Three identical conducting rods are arranged in series between two heat reservoirs at 100°C and 0°C. At steady state, what is the temperature at the junction between the second and third rod?
Answer: A
In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = 3100 = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C
Q.2Hard
A gas undergoes a cyclic process ABCA where AB is isothermal, BC is adiabatic, and CA is isochoric. If work is done on the gas in the cycle, what can be concluded?
Answer: A
If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out
Q.3Hard
Two bodies at temperatures T₁ = 400 K and T₂ = 300 K are brought into thermal contact. If entropy change of universe is 0.575 J/K and heat capacity of both bodies is 1000 J/K, what is the final equilibrium temperature? (Assume no heat loss to surroundings)
Answer: B
Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K
Q.4Hard
If a reversible process occurs in an isolated system, the entropy of the system:
Answer: C
For a reversible process in an isolated system, ΔS_total = ΔS_sys + ΔS_surr = 0 (no heat exchange with surroundings). Therefore entropy remains constant at its initial value.
Q.5Hard
An ideal gas undergoes a process where PV^n = constant. If n = 1, this process is _____ and if n = γ, this process is _____
Answer: A
When n = 1: PV = constant (isothermal). When n = γ = Cp/Cv: PV^γ = constant (adiabatic process).
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Q.6Hard
Consider a system where entropy decreases by 50 J/K. Which statement must be true?
Answer: B
By second law, ΔS_total ≥ 0. If ΔS_sys = -50 J/K, then ΔS_surr ≥ +50 J/K to maintain ΔS_total ≥ 0.
Q.7Hard
A polytropic process with polytropic index n = 1.3 is performed on 1 mole of air (diatomic). The work done when volume changes from 1 m³ to 0.5 m³ at initial pressure 100 kPa is:
Answer: A
For polytropic process: W = [P₁V₁ - P₂V₂]/(n-1). Using PVⁿ = const: P₂ = 100 × (01.5)^1.3 ≈ 245.7 kPa. W = [100×1 - 245.7×0.5]/0.3 ≈ 81.2 kJ
Q.8Hard
For a van der Waals gas with equation (P + a/V²m)(Vm - b) = RT, at the critical point, which relationship is valid?
Answer: A
At the critical point, both first and second derivatives of pressure with respect to volume (at constant T) are zero: (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. This defines the critical point.
Q.9Hard
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.10Hard
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.11Hard
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.
Q.12Hard
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.13Hard
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:
For a van der Waals gas, the critical point is characterized by:
Answer: A
At critical point, both first and second derivatives of pressure with respect to volume are zero, marking the boundary of liquid-gas phase transition
Q.15Hard
A reversible process has entropy change ΔS_sys = -100 J/K. The entropy change of universe is:
Answer: C
For reversible process: ΔS_universe = ΔS_sys + ΔS_surr = 0. Since ΔS_sys = -100, ΔS_surr = +100, making total change zero
Q.16Hard
In a free expansion of ideal gas into vacuum, the entropy change of system is:
Answer: B
Free expansion is irreversible with ΔU = 0 and W = 0, so Q = 0. Volume increases, so S = nR ln(V_f/V_i) > 0
Q.17Hard
A monatomic ideal gas undergoes a cyclic process ABCA where: A→B is isothermal expansion, B→C is isochoric process, C→A is adiabatic compression. If at point A, P = 1 atm, V = 1 L, and T = 300 K, and the volume doubles from A to B, find the heat absorbed during the isothermal process.
Answer: C
For isothermal process of ideal gas: Q = nRT ln(V_f/V_i) = W. n = PV/RT = (101325 × 0.001)/(8.314 × 300) ≈ 0.0405 mol. Q = nRT ln(2) = 0.0405 × 8.314 × 300 × ln(2) ≈ 600 ln(2) J
Q.18Hard
For an ideal gas undergoing a polytropic process (PV^n = constant), the heat capacity is C = C_v + R/(1-n). For which value of n does the polytropic process become adiabatic?
Answer: B
For adiabatic process, Q = 0, so C = 0. This occurs when 1-n approaches infinity, which happens when n = γ. At n = γ, PV^γ = constant (adiabatic relation).
Q.19Hard
A gas sample at 300 K has an entropy of 200 J/K. When heated at constant pressure to 600 K, its entropy becomes:
Which thermodynamic process results in maximum work extraction from an ideal gas expanding from the same initial to final states?
Answer: B
For expansion between the same P-V states, isothermal process produces maximum work because W = nRT ln(V_f/V_i) is maximum when temperature is highest throughout the process.
