Cambridge International A-Level Physics · topic 16
Thermodynamics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A gas cycle with a sign-convention trap
A fixed amount of gas follows the rectangular cycle A→B→C→D→A on the supplied pressure–volume data: A=(1 litre,100 kPa), B=(3 litre,100 kPa), C=(3 litre,300 kPa), D=(1 litre,300 kPa). Straight graph segments join the points. Use ΔU = Q−Wby, with work by the gas positive in expansion. For A→B the internal-energy increase is 300 J. Supplied data (Volume / litre; Pressure / kPa): (1, 100), (3, 100), (3, 300), (1, 300), (1, 100). (a) Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle. [3 marks] (b) Find net heat over the full cycle and assess whether this direction of operation is a heat engine. [3 marks]
Rectangular gas cycle
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Pressure / kPa
Horizontal axis (x): Volume / litre
Read the exact data as a table
| Volume / litre | Pressure / kPa |
|---|---|
| 1 | 100 |
| 3 | 100 |
| 3 | 300 |
| 1 | 300 |
| 1 | 100 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle.
Reveal the worked solution
- At 100 kPa, expansion by 2 litres gives Wby = pΔV = 100000(0.002) = 200 J. Note that 1 kPa·litre = 1 J.
- The first law gives Q = ΔU+Wby = 300+200 = 500 J into the gas.
- On C→D, Wby = 300000(−0.002) = −600 J; vertical legs do no boundary work. Net work is 200−600 = −400 J, so 400 J is done on the gas.
- Internal energy is a state function, so returning to A gives ΔUcycle = 0.
- The first law then gives Qnet = Wby,net = −400 J: net energy leaves as heat while net work is supplied.
- A net-output heat engine would require the opposite orientation, but its efficiency cannot be found from the enclosed area alone; heat input on the relevant legs must also be known.
Answer (a) A→B: Wby = +200 J and Q = +500 J; net cycle Wby = −400 J. (b) Net Q = −400 J; this direction consumes net work rather than delivering heat-engine work.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
An engine's advertised efficiency exceeds its thermal limit
A cyclic engine absorbs 1200 J per cycle from a 600 K reservoir and rejects heat to a 300 K reservoir. The manufacturer claims 800 J of work per cycle. The supplied graph shows the maximum reversible efficiency 1−Tc/Th as cold-reservoir temperature varies for Th = 600 K. Supplied model: For heat Q received reversibly by a reservoir at absolute temperature T, its entropy change is ΔS = Q/T, with Q negative for energy leaving that reservoir. Supplied data (Cold reservoir / K; Maximum efficiency): (150, 0.75), (300, 0.5), (450, 0.25), (600, 0). (a) Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle. [3 marks] (b) A real engine instead delivers 450 J per cycle. Find its efficiency and total reservoir entropy change; explain why waste heat cannot all be recycled into work in the same cycle. [3 marks]
Reversible efficiency limit
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Maximum efficiency
Horizontal axis (x): Cold reservoir / K
Read the exact data as a table
| Cold reservoir / K | Maximum efficiency |
|---|---|
| 150 | 0.75 |
| 300 | 0.5 |
| 450 | 0.25 |
| 600 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
Reveal the worked solution
- Energy conservation alone allows Qout = 1200−800 = 400 J, so a first-law calculation does not expose the problem.
- The reversible limit is 1−300/600 = 0.50. Maximum work is 0.50(1200) = 600 J, and at least 600 J must be rejected.
- At the claimed output the reservoir entropy change would be −1200/600+400/300 = −2/3 J K⁻¹ per cycle, violating the nondecrease of total entropy.
- The actual rejected heat is 1200−450 = 750 J. Efficiency is 450/1200 = 0.375.
- Reservoir entropy change is −1200/600+750/300 = −2+2.5 = +0.50 J K⁻¹ per cycle, consistent with irreversibility.
- The working substance returns to its initial state each cycle. Converting all heat from a single reservoir into cyclic work would violate the second law; reducing losses cannot remove the need for heat rejection at these temperatures.
Answer (a) Claimed efficiency = 66.7% exceeds the 50% limit; Wmax = 600 J and Qrejected,min = 600 J. (b) Efficiency = 37.5%; reservoir entropy increase = 0.50 J K⁻¹ per cycle.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.