Thermodynamics: A Level Physics questions with answers
Free Thermodynamics questions for A Level Physics (9702), part of Thermal & modern physics. Each question has the answer and the marking points an examiner would look for.
An ideal gas is taken round a cycle ABCD. A→B: pressure increases from 1.00 × 105 Pa to 3.00 × 105 Pa at constant volume 2.00 × 10−3 m3. B→C: expands at constant pressure to 5.00 × 10−3 m3. C→D: pressure falls to 1.00 × 105 Pa at constant volume. D→A: compressed at constant pressure back to A.
(a)
Calculate the work done by the gas from B to C.
[2]
(b)
Calculate the net work done by the gas in one cycle.
[2]
(c)
State the change in internal energy of the gas over one complete cycle, with a reason.
[1]
(d)
For the change A→B, explain why the thermal energy supplied equals the increase in internal energy.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
900 J
W = pΔV C1 = 900 J A1
2
(b)
600 J
net = area enclosed = (pB − pA)(VC − VB) C1 = 600 J A1
2
(c)
zero
zero, because the gas returns to its original state (same temperature) B1
1
(d)
no work at constant volume
volume is constant so no work is done (w = 0) M1 so ΔU = q A1
2
Question 2Structured · 10 marksA* challenge
0.38 mol of an ideal gas at a constant pressure of 1.0 × 105 Pa expands from 0.01 m3 to 0.015 m3. 1500 J of thermal energy is supplied to the gas.
(a)
Calculate the initial and final temperatures of the gas.
[2]
(b)
Calculate the work done by the gas and the increase in internal energy, using the first law of thermodynamics.
[3]
(c)
Calculate the root-mean-square speed of the molecules at the initial temperature if the gas is helium (molar mass 4.0 g mol−1).
[3]
(d)
Explain why the internal energy of an ideal gas depends only on its temperature.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
317 K, 475 K
pV = nRT C1 both A1
2
(b)
500 J; 1000 J
W = pΔV C1 ΔU = q + W with W on gas negative C1 1000 A1
3
(c)
1400 m s−1
½m⟨c2⟩ = 3/2 kT C1 m = 0.004 ÷ NAC1 1400 A1
3
(d)
no intermolecular forces so no potential energy; internal energy = total KE which ∝ T
no intermolecular forces / no PE B1 KE ∝ thermodynamic temperature B1
2
Question 3Structured · 12 marks
A fixed mass of ideal gas is taken round the cycle A → B → C → D → A shown in Fig. 3.1.
Fig. 3.1
(a)
State the first law of thermodynamics, defining each symbol.
[3]
(b)
Calculate the work done by the gas during B → C.
[2]
(c)
Calculate the net work done by the gas in one complete cycle.
[2]
(d)
For the change A → B, state the work done on the gas and explain whether thermal energy is supplied to or removed from the gas.
[3]
(e)
State the change in internal energy of the gas for the complete cycle. Explain your answer.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
ΔU = q + W; q heat supplied to gas; W work done on gas
ΔU = q + W B1 q = thermal energy supplied to the system B1 W = work done on the system B1
W = 0 (constant volume); pressure rises so T rises, U increases; q positive, supplied
no work is done (volume constant) B1 p increases at constant V so T and internal energy increase B1 so q is positive: thermal energy is supplied B1
3
(e)
zero; returns to same state/temperature
zero B1 the gas returns to its original state (same temperature), and U depends only on the state B1
2
Question 4Structured · 8 marks
An electric kettle of power 2000 W contains 0.50 kg of water at 30 °C.
(a)
Define specific heat capacity.
[1]
(b)
Calculate the time to bring the water to 100 °C, assuming no losses. (c = 4200 J kg−1 K−1)
[2]
(c)
The kettle is left on for a further 60 s after boiling. Calculate the mass of water boiled away. (L = 2.26 × 106 J kg−1)
[2]
(d)
Explain, in terms of molecules, why the specific latent heat of vaporisation is much larger than that of fusion.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
energy per unit mass per unit temperature rise
energy needed per unit mass to raise the temperature by one degree B1
1
(b)
73.5 s
E = mcΔT = 1.47 × 105 J C1 t = E/P = 73.5 s A1
2
(c)
53.1 g
m = Pt/L C1 53.1 g A1
2
(d)
vaporisation separates molecules completely (to large separation); also work against atmosphere; fusion only loosens bonds
in vaporisation molecules are separated completely, so all intermolecular bonds are broken B1 work is also done pushing back the atmosphere B1 in melting only some bonds are broken and separation hardly changes B1
3
Question 5Structured · 4 marks
A gas is heated at a constant pressure of 1.5 × 105 Pa. Its volume increases by 3.0 × 10−3 m3 and 600 J of thermal energy is supplied.
(a)
State the first law of thermodynamics with the sign convention used in 9702.
[2]
(b)
Calculate the change in internal energy of the gas.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
ΔU = q + W, W done on the system
ΔU = q + W B1 q is heating of system, W is work done on the system B1
2
(b)
150 J
W = −pΔV = −450 J C1 ΔU = 150 J A1
2
Question 6Structured · 12 marks
An electric heater of power 60 W is used to melt ice at 0 °C in a funnel. In 150 s, 30.4 g of water collects. In a control experiment with the heater off, 3.6 g of water collects from an identical funnel of ice in the same time.
(a)
Define specific latent heat of fusion.
[2]
(b)
Explain why the control experiment is needed.
[2]
(c)
Calculate the specific latent heat of fusion of ice from these results.
[3]
(d)
State the first law of thermodynamics, explaining the sign convention for each term.
[3]
(e)
Explain, in terms of molecules, why the internal energy of water increases as ice melts at 0 °C even though its temperature does not change.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
energy to change unit mass from solid to liquid at constant temperature
energy required per unit mass B1 to change a substance from solid to liquid without a change in temperature B1
2
(b)
ice also melts from energy from surroundings; allows this to be accounted for
ice melts because of thermal energy from the surroundings as well B1 the control lets this mass be subtracted B1
2
(c)
3.36 × 105 J kg−1
mass melted by the heater = 30.4 − 3.6 = 26.8 g C1 L = Pt/m C1 3.36 × 105 J kg−1A1
3
(d)
ΔU = q + W; q heat supplied to system; W work done on system
ΔU = q + W B1 q is thermal energy supplied to the system B1 W is work done on the system B1
3
(e)
potential energy of molecules increases as bonds broken; kinetic energy unchanged
the energy breaks bonds / increases the separation of molecules, increasing their potential energy B1 the mean kinetic energy (temperature) is unchanged B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in