Which process results in zero change of entropy for an ideal gas?
Answer: B
For a reversible adiabatic process, dQ = 0, so dS = dQ/T = 0, hence ΔS = 0. This is also called isentropic process.
Q.142Medium
In a throttling process (Joule-Thomson expansion), which of the following remains constant?
Answer: D
Throttling is an irreversible, adiabatic process where enthalpy remains constant (H_initial = H_final). Temperature and pressure both change, and internal energy remains nearly constant only for ideal gases.
Q.143Hard
Two moles of an ideal diatomic gas undergo a process where temperature changes from 400 K to 800 K. If the process is such that PV^(1.4) = constant, then the work done by the gas is approximately (R = 8.314 J/mol·K):
Answer: C
For polytropic process: W = nR(T_i − T_f)/(γ − n) = 2 × 8.314 × (400 − 800)/(1.4 − 1.4). Since n = γ, this is adiabatic: W = nCvΔT = 2 × (25) × 8.314 × (400 − 800) = 5 × 8.314 × (−400) ≈ −16,628 J. But if heating occurs, work done is positive. Actual calculation needs clarification on process direction.
Q.144Hard
A system undergoes a process where both pressure and volume increase. Which thermodynamic quantity must increase?
Answer: A
If both P and V increase for an ideal gas (PV = nRT), then T must increase, hence internal energy U ∝ T increases. Work done by system depends on process path; entropy and heat depend on specific process.
Q.145Hard
Four moles of an ideal gas are heated from 250 K to 350 K at constant volume. Simultaneously, it is allowed to expand at constant temperature. Which process involves greater entropy change?
Answer: B
Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln(250350). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.
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Q.146Easy
A heat pump delivers 5000 J of heat to a house while consuming 1500 J of work. Its coefficient of performance is:
Answer: A
For heat pump: COP = Q_h/W = 15005000 = 3.33. This indicates the pump delivers 3.33 J of heat for every 1 J of work input.
Q.147Easy
A diatomic ideal gas expands isothermally from volume V₁ to 2V₁. If the initial pressure is P₀, what is the work done by the gas?
Answer: A
For isothermal process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = P₀V₁ ln(2)
Q.148Easy
An ideal gas undergoes adiabatic compression from state (P₁, V₁, T₁) to (P₂, V₂, T₂). Which relation is correct?
Answer: A
For adiabatic process: TV^(γ-1) = constant, therefore T₁V₁^(γ-1) = T₂V₂^(γ-1)
Q.149Easy
The internal energy of an ideal gas depends on:
Answer: C
Internal energy U of ideal gas is U = nCᵥT, which depends only on temperature, not on pressure or volume individually
Q.150Easy
A Carnot engine operates between 400 K and 300 K. Its maximum efficiency is:
Two kilograms of water at 100°C is converted to steam at 100°C at 1 atm pressure. The change in entropy is (Latent heat of vaporization = 2.26 × 10⁶ J/kg):
A real gas deviates from ideal behavior. Which condition favors ideal behavior?
Answer: B
At low pressure, molecules are far apart (negligible intermolecular forces), and at high temperature, kinetic energy dominates, making gases behave ideally
Q.156Medium
An open system differs from a closed system in that:
Answer: A
An open system allows both mass and energy exchange with surroundings (e.g., a boiling kettle), while a closed system allows only energy exchange
Q.157Hard
Three moles of ideal gas undergo polytropic process with n = 1.5. If temperature increases from 300 K to 450 K, the work done by gas is:
Answer: B
W = nR(T₂-T₁)/(1-n) = 3 × 8.314 × 150/(1-1.5) = 3,741/(-0.5) = -3,741 J (compression), |W| ≈ 3,372 J accounting for polytropic work formula
Q.158Hard
A heat engine operates between 600 K and 300 K reservoirs. It absorbs 5000 J from hot reservoir. For a Carnot engine operating between same temperatures, maximum work output would be:
