pressure is throttled through a valve, it cools if the temperature is below the inversion temperature. This cooling effect is used to liquefy air, nitrogen, oxygen, and other gases.

Explanation:
The third law of thermodynamics (Nernst's theorem) states that the entropy of a perfect crystalline substance approaches zero as the absolute temperature approaches zero (0 K). This provides an absolute reference point for entropy calculations. It also implies that absolute zero temperature (0 K) cannot be reached in a finite number of steps.

Explanation:
For an ideal gas in an isothermal process, T = constant, so ΔU = mCv ΔT = 0 and ΔH = mCp ΔT = 0. From the first law: Q = ΔU + W = 0 + W = W. Therefore, all the heat added is converted to work (or all work done on the gas is rejected as heat). Q = W = mRT ln(V2/V1) = mRT ln(p1/p2).

Explanation:
The dead state is the condition at which the system is in complete thermodynamic equilibrium with the environment (at T0, p0). At the dead state, the system has zero exergy — it cannot do any useful work. All exergy calculations are made relative to the dead state. Typically T0 = 25°C = 298 K and p0 = 101.325 kPa (standard atmospheric conditions).

Explanation:
For a monatomic ideal gas (He, Ar, Ne), the degrees of freedom f = 3 (only translational). Cv = (3/2)R, Cp = (5/2)R, so γ = Cp/Cv = 5/3 ≈ 1.67. For diatomic gases (N2, O2, air), f = 5, γ = 7/5 = 1.4. For triatomic gases (CO2, H2O), f = 6, γ = 8/6 ≈ 1.33.

Explanation:
In a throttling calorimeter, wet steam is throttled to a lower pressure where it becomes superheated. The dryness fraction is found using enthalpy equality: h1 = h2. h1 = hf1 + x·hfg1 (wet steam before throttling at p1). h2 = hg2 + Cp(T2 - Tsat2) (superheated steam after throttling at p2, T2). Measuring p2 and T2 after throttling gives h2, from which x is calculated.

Explanation:
For a reversible adiabatic process, Q = 0. From the first law: W = Q - ΔU = 0 - mCv(T2 - T1) = mCv(T1 - T2). This is positive (work output) when T1 > T2 (expansion). Alternatively, W = (p1V1 - p2V2)/(γ-1) = mR(T1-T2)/(γ-1). Since Cv = R/(γ-1), both expressions are equivalent.

Explanation:
The Rankine scale is the absolute temperature scale corresponding to the Fahrenheit scale: R = F + 459.67, where R is temperature in Rankine and F is temperature in Fahrenheit. The ice point is 491.67 R (32°F + 459.67) and the steam point is 671.67 R (212°F + 459.67). The Kelvin scale corresponds to Celsius: K = °C + 273.15.