MYP Physics · topic MYP-18.04
First law and gas processes
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.