RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Thermodynamics is the bookkeeping of energy for systems made of enormous numbers of particles, and it rests on a small number of laws that have never been violated. You will learn to track heat, work and internal energy through the first law, and then meet the second law, which explains the one thing the first law cannot: why heat refuses to flow uphill on its own. Along the way you will calculate the work done in different processes and find the ceiling that nature places on the efficiency of every engine ever built.
In an adiabatic expansion of an ideal gas, which of the following is true?
Answer
The temperature falls and the internal energy decreases — in an adiabatic process ΔQ = 0, so the first law gives ΔU = −ΔW. During expansion the gas does positive work on its surroundings, and since no heat enters to pay for it, that energy must come out of the gas's own internal energy store. The internal energy therefore drops, and because U depends only on temperature for an ideal gas, the temperature falls too. This is why compressed air escaping from a cylinder feels cold.
A gas is taken around a closed cycle on a P-V diagram. Over one complete cycle:
Answer
ΔU = 0 and ΔQ = ΔW — internal energy is a state function, so returning the gas to its starting state must return U to its starting value, making ΔU zero over any complete cycle. The first law then reduces to ΔQ = ΔW. Both the net heat absorbed and the net work done are generally non-zero and equal to the area enclosed by the loop.
A Carnot engine operates between a source at 600 K and a sink at 300 K. Its efficiency is:
Answer
50% — for a Carnot engine η = 1 − T₂/T₁ = 1 − 300/600 = 1 − 0.5 = 0.5, that is 50%. Note that the temperatures must be in kelvin; using Celsius here would give a completely wrong answer. No engine working between these two reservoirs, however cleverly designed, can beat this figure.
For one mole of an ideal gas, the relation between the molar specific heats is:
Answer
C_p − C_v = R — this is Mayer's relation. Heating a gas at constant volume raises only its internal energy, but heating it at constant pressure must also supply the work PΔV = RΔT done as the gas pushes back the surroundings. The extra heat needed per kelvin per mole is exactly R, so C_p always exceeds C_v.
Assertion (A): It is impossible to build a heat engine, however well made, that converts all of the heat it absorbs into work in a cyclic process. Reason (R): Some heat must always be rejected to a sink at a lower temperature for the working substance to return to its initial state.
Answer
Both A and R are true and R is the correct explanation of A — this pair is the Kelvin-Planck statement of the second law together with its physical reason. The working substance must be brought back to its starting state at the end of each cycle, and the only way to restore its low-entropy condition is to dump some heat Q₂ into a colder reservoir. Since η = 1 − Q₂/Q₁ and Q₂ can never be zero, the efficiency is always less than one. Note that this is not a matter of friction or poor engineering, which the first law would already penalise; it is a fundamental limit.
A gas absorbs 850 J of heat and expands, doing 340 J of work on its surroundings. Find the change in its internal energy and state whether the gas has warmed or cooled.
Answer
Apply the first law: ΔQ = ΔU + ΔW. Here ΔQ = +850 J (heat absorbed by the gas) and ΔW = +340 J (work done by the gas during expansion). ΔU = ΔQ − ΔW = 850 − 340 = +510 J. The internal energy has increased by 510 J. For an ideal gas the internal energy depends only on temperature, so the gas has warmed up: it kept 510 J of the incoming heat and spent the other 340 J pushing back its surroundings.
Two moles of an ideal gas expand isothermally and reversibly at 350 K from a volume of 6 litres to 18 litres. Calculate the work done by the gas and the heat absorbed. Take R = 8.3 J/mol·K and ln 3 ≈ 1.10.
Answer
For a reversible isothermal expansion: ΔW = nRT ln(V₂/V₁). n = 2 mol, T = 350 K, V₂/V₁ = 18/6 = 3. ΔW = 2 × 8.3 × 350 × ln 3 = 5810 × 1.10 ≈ 6391 J ≈ 6.4×10³ J. Since the process is isothermal and the gas is ideal, the temperature is unchanged, so ΔU = 0. By the first law ΔQ = ΔU + ΔW = 0 + 6391 ≈ 6.4×10³ J. So the gas absorbs about 6.4 kJ of heat and converts every joule of it into work — which it can do only because it ends up in a different state, with a larger volume, and not in a cycle.
(a) State the first law of thermodynamics and explain the sign convention for each term. (b) Derive an expression for the work done by one mole of an ideal gas during a reversible isothermal expansion from volume V₁ to V₂. (c) Derive Mayer's relation C_p − C_v = R for one mole of an ideal gas.
Answer
(a) Statement: when heat ΔQ is supplied to a system, part of it increases the internal energy of the system by ΔU and the rest is used by the system to do external work ΔW: ΔQ = ΔU + ΔW. Sign convention: ΔQ is positive when heat is absorbed by the system and negative when heat is given out. ΔW is positive when work is done by the system, that is during expansion, and negative when work is done on the system. ΔU is positive when the internal energy rises. The law is nothing more than the conservation of energy applied to a thermodynamic system, extended to include heat as a form of energy transfer. Note that ΔU is a state function, depending only on the initial and final states, whereas ΔQ and ΔW both depend on the path taken. (b) Work done in a reversible isothermal expansion. The small work done when the gas expands by dV against pressure P is dW = P dV. For one mole of an ideal gas, PV = RT, so P = RT/V. Total work from V₁ to V₂: W = ∫ from V₁ to V₂ of (RT/V) dV. Since the process is isothermal, T is a constant and comes outside the integral: W = RT ∫ from V₁ to V₂ of dV/V W = RT [ln V] from V₁ to V₂ W = RT ln(V₂/V₁) = 2.303 RT log₁₀(V₂/V₁). Because PV = constant along the isotherm, this can equally be written W = RT ln(P₁/P₂). For an expansion V₂ > V₁, so W is positive, and since ΔU = 0 for an isothermal change in an ideal gas, all of this work is supplied by heat drawn in from the reservoir. (c) Derivation of Mayer's relation. Consider one mole of an ideal gas heated through a small temperature rise dT. At constant volume, no work is done because dV = 0, so the first law gives dQ_v = dU = C_v dT. ...(i) At constant pressure, the gas expands by dV and does work P dV, so dQ_p = dU + P dV. The internal energy of an ideal gas depends only on temperature, so for the same rise dT the change dU is the same as in (i), namely C_v dT. Also dQ_p = C_p dT by definition. Hence C_p dT = C_v dT + P dV. ...(ii) Now differentiate the ideal gas equation PV = RT at constant pressure: P dV = R dT. ...(iii) Substituting (iii) into (ii): C_p dT = C_v dT + R dT. Dividing throughout by dT: C_p − C_v = R. The physical meaning is direct: the extra heat needed at constant pressure, over and above that needed at constant volume, is exactly the work the gas performs in pushing back its surroundings, and for one mole that work is R joules per kelvin.
A heat engine takes in 2400 J of heat per cycle from a source at 500 K and rejects 1560 J to a sink at 300 K. (a) Find the work done per cycle and the actual efficiency. (b) Find the maximum possible efficiency for these reservoirs and comment on the engine's performance. (c) If the engine completes 25 cycles per second, find its power output. (d) The same machine is now run backwards as a refrigerator between the same reservoirs, operating ideally. Find its coefficient of performance and the work needed per cycle to remove 1560 J from the cold space.
Answer
(a) Work done per cycle. Over a complete cycle ΔU = 0, so by the first law the net work equals the net heat: W = Q₁ − Q₂ = 2400 − 1560 = 840 J. Actual efficiency: η = W/Q₁ = 840/2400 = 0.35, that is 35%. (b) Maximum possible efficiency. The best any engine can do between two reservoirs is the Carnot value: η_max = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.40, that is 40%. Comment: the engine achieves 35% against a ceiling of 40%, so it reaches 35/40 = 87.5% of the ideal. This is a good but entirely possible result, since 35% < 40%. Had the stated figures given an efficiency above 40%, the engine would have violated the second law and the data would have to be rejected as impossible. The shortfall of 5 percentage points represents the irreversibilities of the real machine: friction, turbulence in the working fluid, and heat conducted across finite temperature differences. (c) Power output. Each cycle delivers 840 J and there are 25 cycles per second, so P = W × f = 840 × 25 = 21000 W = 21 kW. (d) Running as an ideal refrigerator between the same reservoirs, the coefficient of performance is β = T₂/(T₁ − T₂) = 300/(500 − 300) = 300/200 = 1.5. Since β = Q₂/W, the work required to extract Q₂ = 1560 J from the cold space is W = Q₂/β = 1560/1.5 = 1040 J. The machine would then dump Q₁ = Q₂ + W = 1560 + 1040 = 2600 J into the hot reservoir. Note that β can exceed 1 without breaking any law, because it is a ratio of heat moved to work supplied, not a fraction of energy converted; a refrigerator pumps heat rather than creating it. Note also that β falls as the two temperatures are pulled further apart, which is why deep-freezing on a hot day is so expensive to run.
Read the following and answer the questions that follow: One mole of an ideal monatomic gas (γ = 5/3, C_v = 12.5 J/mol·K) is taken around the cycle A → B → C → A. In step A → B the gas is heated at a constant volume of 8×10⁻³ m³ from a pressure of 1.0×10⁵ Pa to 3.0×10⁵ Pa. In step B → C it expands at constant pressure to a volume of 24×10⁻³ m³. In step C → A it is compressed back to state A along a straight line on the P-V diagram. Take R = 8.3 J/mol·K. (a) Find the temperature at each of the three states. (b) Find the work done in each of the three steps. (c) Find the heat absorbed in step A → B. (d) Find the net work done per cycle and hence the heat absorbed per cycle, and state what the sense of the loop tells you about the device.
Answer
(a) Temperatures from PV = nRT with n = 1 mol. State A: T_A = PV/R = (1.0×10⁵ × 8×10⁻³)/8.3 = 800/8.3 ≈ 96.4 K. State B: T_B = (3.0×10⁵ × 8×10⁻³)/8.3 = 2400/8.3 ≈ 289.2 K. State C: T_C = (3.0×10⁵ × 24×10⁻³)/8.3 = 7200/8.3 ≈ 867.5 K. (b) Work done in each step. A → B is isochoric, so ΔV = 0 and W_AB = 0. B → C is isobaric, so W_BC = PΔV = 3.0×10⁵ × (24×10⁻³ − 8×10⁻³) = 3.0×10⁵ × 16×10⁻³ = 4800 J, done by the gas. C → A is a straight line on the P-V diagram from (24×10⁻³ m³, 3.0×10⁵ Pa) to (8×10⁻³ m³, 1.0×10⁵ Pa). The work is the area of the trapezium beneath it, taken negative because the gas is compressed: W_CA = −½(P_C + P_A)(V_C − V_A) = −½(3.0×10⁵ + 1.0×10⁵)(16×10⁻³) = −½ × 4.0×10⁵ × 16×10⁻³ = −3200 J. (c) Heat in step A → B. No work is done, so all the heat goes into internal energy: Q_AB = nC_vΔT = 1 × 12.5 × (289.2 − 96.4) = 12.5 × 192.8 ≈ 2410 J, absorbed by the gas. (d) Net work per cycle: W_net = W_AB + W_BC + W_CA = 0 + 4800 − 3200 = +1600 J. Since the gas returns to state A, the internal energy is unchanged over the cycle, ΔU = 0, so the first law gives Q_net = W_net = 1600 J. The net work is positive, which means the cycle is traversed clockwise on the P-V diagram and the enclosed area of 1600 J is delivered as useful work each time round. The device is therefore acting as a heat engine: it absorbs heat at the higher temperatures, converts 1600 J of it into work per cycle, and rejects the rest. Had the loop been traced anticlockwise, W_net would have been negative and the machine would have been consuming work to pump heat, that is acting as a refrigerator or heat pump.
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