In an adiabatic compression process for an ideal gas, which statement is correct?
A Temperature remains constant B Work done equals change in internal energy C Entropy increases D Heat capacity is zero
For adiabatic process, Q = 0, so ΔU = W (first law). Temperature increases during compression.
The Joule-Thomson coefficient μ_JT is negative for most gases at room temperature. This means:
A Temperature increases during expansion B Temperature decreases during expansion C No temperature change occurs D Entropy decreases during expansion
μ_JT = (∂T/∂P)_H. Negative coefficient means T increases with pressure drop (cooling requires very low T or high P).
For a binary ideal solution, Raoult's law states that:
A Partial pressure of each component = mole fraction × vapor pressure of pure component B Total pressure = sum of mole fractions × atmospheric pressure C Activity coefficient of each component = 1 D Vapor pressure is independent of composition
Raoult's law: P_i = x_i × P_i° for ideal solutions. Both statements A and C are equivalent for ideal solutions.
A system absorbs 500 J of heat and does 300 J of work. The change in internal energy is:
A 200 J B 800 J C -200 J D 500 J
First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).
The compressibility factor Z for a real gas at high pressures is typically:
A Greater than 1 (repulsive forces dominate) B Less than 1 (attractive forces dominate) C Equal to 1 D Always greater than 2
At high pressures, molecular volume effect (b term) dominates, making Z > 1. At moderate pressures, Z < 1 due to intermolecular attractions.
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For a constant pressure process, the heat absorbed equals:
A Change in internal energy B Change in enthalpy C Change in entropy D Work done by system
At constant pressure: ΔH = Q_p (definition of enthalpy). From ΔU = Q - W and W = PΔV, we get Q = ΔH.
A reversible process between two states A and B will have entropy change:
A Greater than that of an irreversible process between same states B Less than that of an irreversible process between same states C Equal to that of any path between same states D Dependent on the type of process
Entropy is a state function. ΔS is path-independent and same for all processes (reversible or irreversible) between fixed states.
For a polytropic process PV^n = constant, if n = γ (heat capacity ratio), the process is:
A Isothermal B Adiabatic C Isobaric D Isochoric
For adiabatic process of ideal gas, PV^γ = constant where γ = Cp/Cv. This is the defining equation for adiabatic polytropic process.
An engineer needs to liquefy natural gas (primarily methane). The gas must be cooled below the inversion temperature because:
A Below inversion temperature, Joule-Thomson coefficient is positive B Below inversion temperature, entropy increases C Gibbs energy becomes negative only below inversion temperature D Molecular interactions become repulsive below inversion temperature
Above inversion temperature (for methane ≈ 625 K), μ_JT < 0 (heating on expansion). Below it, μ_JT > 0 (cooling on expansion), enabling liquefaction.
For a binary ideal solution at constant T and P, if we mix 1 mole of component A and 1 mole of component B, the entropy of mixing is:
A ΔS_mix = -R(ln 0.5 + ln 0.5) = R ln 4 > 0 B ΔS_mix = 0 (ideal solution) C ΔS_mix = -R(x_A ln x_A + x_B ln x_B) < 0 D ΔS_mix depends on temperature
ΔS_mix = -nR Σx_i ln x_i. For equal moles: ΔS_mix = -2R(0.5 ln 0.5 + 0.5 ln 0.5) = R ln 4 > 0, always positive.
In a desalination plant using reverse osmosis, work must be applied because:
A ΔG > 0 for the desalination process B The process is adiabatic C Heat must be removed from the system D Entropy of the system decreases
Desalination (salt separation) is non-spontaneous: ΔG > 0. External work must be supplied to drive the process. This applies to RO and most separation processes.
