IB · B · extended response
A gas cycle with a sign-convention trap
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
At 100 kPa, expansion by 2 litres gives Wby = pΔV = 100000(0.002) = 200 J. Note that 1 kPa·litre = 1 J.
- 2
The first law gives Q = ΔU+Wby = 300+200 = 500 J into the gas.
- 3
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.
(b)
Find net heat over the full cycle and assess whether this direction of operation is a heat engine.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Internal energy is a state function, so returning to A gives ΔUcycle = 0.
- 2
The first law then gives Qnet = Wby,net = −400 J: net energy leaves as heat while net work is supplied.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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