Specification: AQA A-level Physics (7408) 3.11.2.2 Engineering physics option | Paper: 3 Section B | Data and formulae booklet: given in the exam (including pVγ = constant, W = pΔV and pV = nRT); practise combining them to find a temperature
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Isothermal change: a change at constant temperature, so for an ideal gas ΔU = 0 and pV = constant. It must happen slowly, so heat can flow in or out to keep the temperature fixed.
Adiabatic change: a change in which no heat is transferred to or from the gas (Q = 0), so pVγ = constant, where γ is the ratio of the principal specific heat capacities of the gas. It happens quickly or in a perfectly insulated container; compression heats the gas and expansion cools it.
Work done by a gas: at constant pressure W = pΔV. In general the work done is the area under the p–V graph, and the net work done in a cycle is the area enclosed by the loop.
The equation
adiabatic change: pVγ = constant, so p₁V₁γ = p₂V₂γ isothermal change: pV = constant work done at constant pressure = pressure × change in volume W = pΔV
| Quantity | Symbol | Unit |
|---|---|---|
| pressure | p | pascals (Pa) |
| volume | V | cubic metres (m³) |
| ratio of specific heat capacities | γ | no unit (1.67 for a monatomic gas, 1.40 for air) |
| work done by the gas | W | joules (J) |
| change in volume | ΔV | cubic metres (m³) |
| absolute temperature (in pV = nRT) | T | kelvin (K) |
Rearranged: p₂ = p₁(V₁ ÷ V₂)γ | V₂ = V₁(p₁ ÷ p₂)1/γ | isothermal: p₂ = p₁V₁ ÷ V₂ | ΔV = W ÷ p | temperature after any change: T₂ = T₁ × (p₂V₂) ÷ (p₁V₁) from pV = nRT | Volumes can stay in cm³ inside a ratio, but convert to m³ (× 10−6) for W = pΔV.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A gas at a constant pressure of 2.0 × 105 Pa expands by 3.0 × 10−4 m³. Calculate the work done by the gas.
Show solution
F: W = pΔV
I: W = 2.0 × 105 × 3.0 × 10−4
F: W is already the subject
A: 60 J
Q2 (Core) Air (γ = 1.40) at 1.0 × 105 Pa is compressed adiabatically to half its original volume. Calculate its new pressure, to 2 significant figures.
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F: p₁V₁γ = p₂V₂γ, so p₂ = p₁(V₁ ÷ V₂)γ
I: p₂ = 1.0 × 105 × 21.40
F: p₂ is already the subject
A: 2.639… × 105 Pa = 2.6 × 105 Pa (2 s.f.)
Watch out: halving the volume does not just double the pressure here. That is the isothermal answer; adiabatic compression also heats the gas.
Q3 (Core) The gas in a syringe stays at a constant pressure of 101 kPa while its volume increases from 50 cm³ to 80 cm³. Calculate the work done by the gas, to 2 significant figures.
Show solution
Convert first: 101 kPa = 1.01 × 105 Pa; ΔV = 80 − 50 = 30 cm³ = 3.0 × 10−5 m³
F: W = pΔV
I: W = 1.01 × 105 × 3.0 × 10−5
F: W is already the subject
A: 3.03 J = 3.0 J (2 s.f.)
Watch out: 1 cm³ = 10−6 m³, not 10−2 m³. Cube the conversion factor.
Q4 (Core) 400 cm³ of air (γ = 1.40) at 1.0 × 105 Pa is compressed adiabatically until its pressure is 8.0 × 105 Pa. Calculate its final volume, to 2 significant figures.
Show solution
F: p₁V₁γ = p₂V₂γ
I: 1.0 × 105 × 4001.40 = 8.0 × 105 × V₂1.40
F: V₂ = V₁(p₁ ÷ p₂)1/γ = 400 × (1 ÷ 8.0)1/1.40 = 400 × 0.2264…
A: 90.57… cm³ = 91 cm³ (2 s.f.)
Watch out: to undo the power γ, raise to the power 1/γ. Dividing by γ is a common error.
Q5 (Exam) A monatomic gas (γ = 1.67) at 1.2 × 105 Pa occupies 3.0 × 10−3 m³. It is compressed to 1.0 × 10−3 m³. Calculate the final pressure, to 2 significant figures, if the compression is (a) isothermal, (b) adiabatic. Explain why the answers differ.
Show solution
Step 1 – (a) isothermal
F: p₁V₁ = p₂V₂
I: 1.2 × 105 × 3.0 × 10−3 = p₂ × 1.0 × 10−3
F: p₂ = 1.2 × 105 × 3
A: 3.6 × 105 Pa
Step 2 – (b) adiabatic
F: p₂ = p₁(V₁ ÷ V₂)γ
I: p₂ = 1.2 × 105 × 31.67 = 1.2 × 105 × 6.263…
A: 7.515… × 105 Pa = 7.5 × 105 Pa (2 s.f.)
Why: in the adiabatic compression no heat can leave, so the work done on the gas increases its internal energy and temperature. The hotter gas exerts a greater pressure at the same volume.
Q6 (Exam) In a diesel engine, air (γ = 1.40) at 1.0 × 105 Pa and 27 °C is compressed adiabatically to 1/18 of its original volume. Calculate the pressure and the temperature of the air after compression, to 3 significant figures.
Show solution
Convert first: T₁ = 27 + 273 = 300 K
Step 1 – pressure
F: p₂ = p₁(V₁ ÷ V₂)γ
I: p₂ = 1.0 × 105 × 181.40 = 1.0 × 105 × 57.19…
A: 5.72 × 106 Pa (3 s.f.)
Step 2 – temperature, using pV = nRT
n is constant, so p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂, giving T₂ = T₁ × (p₂ ÷ p₁) × (V₂ ÷ V₁) = 300 × 57.19… ÷ 18
A: 953.3… K = 953 K (3 s.f.), about 680 °C, hot enough to ignite the fuel without a spark
Watch out: pVγ = constant does not contain T. Find p₂ first, then use pV = nRT with temperatures in kelvin.
Q7 (Exam) 0.12 mol of an ideal gas is heated at a constant pressure of 1.0 × 105 Pa from 290 K to 390 K. Calculate the work done by the gas, to 3 significant figures. (R = 8.31 J mol⁻¹ K⁻¹)
Show solution
Step 1 – volumes from pV = nRT
V₁ = nRT₁ ÷ p = 0.12 × 8.31 × 290 ÷ 1.0 × 105 = 2.89188 × 10−3 m³
V₂ = nRT₂ ÷ p = 0.12 × 8.31 × 390 ÷ 1.0 × 105 = 3.88908 × 10−3 m³
Step 2 – work done
F: W = pΔV
I: W = 1.0 × 105 × (3.88908 − 2.89188) × 10−3 = 1.0 × 105 × 9.972 × 10−4
A: 99.72 J = 99.7 J (3 s.f.)
Shortcut: at constant pressure W = pΔV = nRΔT = 0.12 × 8.31 × 100 = 99.72 J
Q8 (Exam) On a p–V diagram, a gas expands along a straight line from A (1.0 × 10−3 m³, 4.0 × 105 Pa) to B (3.0 × 10−3 m³, 2.0 × 105 Pa). It is then compressed at a constant 2.0 × 105 Pa back to 1.0 × 10−3 m³ (point C), and finally heated at constant volume back to A. Calculate (a) the work done by the gas from A to B, (b) the net work done by the gas in one cycle.
Show solution
Step 1 – (a) area under AB (a trapezium)
W = ½ × (4.0 × 105 + 2.0 × 105) × (3.0 − 1.0) × 10−3 = 3.0 × 105 × 2.0 × 10−3
A: 600 J
Step 2 – work done on the gas from B to C
W = pΔV = 2.0 × 105 × (−2.0 × 10−3) = −400 J; from C to A the volume is constant, so W = 0
Step 3 – (b) net work
net W = 600 − 400 + 0 = 200 J, which is the area of the triangle ABC: ½ × 2.0 × 10−3 × 2.0 × 105
A: 200 J per cycle
Watch out: the net work in a cycle is the area enclosed by the loop, not the area under the top line.
Q9 (Stretch) 0.050 mol of helium (γ = 1.67) at 4.0 × 105 Pa and 3.0 × 10−4 m³ expands adiabatically to 6.0 × 10−4 m³. Calculate (a) the final pressure, (b) the initial and final temperatures, (c) the work done by the gas. Give (a) and (b) to 3 significant figures and (c) to 2 significant figures. (For a monatomic ideal gas ΔU = 3⁄2nRΔT; R = 8.31 J mol⁻¹ K⁻¹)
Show solution
Step 1 – (a) final pressure
p₂ = p₁(V₁ ÷ V₂)γ = 4.0 × 105 × 0.51.67 = 1.257… × 105 Pa
A: 1.26 × 105 Pa (3 s.f.)
Step 2 – (b) temperatures from T = pV ÷ nR
nR = 0.050 × 8.31 = 0.4155 J K⁻¹
T₁ = 4.0 × 105 × 3.0 × 10−4 ÷ 0.4155 = 120 ÷ 0.4155 = 288.8… K, so T₁ = 289 K
T₂ = 1.257… × 105 × 6.0 × 10−4 ÷ 0.4155 = 75.42… ÷ 0.4155 = 181.5… K, so T₂ = 182 K
Step 3 – (c) work done, from the first law
ΔU = 3⁄2nRΔT = 1.5 × (p₂V₂ − p₁V₁) = 1.5 × (75.42… − 120) = −66.86… J
adiabatic, so Q = 0 and W = −ΔU
A: 67 J (2 s.f.)
Watch out: you cannot use W = pΔV here because the pressure is not constant. Use Q = ΔU + W with Q = 0.
Q10 (Stretch) Air (γ = 1.40) at 1.0 × 105 Pa, 300 K and 2.4 × 10−3 m³ is compressed adiabatically to 6.0 × 10−4 m³. It is then cooled at constant volume back to 300 K, and then expands at constant pressure until its volume is 1.2 × 10−3 m³. Calculate (a) the pressure and temperature after the compression, to 3 significant figures, (b) the pressure after cooling, (c) the work done by the air and its temperature at the end.
Show solution
Step 1 – (a) adiabatic compression
V₁ ÷ V₂ = 2.4 × 10−3 ÷ 6.0 × 10−4 = 4.0
p₂ = 1.0 × 105 × 4.01.40 = 6.964… × 105 Pa, so p₂ = 6.96 × 105 Pa (3 s.f.)
T₂ = T₁ × (p₂ ÷ p₁) × (V₂ ÷ V₁) = 300 × 6.964… ÷ 4.0 = 522.3… K, so T₂ = 522 K (3 s.f.)
Step 2 – (b) cooling at constant volume
the air is back at 300 K, so p₃V₃ = p₁V₁: p₃ = 1.0 × 105 × 2.4 × 10−3 ÷ 6.0 × 10−4
A: 4.0 × 105 Pa
Step 3 – (c) expansion at constant pressure
W = pΔV = 4.0 × 105 × (1.2 × 10−3 − 6.0 × 10−4) = 240 J
at constant pressure V ∝ T, and the volume doubles, so T = 2 × 300 = 600 K
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