Thermodynamics — Practice Quiz

Temperature increases, internal energy increases.

Step 1: In an adiabatic process, ΔQ = 0 (no heat exchange). Step 2: Applying the first law: ΔU = ΔQ − ΔW = 0 − ΔW. Step 3: During compression, work is done ON the gas, so ΔW is negative (work done BY the gas is negative). Therefore ΔU = −(negative value) = positive. Step 4: Since internal energy increases, the temperature of the gas also increases. The compression squeezes molecules together, increasing their kinetic energy. Why other options are wrong: Option A describes adiabatic expansion; Option C is for isothermal process; Option D contradicts the relationship between internal energy and

1600 K

Step 1: Efficiency between 800 K and 400 K: η₁ = 1 − (400/800) = 1 − 0.5 = 0.5. Step 2: For the second case (source T, sink 800 K): η₂ = 1 − (800/T). Step 3: Set η₁ = η₂: 0.5 = 1 − (800/T). Step 4: 800/T = 0.5, so T = 800/0.5 = 1600 K. Why other options are wrong: 400 K is the sink temperature in the first case; 1200 K would give η = 1 − 800/1200 = 0.33 ≠ 0.5; 200 K would be lower than the sink, which is physically meaningless for a source.

0 J

Step 1: In an isochoric process, volume remains constant (ΔV = 0). Step 2: Work done by a gas is given by ΔW = P·ΔV. Step 3: Since ΔV = 0, ΔW = P × 0 = 0 J. Step 4: All the heat supplied goes entirely into increasing the internal energy: ΔU = ΔQ = 800 J. Why other options are wrong: 800 J would be the work done if all heat were converted to work (impossible here since ΔV = 0); 400 J and −800 J have no physical basis for this constant-volume scenario.

The net work done by the system in one complete cycle

Step 1: In a P-V diagram, the area of a small strip = P·ΔV = work done for a small volume change. Step 2: Integrating over the entire path, the area under a curve from V₁ to V₂ gives total work done. Step 3: For a closed loop (cyclic process), the area enclosed = net work done in the complete cycle = W_expansion − W_compression. Step 4: Also, for a cyclic process, ΔU = 0, so net heat = net work (ΔQ = ΔW). Why other options are wrong: Total heat absorbed is not directly given by the loop area; ΔU = 0 for a cyclic process so option C is zero; heat rejected is only part of the energy exchange, no