AP · AP 2 · free response
AP Physics 2: Algebra-Based · Question 7
AP Physics 2: Algebra-Based · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 12
- Topics
- 1
- Answer
- Complete
A fixed quantity of a monatomic ideal gas is taken around the closed cycle A → B → C → D → A. State A is at pressure 2.0 × 10⁵ Pa and volume 1.0 × 10⁻³ m³. The gas expands at constant pressure to state B at 3.0 × 10⁻³ m³. It is then cooled at constant volume to state C at 1.0 × 10⁵ Pa. It is compressed at constant pressure to state D at 1.0 × 10⁻³ m³, and finally warmed at constant volume back to state A.
- (a)
Calculate Calculate the work done by the gas during the process A → B.
2 marks - (b)
Determine Determine the change in internal energy of the gas during A → B, and hence the thermal energy added to the gas.
4 marks - (c)
Determine Determine the net work done by the gas in one complete cycle.
3 marks - (d)
Explain Explain why the change in internal energy of the gas over one complete cycle is zero, and state what this implies about the net thermal energy exchanged with the surroundings.
3 marks
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(a)
Calculate Calculate the work done by the gas during the process A → B.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: W = PΔV for a constant-pressure process
- 2
1 point: W = 2.0 × 10⁵ × 2.0 × 10⁻³ = 400 J, done by the gas
(b)
Determine Determine the change in internal energy of the gas during A → B, and hence the thermal energy added to the gas.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: for a monatomic ideal gas ΔU = (3/2)nRΔT = (3/2)Δ(PV)
- 2
1 point: Δ(PV) = 2.0 × 10⁵ (3.0 × 10⁻³ − 1.0 × 10⁻³) = 400 J, so ΔU = 600 J
- 3
1 point: applies the first law, Q = ΔU + W_by
- 4
1 point: Q = 600 + 400 = 1000 J added to the gas
(c)
Determine Determine the net work done by the gas in one complete cycle.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: net work is the area enclosed by the cycle on the PV diagram
- 2
1 point: area = (2.0 × 10⁵ − 1.0 × 10⁵)(3.0 × 10⁻³ − 1.0 × 10⁻³)
- 3
1 point: net work = 200 J, done by the gas because the cycle is traversed clockwise
(d)
Explain Explain why the change in internal energy of the gas over one complete cycle is zero, and state what this implies about the net thermal energy exchanged with the surroundings.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: internal energy is a function of state and depends only on the temperature
- 2
1 point: the gas returns to state A, so its temperature — and hence its internal energy — is the same as at the start, giving ΔU = 0 for the cycle
- 3
1 point: by the first law the net thermal energy added must equal the net work done by the gas, 200 J
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