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.107Easy
What is the Clausius statement of the second law of thermodynamics?
Answer: A
Clausius statement: Heat cannot spontaneously transfer from a colder body to a hotter body without external work being done on the system
Q.108Hard
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.109Medium
For a van der Waals gas, which statement is correct?
Answer: A
Van der Waals equation (P + a/V²)(V - b) = RT accounts for molecular volume (b term) and intermolecular attractive forces (a term)
Q.110Medium
In an expansion process, a gas does 500 J of work and absorbs 300 J of heat. What is the change in internal energy?
Answer: A
Using first law: ΔU = Q - W = 300 - 500 = -200 J (internal energy decreases)
Q.111Easy
For a diatomic ideal gas at room temperature, what is the ratio γ = Cₚ/Cᵥ?
Answer: A
For diatomic gas: Cᵥ = (25)R and Cₚ = (27)R. γ = Cₚ/Cᵥ = 57 = 1.40
Q.112Hard
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.113Easy
A thermodynamic system undergoes a process where internal energy increases by 150 J while the system does 100 J of work on surroundings. What is the heat absorbed by the system?
Answer: A
By first law: ΔU = Q - W. Here ΔU = 150 J, W = 100 J (work done by system). So Q = ΔU + W = 150 + 100 = 250 J
Q.114Easy
For one mole of an ideal monatomic gas, the ratio Cp/Cv is:
Answer: B
For monatomic gas: Cv = (23)R and Cp = (25)R. Therefore Cp/Cv = (25)/(23) = 35 ≈ 1.67
Q.115Easy
In an adiabatic process, if a gas is compressed, which statement is correct?
Answer: B
In adiabatic compression, no heat exchange occurs (Q=0). Work is done on the gas, so ΔU = W (positive). Since ΔU increases, temperature must increase.
Q.116Easy
A carnot engine operates between temperatures 500 K and 300 K. What is its maximum efficiency?
During an isobaric expansion of an ideal gas, the work done by the gas is 400 J. If pressure is constant at 2 atm, what is the change in volume? (1 atm = 101325 Pa)
Answer: B
W = PΔV. Here W = 400 J, P = 2 × 101325 = 202650 Pa. So ΔV = W/P = 202650400 ≈ 0.00197 m³
Q.118Medium
For a diatomic ideal gas undergoing an isothermal process, which quantity remains constant?
Answer: C
In an isothermal process, temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. Pressure and volume change according to PV = constant.
Q.119Medium
Two samples of the same ideal gas at the same temperature have volumes V and 2V respectively. The ratio of their internal energies is:
Answer: D
Internal energy U = nCvT. Without knowing the number of moles in each sample, the ratio cannot be determined. Same temperature doesn't mean same internal energy.
Q.120Medium
A heat engine absorbs 1000 J of heat and rejects 600 J to the cold reservoir in one cycle. What is its efficiency?
