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Questions on the first law and its sign convention
Heat supplied to a gas raises its internal energy, leaves as work the gas does, or both: this is the first law of thermodynamics, Q = ΔU + W. Compression makes W negative and heat lost makes Q negative, so the signs settle whether the gas warmed or cooled. At constant pressure the work is pΔV.
17 original questions · 51 marks · the first law and its sign convention notes · Engineering physics
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These questions cover both lessons on the first law of thermodynamics: The first law and its sign convention and Non-flow processes and p-V diagrams.
This topic is tagged for AQA, CIE, OCR A.
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State the first law of thermodynamics, and state what Q and W each stand for in the form used at A-level.
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Q = ΔU + W (1). Q is the heat supplied to the gas, ΔU the increase in its internal energy, and W the work done by the gas on its surroundings (1).State what is meant by the internal energy of an ideal gas, and state the one property of the gas that fixes it.
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It is the total kinetic energy of the random motion of its molecules (1). An ideal gas has no intermolecular potential energy, so its internal energy depends only on its absolute temperature (1).State which term of the first law is zero for a change at constant volume, and which is zero for an isothermal change, giving a reason in each case.
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Constant volume: W = 0, because the gas does not move its surroundings, so no work is done (1). Isothermal: ΔU = 0, because the temperature is unchanged and the internal energy of an ideal gas depends only on temperature, so Q = W (1).State what is meant by an adiabatic change, and give one practical example.
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A change in which no heat enters or leaves the gas, Q = 0, achieved by good insulation or by acting too quickly for heat to flow (1). Examples: the compression stroke of a diesel engine, or the barrel of a bicycle pump warming as you pump (1).Explain why the work done by an expanding gas is equal to the area under the curve on a p-V diagram.
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For a small increase in volume δV the gas does work p δV, the area of a thin strip under the curve (1). Adding all such strips between the initial and final volumes gives the total work, which is the area under the curve (1). Only at constant pressure does this reduce to the single rectangle pΔV.State the sign of W in Q = ΔU + W when a gas is compressed.
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W is negative (1), because W means the work done by the gas and here the surroundings do work on it instead.A gas absorbs 850 J of heat while expanding and does 320 J of work on its surroundings. Calculate the change in its internal energy and state whether its temperature rises or falls.
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ΔU = Q − W (1) = 850 − 320 = +530 J (1). The internal energy has risen, and for an ideal gas that means the temperature rises (1), though by less than 850 J of heating alone would give, because 320 J left as work.The area under the p-V curve for an expanding gas is estimated by counting squares: 34 squares, each representing 25 J. During the expansion 1200 J of heat is supplied to the gas. Calculate the work done by the gas and the change in its internal energy.
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W = area under the curve = 34 × 25 = 850 J (1). ΔU = Q − W (1) = 1200 − 850 = +350 J (1). Counting squares is an accepted method here; pΔV would only serve if the pressure were constant.A gas is compressed adiabatically, 500 J of work being done on it. State the value of Q, calculate the change in internal energy, and state what happens to the temperature of the gas.
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Adiabatic means no heat flow, so Q = 0 (1). Work is done on the gas, so in this convention W = −500 J, and ΔU = Q − W = 0 − (−500) = +500 J (1). The internal energy rises, so the temperature rises (1). Writing W as +500 J reverses the sign of ΔU and predicts cooling.A gas at a pressure of 2.4 × 105 Pa occupies 3.0 × 10−3 m3. It expands isothermally to 8.0 × 10−3 m3. Calculate the final pressure and state, with a reason, the direction of the heat flow.
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Isothermal, so pV = constant (1): p2 = 2.4 × 105 × 3.0/8.0 = 9.0 × 104 Pa (1). ΔU = 0 for an isothermal change, so Q = W; the gas expands so W is positive, so heat flows into the gas at the same rate as it does work (1). pVγ = constant describes an adiabatic change and does not apply here.Air at 1.0 × 105 Pa fills 6.0 × 10−4 m3 of a cylinder and is compressed adiabatically to 1.2 × 10−4 m3. Calculate the final pressure. Take γ = 1.4.
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p1V1γ = p2V2γ, so p2 = p1(V1/V2)γ (1). V1/V2 = 5.0, and 5.01.4 = 9.52 (1), so p2 = 1.0 × 105 × 9.52 = 9.5 × 105 Pa (1). Note the pressure rises by more than the volume ratio of 5, because the gas heats as it is squeezed.A gas is sealed in a rigid container and supplied with 1200 J of heat. It is then cooled, losing 400 J of heat. For each stage state the work done by the gas and calculate the change in internal energy, and give the overall change.
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Rigid means constant volume, so W = 0 in both stages (1). Stage 1: ΔU = Q = +1200 J (1). Stage 2: Q is negative because heat leaves, so ΔU = −400 J. Overall ΔU = +800 J (1), so the gas ends hotter than it began.A gas expands at a constant pressure of 8.0 × 104 Pa, its volume increasing from 1.5 × 10−3 m3 to 6.5 × 10−3 m3, while 620 J of heat is supplied. Calculate the work done by the gas and the change in its internal energy.
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ΔV = 6.5 × 10−3 − 1.5 × 10−3 = 5.0 × 10−3 m3 (1). W = pΔV = 8.0 × 104 × 5.0 × 10−3 = 400 J (1). ΔU = Q − W = 620 − 400 = +220 J (1). W = pΔV takes the change in volume: using the final volume alone gives 8.0 × 104 × 6.5 × 10−3 = 520 J instead of 400 J.A fixed mass of gas is taken round the following four-stage cycle. From A (1.0 × 105 Pa, 1.0 × 10−3 m3) it is heated at constant volume to B (3.0 × 105 Pa). From B it expands at constant pressure to C (3.0 × 10−3 m3). From C it is cooled at constant volume to D (1.0 × 105 Pa). From D it is compressed at constant pressure back to A. Calculate the work done by the gas in each stage, the net work per cycle, and the net heat supplied per cycle.
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A to B: constant volume, so W = 0 J (1). B to C: W = pΔV = 3.0 × 105 × 2.0 × 10−3 = +600 J (1). C to D: constant volume, so W = 0 J (1). D to A: the gas is compressed, so W = 1.0 × 105 × (−2.0 × 10−3) = −200 J (1). Net work per cycle = 600 − 200 = +400 J, which is exactly the area enclosed by the rectangle, Δp × ΔV = 2.0 × 105 × 2.0 × 10−3 (1). The gas returns to state A, so its temperature and hence ΔU return to their starting values and ΔU = 0 over the cycle; the first law then gives net Q = net W = +400 J (1). The compression contributes a negative W, so adding the four works with every sign positive gives 800 J instead.In a diesel engine, air at 1.0 × 105 Pa and 300 K fills 9.0 × 10−4 m3 of a cylinder and is compressed adiabatically to 5.0 × 10−5 m3. Take γ = 1.4. Calculate the final pressure and the final temperature, and explain why no spark plug is needed.
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Compression ratio V1/V2 = 18 (1). p2 = p1(V1/V2)γ = 1.0 × 105 × 181.4 = 1.0 × 105 × 57.2 = 5.72 × 106 Pa (1). Then pV/T is constant, so T2/T1 = p2V2/p1V1 = 57.2/18 = 3.18 (1), giving T2 = 300 × 3.18 = 953 K (1). No heat was supplied at all: the piston did work on the molecules, raising their mean kinetic energy and so the temperature, and about 950 K is above the ignition temperature of the fuel, so it ignites on injection (1). The temperature ratio must be taken in kelvin, and pV is not constant in this change.A gas expands from the same initial state to the same final volume, once isothermally and once adiabatically. Explain which expansion does more work, and account for the difference using the first law.
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The isothermal expansion does more work (1). On the p-V diagram the adiabat falls more steeply than the isotherm, so less area lies beneath it between the same two volumes, and work is that area (1). Physically, the isothermal gas is fed heat continuously, and with ΔU = 0 every joule of that heat leaves as work (1). The adiabatic gas receives no heat at all, so with Q = 0 it must pay for its work out of its own internal energy, ΔU = −W, and it cools as it expands, which lowers its pressure further and limits the work available (1).A fixed mass of gas undergoes three changes in succession. First it absorbs 500 J of heat at constant volume. Then it expands isothermally, doing 380 J of work. Finally it is compressed adiabatically, 260 J of work being done on it. Calculate the change in internal energy in each change and the overall change, and verify that Q = ΔU + W for the complete sequence.
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Constant volume: W = 0, so ΔU = Q = +500 J (1). Isothermal: ΔU = 0, and Q = W = +380 J (1). Adiabatic: Q = 0 and W = −260 J because the work is done on the gas, so ΔU = −W = +260 J (1). Overall ΔU = 500 + 0 + 260 = +760 J. Check: total Q = 500 + 380 + 0 = 880 J and total W = 0 + 380 − 260 = 120 J, so the totals satisfy 880 J = 760 J + 120 J (1).
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