Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
Internal energy and the ideal gas
40 free, high-quality past-paper-style questions on Internal energy and the ideal gas for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 143 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U5-2-A-01Unit 5 · A2Easy1 mark
Which statement about the internal energy of a fixed mass of an ideal gas is correct?
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Answer
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Notes
In an ideal gas there are no forces between the molecules except during collisions, so the molecules have no potential energy. The internal energy is the total random kinetic energy of the molecules, which is proportional to the absolute (kelvin) temperature.
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Answer A
Answer: A
Exam tip: Internal energy = sum of the randomly distributed kinetic and potential energies of the molecules. For an ideal gas the potential energy term is zero.
A sealed flask of volume 2.0 × 10−3 m3 contains air at a pressure of 1.0 × 105 Pa and a temperature of 27 °C. k = 1.38 × 10−23 J K−1. How many molecules are in the flask?
A block of ice at 0 °C is heated and melts, forming water at 0 °C. Which row describes the changes in the mean kinetic energy and in the potential energy of the molecules during the melting?
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The temperature stays at 0 °C, so the mean kinetic energy of the molecules is unchanged. The energy supplied separates the molecules against their attractive forces, so their potential energy increases and the internal energy increases.
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Answer A
Answer: A
Exam tip: Constant temperature means constant mean kinetic energy. During a change of state, the energy supplied goes into molecular potential energy.
A sealed plastic bottle of air is carried in a moving car from the coast to a mountain village. The temperature of the air in the bottle stays the same throughout. Treat the air as an ideal gas.
(a)
State what is meant by the internal energy of a body.
(2)
(b)
Explain why the internal energy of the air in the bottle does not change during the journey.
(2)
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Part (a)
the sum of the kinetic and potential energies of the molecules
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which are randomly distributed among the molecules
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Part (b)
an ideal gas has no forces between its molecules, so no molecular potential energy; its internal energy is the total random kinetic energy, which depends only on the temperature
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the kinetic and gravitational potential energy of the bottle as a whole are not part of internal energy, and the temperature is unchanged
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Common error: Counting the speed or height of the whole object as internal energy. Only the random, molecular-scale energies count.
A constant-volume gas thermometer consists of a fixed mass of gas in a rigid bulb connected to a pressure gauge. The pressure of the gas is measured at two fixed points: 0 °C: 92.0 kPa; 100 °C: 125.7 kPa. The pressure varies linearly with temperature.
Figure: pressure of the gas at the two fixed points
(a)
State what is meant by absolute zero.
(1)
(b)
Use the readings to estimate the value of absolute zero in °C.
(2)
(c)
Calculate the pressure of the gas when the bulb is at −40 °C.
(2)
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Answer
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Part (a)
the lowest possible temperature, at which the molecules have minimum (kinetic) energy / at which the pressure of an ideal gas would be zero
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Part (b)
pressure rises by 125.7 − 92.0 = 33.7 kPa per 100 °C, i.e. 0.337 kPa per °C
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zero pressure at 0 − 92.0 ÷ 0.337 = −273 °C
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Part (c)
p ∝ T: T = 233 K, or 92.0 − 40 × 0.337
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p = 92.0 × 233 ÷ 273 = 78.5 kPa
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Model sketch
Figure: extrapolating the line to zero pressure
Exam tip: A straight-line extrapolation to zero pressure needs two points: find the gradient first, then how far below 0 °C the line meets the axis.
Which graph shows how the mean kinetic energy of the molecules of an ideal gas varies with the Celsius temperature θ of the gas?
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Answer
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½m⟨c2⟩ = 3/2kT, so the mean kinetic energy is proportional to the kelvin temperature T = θ + 273. Against θ this is a straight line that reaches zero at θ = −273 °C (absolute zero): graph A.
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Answer A
Answer: A
Exam tip: A quantity proportional to T plotted against θ in °C gives a straight line that is not through the origin: it cuts the θ axis at −273 °C.
The graph shows the distribution of molecular speeds in two samples, X and Y, of the same gas. Which statement is correct?
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The molecules in Y are spread to higher speeds, so their r.m.s. speed and mean kinetic energy are greater: Y is at the higher temperature. The area under each curve represents the total number of molecules, and the two areas are equal, so the samples contain the same number of molecules. Y's peak is lower because the same number of molecules is spread over a wider range of speeds.
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Answer D
Answer: D
Exam tip: On a speed-distribution graph, the area under the curve is the number of molecules. Heating the gas moves the peak to a higher speed and makes it lower and broader; the area stays the same.
A thermometer is left inside a freezer until its reading is steady. Figure 1 shows the thermometer. Treat the air in the freezer as an ideal gas. k = 1.38 × 10−23 J K−1
(a)
Read the temperature of the air in the freezer from Figure 1 and convert it to kelvin.
(2)
(b)
Calculate the mean kinetic energy of a molecule of the air in the freezer.
(2)
(c)
State the temperature, in °C, at which the mean kinetic energy of the air molecules would be twice its value in the freezer.
(1)
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Part (a)
θ = −18 °C
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T = −18 + 273 = 255 K
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Part (b)
mean kinetic energy = 3/2kT = 1.5 × 1.38 × 10−23 × 255
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= 5.3 × 10−21 J (5.28 × 10−21 J)
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Part (c)
T = 2 × 255 = 510 K, so θ = 237 °C
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Common error: Doubling the Celsius reading (−18 °C → −36 °C). Mean kinetic energy is proportional to the kelvin temperature, so double the kelvin value and convert back.
A block of ice at −20 °C is heated by a 70 W heater. All the energy from the heater is transferred to the ice, which melts, and then to the water formed. Figure 1 shows how the temperature θ varies with time t.
(a)
Describe what happens to the mean kinetic energy and to the potential energy of the molecules between t = 1.0 min and t = 9.0 min.
(2)
(b)
Calculate the increase in the internal energy of the ice while it is melting.
(2)
(c)
Calculate the total increase in internal energy from t = 0 to t = 11.0 min.
(1)
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Answer
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Part (a)
the temperature is constant, so the mean kinetic energy of the molecules does not change
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the potential energy of the molecules increases, as they are separated against the attractive forces between them (bonds are broken)
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Part (b)
melting time = 9.0 − 1.0 = 8.0 min = 480 s
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ΔU = 70 × 480 = 3.4 × 104 J (33 600 J)
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Part (c)
70 × 11.0 × 60 = 4.6 × 104 J (46 200 J)
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Exam tip: During a change of state the temperature is constant, so the energy supplied goes into molecular potential energy. Internal energy still increases.
A fixed mass of gas in a cylinder has a volume of 450 cm3, a pressure of 1.2 × 105 Pa and a temperature of 17 °C. A piston compresses it to a volume of 180 cm3, and its temperature rises to 77 °C. What is the new pressure of the gas?
Figure: the gas before and after compression (not to scale)
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Answer
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Notes
For a fixed mass, pV/T is constant: p2 = 1.2 × 105 × (450 ÷ 180) × (350 ÷ 290) = 3.6 × 105 Pa.
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Answer C
Answer: C
Exam tip: Write p1V1/T1 = p2V2/T2, then check each ratio: a smaller volume and a higher temperature must both increase the pressure.
A container holds a mixture of hydrogen molecules (mass 2.0 u) and oxygen molecules (mass 32 u) at the same temperature. What is the ratio (r.m.s. speed of the hydrogen molecules) ÷ (r.m.s. speed of the oxygen molecules)?
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At the same temperature the mean kinetic energy ½m⟨c2⟩ = 3/2kT is the same, so crms ∝ 1/√m. Ratio = √(32 ÷ 2.0) = 4.0.
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Answer C
Answer: C
Exam tip: Same temperature means the same mean kinetic energy, not the same speed. The lighter molecules move faster.
In an experiment on a fixed mass of gas at constant temperature, the volume of air trapped in a uniform tube is found from V = πd2l/4. The internal diameter is d = 1.00 ± 0.02 cm and the length of the air column is l = 25.0 ± 0.1 cm. What is the percentage uncertainty in V?
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%d = 0.02 ÷ 1.00 × 100 = 2.0%, which is doubled because d is squared: 4.0%. %l = 0.1 ÷ 25.0 × 100 = 0.4%. %V = 4.0 + 0.4 = 4.4%.
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Answer D
Answer: D
Exam tip: When quantities are multiplied, add their percentage uncertainties; a quantity raised to a power has its percentage uncertainty multiplied by that power.
A weather balloon is filled with helium at ground level, where the pressure is 101 kPa and the temperature is 15 °C. The volume of the balloon at ground level is 4.0 m3. k = 1.38 × 10−23 J K−1.
(a)
Calculate the number of helium atoms in the balloon.
(2)
(b)
The balloon rises to a height where the pressure is 25 kPa and the temperature is −45 °C. The balloon fabric is slack, so the pressure inside equals the pressure outside. Calculate the new volume of the balloon.
(3)
(c)
State one assumption you made in your calculation in (b).
no helium escapes (the number of atoms is constant) / helium behaves as an ideal gas
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Exam tip: Fall in pressure and fall in temperature act in opposite directions on the volume. Check that your answer is larger than 4.0 m3 but smaller than 4.0 × 101 ÷ 25.
A steel cylinder of volume 5.0 × 10−3 m3 contains 2.4 mol of argon at 20 °C. The Avogadro constant NA = 6.02 × 1023 mol−1. k = 1.38 × 10−23 J K−1. The mass of an argon atom is 40 u, where 1 u = 1.66 × 10−27 kg.
(a)
Calculate the pressure of the argon.
(2)
(b)
Argon is used from the cylinder until the pressure has fallen to 4.0 × 105 Pa, with the temperature unchanged. Calculate the number of argon atoms that have been used.
A student investigates how the pressure of a fixed mass of air depends on its volume. The air is trapped in a gas syringe connected by a very short tube to a pressure sensor and data logger. She pushes the plunger in, in steps, and records the volume from the syringe scale and the pressure. The room temperature is 21 °C. Her results are: V / cm3: 60, 50, 40, 30, 24, 20 — p / kPa: 101, 121, 152, 202, 252, 303. k = 1.38 × 10−23 J K−1.
Figure: the student's readings of pressure against volume
(a)
Explain why she should wait for about a minute after each movement of the plunger before recording the pressure.
(2)
(b)
Show that these results are consistent with Boyle's law.
(2)
(c)
Calculate the number of air molecules in the syringe.
(2)
(d)
Suggest one advantage of using a pressure sensor and data logger rather than a dial gauge.
(1)
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Part (a)
compressing the air does work on it, so its temperature rises
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waiting allows the air to return to room temperature, so the temperature is the same (constant) for every reading
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Part (b)
calculates pV for at least three readings, e.g. 101 × 60 = 6060, 152 × 40 = 6080, 303 × 20 = 6060 (kPa cm3)
higher resolution / no parallax error in reading a pointer / readings recorded automatically at a chosen moment
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Exam tip: To test an inverse proportion from a table, calculate the product of the two variables for several rows. If the products agree within the precision of the data, the relationship holds.
Air breathed out of the lungs is at body temperature, 37 °C. An oxygen molecule has a mass of 32 u, where 1 u = 1.66 × 10−27 kg. k = 1.38 × 10−23 J K−1.
(a)
Explain what is meant by the root-mean-square (r.m.s.) speed of the molecules of a gas.
(1)
(b)
Calculate the r.m.s. speed of the oxygen molecules in the exhaled air.
(3)
(c)
Calculate the mean kinetic energy of an oxygen molecule in the exhaled air.
(1)
(d)
Exhaled air also contains carbon dioxide molecules, of mass 44 u. State how the r.m.s. speed of the carbon dioxide molecules compares with that of the oxygen molecules, and why.
(1)
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Part (a)
the square root of the mean of the squares of the speeds of all the molecules
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Part (b)
T = 310 K and m = 32 × 1.66 × 10−27 = 5.31 × 10−26 kg
A sealed glass lamp contains xenon gas. When the lamp is off the xenon is at 25 °C; when it is running the xenon reaches 600 °C. The mass of a xenon atom is 131 u, where 1 u = 1.66 × 10−27 kg. k = 1.38 × 10−23 J K−1.
(a)
Calculate the ratio (mean kinetic energy of a xenon atom when the lamp is running) ÷ (mean kinetic energy when it is off).
(1)
(b)
Calculate the r.m.s. speed of the xenon atoms when the lamp is running.
(2)
(c)
The volume of the lamp does not change. State and explain the factor by which the pressure of the xenon increases.
with N and V constant, pV = NkT gives p ∝ T (in kelvin)
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Common error: Taking the ratio of Celsius temperatures (600 ÷ 25 = 24). Mean kinetic energy and pressure at constant volume are proportional to the kelvin temperature.
The bar charts show the number of molecules N and the Celsius temperature θ of four samples, W, X, Y and Z, of the same ideal gas. Which sample has the greatest internal energy?
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Answer
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For an ideal gas the internal energy is the total random kinetic energy of the molecules, N × 3/2kT, so it is proportional to NT with T in kelvin. NT (in units of 1023 K): W 1.0 × 573 = 573; X 2.0 × 433 = 866; Y 4.0 × 213 = 852; Z 3.0 × 300 = 900. The greatest is Z.
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Answer D
Answer: D
Common error: Comparing temperatures in °C. The mean kinetic energy of a molecule is proportional to the kelvin temperature, so convert every temperature before multiplying by the number of molecules.
The graph shows how the pressure p of a fixed mass of ideal gas varies with its volume V at two different constant temperatures. The lower curve is for a temperature of 300 K. What is the temperature T of the upper curve?
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For a fixed mass, pV ∝ T. At P: pV = 100 × 103 × 3.0 × 10−3 = 300 J (at 300 K). At Q: pV = 300 × 103 × 1.5 × 10−3 = 450 J. T = 300 × 450 ÷ 300 = 450 K.
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Answer C
Answer: C
Exam tip: Points on different isotherms usually have different volumes and pressures. Compare the product pV, not p or V alone.
A fixed mass of gas is trapped in a vertical cylinder of uniform cross-section by a piston that moves freely without friction, so the pressure of the gas stays constant. The cylinder stands in a water bath. The diagram shows the apparatus when the water is at 17 °C; the ruler measures the length of the gas column. The water is heated to 107 °C and the gas reaches this temperature. What is the new length of the gas column?
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Read the length: 12.0 cm. At constant pressure V ∝ T, and V ∝ length for a uniform cylinder. Length = 12.0 × (107 + 273) ÷ (17 + 273) = 12.0 × 380 ÷ 290 = 15.7 cm.
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Answer B
Answer: B
Common error: Using the ratio of Celsius temperatures, 107 ÷ 17. Charles's law needs kelvin temperatures.
Exam tip: r.m.s. means: square every speed, take the mean over all the molecules, then take the square root. Each speed counts once for every molecule that has it.
The graph shows how the product pV varies with the kelvin temperature T for two samples, X and Y, of an ideal gas. What is the ratio (number of molecules in X) ÷ (number of molecules in Y)?
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pV = NkT, so the gradient of each line is Nk. Gradient of X = 6.0 ÷ 300 = 0.020 J K−1; gradient of Y = 2.0 ÷ 400 = 0.0050 J K−1. Ratio = 0.020 ÷ 0.0050 = 4.0.
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Answer D
Answer: D
Exam tip: On a graph of pV against T, compare gradients, not single readings: the gradient is Nk, so it is proportional to the number of molecules.
A gas molecule of mass 4.6 × 10−26 kg travelling at 500 m s−1 collides elastically with a flat wall, as shown in the diagram. What is the magnitude of the change in momentum of the molecule?
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Only the component of velocity at right angles to the wall reverses. The path is at 60° to the wall, so this component is 500 sin 60° = 433 m s−1. Δp = 2mv sin 60° = 2 × 4.6 × 10−26 × 433 = 4.0 × 10−23 N s, directed away from the wall.
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Answer C
Answer: C
Common error: Using the angle to the wall as if it were the angle to the normal. Draw the normal and find the component of velocity along it before doubling.
Figure 1 shows how the pressure p and volume V of a fixed mass of ideal gas change as the gas is taken from state A to state B and then to state C. The temperature of the gas at A is 300 K. k = 1.38 × 10−23 J K−1
(a)
Show that the temperature of the gas at B is the same as at A.
(2)
(b)
Calculate the number of molecules of gas.
(2)
(c)
Calculate the temperature of the gas at C.
(1)
(d)
Calculate the ratio (r.m.s. speed of the molecules at C) ÷ (r.m.s. speed of the molecules at A).
(1)
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Answer
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Part (a)
pV at A = 240 × 103 × 1.0 × 10−3 = 240 J and at B = 80 × 103 × 3.0 × 10−3 = 240 J
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pV = NkT with N constant, so equal pV means equal T (300 K)
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Part (b)
N = pV/kT = 240 ÷ (1.38 × 10−23 × 300)
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= 5.8 × 1022 (5.80 × 1022)
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Part (c)
at constant volume p ∝ T: T = 300 × 160 ÷ 80 = 6.00 × 102 K (600 K)
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Part (d)
crms ∝ √T: √(600 ÷ 300) = 1.41
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Exam tip: Read the volume axis label: 1.0 on an axis marked V / 10−3 m3 means 1.0 × 10−3 m3, and kPa must become Pa before you use pV = NkT.
Figure 1 shows how the air temperature θ varies with height above sea level in the lower atmosphere. Treat the air as an ideal gas consisting of nitrogen molecules, each of mass 28 u. 1 u = 1.66 × 10−27 kg k = 1.38 × 10−23 J K−1
(a)
Use Figure 1 to find the kelvin temperature of the air at a height of 10 km.
(1)
(b)
Calculate the r.m.s. speed of the molecules at a height of 10 km.
(3)
(c)
The air pressure at a height of 10 km is 26 kPa. Calculate the number of molecules in each cubic metre of air at this height.
Common error: Putting the Celsius reading (−50) into √(3kT/m). A negative number under the square root is a sure sign that the temperature has not been converted to kelvin.
Figure 1 shows a hand pump used to pump air into a rigid tank of volume 5.0 × 10−3 m3. On each stroke the pump fills with air at atmospheric pressure, 1.0 × 105 Pa, and then pushes all of this air through the one-way valve into the tank. The dimensions in Figure 1 are the internal diameter of the pump barrel and the internal length swept by the piston. Assume the temperature of the air stays at 20 °C. k = 1.38 × 10−23 J K−1
(a)
Calculate the number of air molecules pushed into the tank on each stroke.
(2)
(b)
The tank initially contains air at 1.0 × 105 Pa. Calculate the pressure in the tank after 20 strokes.
(2)
(c)
In practice the temperature of the air in the tank rises during pumping. Suggest why.
A computer model of an ideal gas calculates the mean kinetic energy of the molecules at different kelvin temperatures. Figure 1 shows the results with a line of best fit.
(a)
Determine the gradient of the line.
(2)
(b)
The model uses ½m⟨c2⟩ = 3/2kT. Use your gradient to find the value of the Boltzmann constant k in the model.
(1)
(c)
Explain why the internal energy of the model gas is equal to the number of molecules multiplied by the mean kinetic energy of a molecule.
(1)
(d)
The model contains 2.0 × 1022 molecules. Use Figure 1 to calculate the internal energy of the model gas at 400 K.
(2)
(e)
Each model molecule has a mass of 6.6 × 10−27 kg. Calculate the r.m.s. speed of the molecules at 400 K.
(1)
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Answer
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Notes
Part (a)
uses a large triangle, e.g. from (0, 0) to (580 K, 12.0 × 10−21 J)
gradient = 3/2k, so k = ⅔ × 2.07 × 10−23 = 1.38 × 10−23 J K−1 (allow 1.35 to 1.41 × 10−23, consistent with the gradient)
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Part (c)
in an ideal gas there are no forces between the molecules (except in collisions), so they have no potential energy and the internal energy is the total random kinetic energy
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Part (d)
mean kinetic energy at 400 K = 8.3 × 10−21 J (allow 8.2 to 8.4)
⟨c2⟩ = 2 × 8.3 × 10−21 ÷ 6.6 × 10−27, so crms = 1.6 × 103 m s−1
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Exam tip: To find a gradient, use a triangle that covers at least half of the line and read its corners at gridline intersections; never use a single data point unless it lies on the line.
Figure 1 shows apparatus used to investigate how the pressure of a fixed mass of air depends on its volume at constant temperature. Air is trapped above oil in a glass tube of uniform cross-section. A pump changes the pressure on the oil, and the Bourdon gauge shows the pressure of the trapped air.
(a)
Record the length ℓ of the air column and the pressure p shown in Figure 1.
(2)
(b)
An earlier reading was ℓ = 35.0 cm at p = 121 kPa. Show that the two readings are consistent with Boyle's law.
(2)
(c)
Explain why the length of the air column can be used in place of its volume.
(1)
(d)
The uncertainty in judging the position of the oil surface is ±0.2 cm. Calculate the percentage uncertainty in ℓ.
(1)
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Answer
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Part (a)
ℓ = 28.0 cm
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p = 150 kPa
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Part (b)
pℓ = 121 × 35.0 = 4.24 × 103 kPa cm and 150 × 28.0 = 4.20 × 103 kPa cm
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the products agree to within about 1%, which is within the uncertainty of the readings, so pℓ, and hence pV, is constant
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Part (c)
the tube has a uniform cross-sectional area A, so V = Aℓ and V ∝ ℓ
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Part (d)
0.2 ÷ 28.0 × 100 = 0.7% (0.71%)
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Exam tip: When two readings should give a constant product, calculate the percentage difference between the products and compare it with the percentage uncertainties in the readings.
A sealed rigid container of an ideal gas is heated from 300 K to 600 K. The pressure doubles. The average momentum change of a molecule in each collision with a wall increases by a factor of √2. By what factor does the number of collisions per second with a wall increase?
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Answer
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Notes
Pressure ∝ (collisions per second) × (momentum change per collision). 2 = (factor for collisions) × √2, so the collision rate increases by 2 ÷ √2 = √2 = 1.4. This is consistent with crms ∝ √T: faster molecules cross the container more often.
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Answer B
Answer: B
Exam tip: Heating a gas at constant volume raises the pressure for two reasons: the molecules hit the walls more often and harder. Each effect contributes a factor of √(T2/T1).
Two flasks are joined by a thin tube, so the gas in them is at the same pressure. Flask X has a volume of 1.0 litre and is kept at 300 K. Flask Y has a volume of 2.0 litres and is kept at 450 K. What fraction of the gas molecules is in flask Y?
Figure: two flasks joined by a thin tube (not to scale)
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Answer
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Notes
N = pV/kT with the same p, so N ∝ V/T. X: 1.0/300; Y: 2.0/450. Fraction in Y = (2.0/450) ÷ (1.0/300 + 2.0/450) = 0.57.
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Answer B
Answer: B
Exam tip: When the pressure is the same, the number of molecules in each part is proportional to V/T. A hotter region holds fewer molecules per unit volume.
A flask of volume 250 cm3 containing dry air is sealed, connected to a pressure gauge by a short narrow tube, and placed in a water bath. The water is heated and the pressure of the air is recorded at several temperatures: θ / °C: 10, 30, 50, 70, 90 — p / kPa: 97.8, 104.7, 111.6, 118.5, 125.4. k = 1.38 × 10−23 J K−1.
Figure: the pressure readings plotted against temperature
(a)
Calculate the gradient of the graph of p against θ, in kPa K−1.
(2)
(b)
Use your gradient to determine the temperature, in °C, at which the pressure would be zero.
(2)
(c)
The gradient of the graph is equal to Nk/V. Calculate the number of air molecules in the flask. The percentage uncertainty in the gradient is 2.0% and in the volume of the flask is 1.2%; calculate the percentage uncertainty in your value of N.
(3)
(d)
Suggest why the tube connecting the flask to the gauge should be short and narrow.
(1)
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Answer
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Notes
Part (a)
gradient = (125.4 − 97.8) ÷ (90 − 10)
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= 0.345 kPa K−1 (a change of 1 °C is a change of 1 K)
the air in the tube is not at the temperature of the water bath, so its volume must be negligible compared with the flask
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Model sketch
Figure: the best-fit line extrapolated to zero pressure
Exam tip: A gradient read in kPa per °C is the same number in kPa per K, because the two scales have degrees of the same size. Only the zero is different.
A gas is sealed in a rigid metal can. The can is placed in boiling water and the pressure of the gas increases.
(a)
State two assumptions made in the kinetic model of an ideal gas.
(2)
(b)
Explain, using the kinetic model of a gas, how the gas exerts a pressure on the walls of the can and why this pressure increases when the gas is heated.
(6)
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Answer
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Notes
Part (a)
any two, one mark each: molecules are in random motion / the volume of the molecules is negligible compared with the volume of the container / there are no forces between molecules except during collisions / collisions are elastic / the duration of a collision is negligible compared with the time between collisions
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Part (b)
Indicative content, one mark for each point:
—
molecules move randomly and collide with the walls
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in each collision a molecule's momentum changes (it rebounds), so the wall exerts a force on it
✓ 1
by Newton's third law the molecule exerts an equal and opposite force on the wall; pressure is the total force per unit area of wall
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heating increases the mean kinetic energy, so the molecules move faster / the r.m.s. speed increases
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each collision involves a larger change of momentum
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and collisions with the walls are more frequent, since the volume is constant, so the rate of change of momentum per unit area, and the pressure, increase
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Exam tip: Build the explanation as a chain: collision → momentum change → force (Newton's second and third laws) → pressure = force/area. Then say what heating changes: harder collisions and more frequent collisions.
The kinetic theory of gases models a gas as N molecules, each of mass m, moving at random in a cube of side L.
Figure: one molecule moving at right angles to a face of the cube (not to scale)
(a)
A molecule moves with a velocity component u at right angles to one face of the cube and collides elastically with it. Show that the average force exerted on that face by this molecule is mu2/L.
(3)
(b)
State why ⟨u2⟩ = ⅓⟨c2⟩ for the molecules of the gas.
(1)
(c)
The theory leads to pV = ⅓Nm⟨c2⟩. Use this and pV = NkT to show that the mean kinetic energy of a molecule is proportional to the absolute temperature.
(1)
(d)
A sample of neon has a pressure of 2.0 × 105 Pa and a density of 1.6 kg m−3. Calculate the r.m.s. speed of its atoms.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
change of momentum in each collision = mu − (−mu) = 2mu
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time between collisions with the same face = 2L/u
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force = rate of change of momentum = 2mu ÷ (2L/u) = mu2/L
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Part (b)
the motion is random, so the mean square velocity components in the three perpendicular directions are equal, and c2 = u2 + v2 + w2
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Part (c)
⅓Nm⟨c2⟩ = NkT, so ½m⟨c2⟩ = 3/2kT, and 3/2k is a constant
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Part (d)
p = ⅓ρ⟨c2⟩, so ⟨c2⟩ = 3 × 2.0 × 105 ÷ 1.6 = 3.75 × 105 m2 s−2
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crms = 610 m s−1 (612 m s−1)
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Exam tip: In a “show that” derivation, write each physical step as a separate line with its reason: momentum change, time between collisions, then force as rate of change of momentum.
A student traps a fixed mass of air at 17 °C and measures its pressure p at different volumes V, keeping the temperature constant. The graph shows p against 1/V, with a line of best fit through the origin. P is a point on the line. k = 1.38 × 10−23 J K−1 How many air molecules are trapped?
The graph shows how the mean square speed ⟨c2⟩ of the molecules of a gas varies with the kelvin temperature T. P is a point on the line. k = 1.38 × 10−23 J K−1 1 u = 1.66 × 10−27 kg What is the mass of one molecule of the gas?
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Answer
Mark
Notes
½m⟨c2⟩ = 3/2kT, so m = 3kT/⟨c2⟩ = 3 × 1.38 × 10−23 × 400 ÷ 2.5 × 106 = 6.62 × 10−27 kg. In u: 6.62 × 10−27 ÷ 1.66 × 10−27 = 4.0 u.
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Answer B
Answer: B
Exam tip: Read the axis label carefully: the graph shows ⟨c2⟩ in units of 106 m2 s−2, so the reading at P is 2.5 × 106 m2 s−2 and needs no squaring.
A student investigates how the pressure p of a fixed mass of air depends on its volume V at a constant temperature of 25 °C. She plots lg (p / kPa) against lg (V / cm3), as shown in Figure 1, and draws a line of best fit. k = 1.38 × 10−23 J K−1
(a)
Determine the gradient of the line.
(2)
(b)
Explain how the gradient shows that the results obey Boyle's law.
(1)
(c)
Use the line to show that pV is about 3.2 × 103 kPa cm3.
(2)
(d)
Calculate the number of air molecules trapped.
(2)
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Answer
Mark
Notes
Part (a)
uses two well-separated points on the line, e.g. (1.00, 2.50) and (2.00, 1.50)
A sealed metal can of volume 4.0 × 10−4 m3 contains a gas. The can is heated slowly and the pressure p of the gas is recorded at different kelvin temperatures T. At 400 K a safety valve opens briefly, lets some gas escape and closes again. Figure 1 shows the results. The volume of the can does not change. k = 1.38 × 10−23 J K−1
(a)
Show that, between 280 K and 400 K, the pressure of the gas is proportional to its kelvin temperature.
(2)
(b)
Calculate the number of gas molecules in the can before the valve opens.
(2)
(c)
Calculate the fraction of the molecules that escaped through the valve.
(2)
(d)
Explain, in terms of the molecules, why the pressure fell when the gas escaped, although the temperature did not change.
(1)
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Answer
Mark
Notes
Part (a)
p/T at two points, e.g. 240 ÷ 300 = 0.80 kPa K−1 and 320 ÷ 400 = 0.80 kPa K−1
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p/T is constant (the line extends through the origin), so p ∝ T
the mean kinetic energy (speed) of the molecules is unchanged, but there are fewer molecules in the same volume, so there are fewer collisions with each unit area of wall per second and a smaller rate of change of momentum (force) per unit area
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Exam tip: At constant volume, pV = NkT gives p/T = Nk/V. The gradient of a p–T line through the origin is a direct measure of the number of molecules.
Figure 1 shows a gas in a cylinder of uniform cross-section fitted with a pressure gauge, before and after the piston is pushed in slowly. The temperature of the gas stays at 20 °C throughout and no gas escapes. Treat the gas as ideal.
(a)
Use Figure 1 to determine the pressure of the gas after the compression.
(1)
(b)
Explain, using the kinetic model of a gas, why the pressure of the gas changes by this factor.
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(c)
The piston is now locked in place and the gas is heated to 80 °C. Calculate the new pressure of the gas.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the length of the gas column, and so the volume, halves (30.0 cm → 15.0 cm) at constant temperature, so p = 2 × 110 = 220 kPa
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Part (b)
Indicative content, one mark for each point:
—
the temperature is constant, so the mean kinetic energy and the r.m.s. speed of the molecules are unchanged
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so the average change of momentum in each collision with a wall is unchanged
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the same number of molecules now occupies half the volume, so the number of molecules per unit volume doubles
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so the number of collisions per second with each unit area of wall doubles
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the rate of change of momentum, and so the force, on each unit area of wall doubles
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pressure = force ÷ area, so the pressure doubles: p ∝ 1/V, Boyle's law
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Part (c)
at constant volume p ∝ T: p = 220 × 353 ÷ 293 = 265 kPa
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Exam tip: In a kinetic-theory explanation, say what stays the same (speed, momentum change per collision) as well as what changes (collision rate per unit area). Both are needed to get the factor right.
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