Thermodynamic State Variable ((P, V, T, U)) and Thermodynamic Processes – isothermal and adiabatic processes: NEET MCQs with Answers

Explanation: The isothermal expansion from A is a falling hyperbola from V to 3V, so its arrow must point down and to the right. The isobaric compression is a horizontal line from 3V back to V, so its arrow must point left, ending at B. The correct diagram has both of these arrow directions; the others have an arrow reversed or draw the second step as a sloping line.

Explanation: Same endpoints ⇒ equal ΔU for all three paths. The uppermost curve does the most work, so it absorbs the most heat: Q₁ > Q₂ > Q₃ with ΔU₁ = ΔU₂ = ΔU₃.

Explanation: Enclosed area = ΔP×ΔV = (2P−P)(3V−V) = 2PV. The loop is anticlockwise so W = −2PV, and for a cycle Q = W, meaning 2PV of heat is rejected.

Explanation: AB is a vertical line on the p–V diagram, so it is isochoric: dV = 0 and hence dW = 0, which also means V_B = V_A = 2 × 10⁻³ m³. Applying the first law to AB: 600 = (U_B − U_A) + 0, so U_B − U_A = 600 J. BC is a horizontal line, so it is isobaric at pressure p_B: 200 = (U_C − U_B) + p_B(V_C − V_B). Here p_B(V_C − V_B) = 8 × 10⁴ × (5 − 2) × 10⁻³ = 240 J, so U_C − U_B = 200 − 240 = −40 J. Adding the two steps: ΔU_AC = (U_B − U_A) + (U_C − U_B) = 600 − 40 = 560 J.

Explanation: The work done along the outward path ACB is the area beneath it, namely ACBB′A′A, and it is positive because the volume increases. The work done along the return path BDA is the area beneath that path, BDAA′B′B, and it is negative because the volume decreases. The net work for the cycle is the difference of these two areas, which is exactly the region enclosed by the closed curve — the area ACBDA. In general, the net work done in any cyclic process equals the area enclosed by the cycle on the p–V diagram.

Explanation: In the isothermal expansion pV = nRT = constant, so the p–V curve is a rectangular hyperbola: as the volume grows from V to 3V, the pressure falls along a smooth downward-sloping arc starting at A. In the isobaric compression that follows, the pressure remains fixed at this new lower value while the volume returns from 3V to V. On the p–V diagram this is a horizontal straight line drawn leftwards, ending at B. So the correct diagram shows a curved hyperbolic arc falling to the right from A, followed by a horizontal line running back to the left at the lower pressure, with B directly below A.

Explanation: Over one complete cycle the system returns to its initial state, so ΔU = 0 and the first law gives Q = W. The closed path is a rectangle of width (3V − V) = 2V and height (2p − p) = p, so the area enclosed is 2pV. The cycle is described in the anticlockwise sense, which makes the net work done by the gas negative: W = −2pV. Hence Q = −2pV; the negative sign means that 2pV of heat is rejected by the gas during the cycle.

Explanation: All three paths start at the same state A and finish at the same state B. Internal energy is a state function, so ΔU₁ = ΔU₂ = ΔU₃. Work done equals the area under each curve on the p–V diagram. From the figure, the area under curve 1 is the greatest and that under curve 3 the least, so W₁ > W₂ > W₃. By the first law Q = ΔU + W. Since ΔU is the same for all three, the ordering of Q follows that of W: Q₁ > Q₂ > Q₃.

Explanation: For the path adc, the first law gives ΔU_adc = δQ₁ − δW₁ = 50 − 20 = 30 J. Internal energy is a state function, so the change between the same two end states a and c is the same along every path: ΔU_abc = 30 J. Applying the first law to the path abc: W_abc = δQ₂ − ΔU_abc = 36 − 30 = 6 J. (The value U_a = 10 J is not needed — only the change in internal energy matters.)

Explanation: Process I is drawn parallel to the pressure axis, so the volume stays constant: it is isochoric. Process IV is drawn parallel to the volume axis, so the pressure stays constant: it is isobaric. Process III runs along one of the given isotherms, so the temperature stays constant: it is isothermal. Process II falls away more steeply than the isotherm through the same point. Since the adiabat is steeper than the isotherm by the factor γ, process II is adiabatic.

Explanation: For an adiabatic process PV^γ = constant. Substituting V = nRT/P gives P^(1–γ)T^γ = constant, while the volume form is TV^(γ–1) = constant.

Explanation: PV² = C with PV = nRT gives TV = constant, so T ∝ 1/V: an increase in volume lowers the temperature (V₂ > V₁ ⇒ T₂ < T₁).

Explanation: For an adiabatic process the work done by the gas is nR(T₁ – T₂)/(γ – 1), not nR(T₂ – T₁)/(γ – 1); the sign is reversed. The other three pairs are correct.

Explanation: Pressure, volume and temperature are state variables that specify an equilibrium state. Heat and work are path-dependent quantities (not properties of a state).

Explanation: AB is the isobaric step at P = 3 × 10² N/m² with ΔV = 3 – 1 = 2 m³. W = PΔV = 600 J.