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IB Physics B.3: Gas laws worked solutions

Worked solutions to every practice question in B.3: Gas laws of Photon Physics: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

Type A1 · n, N, R and kB / P1A[1 mark]

A student writes three statements about a sample of ideal gas containing N molecules.

I.; The amount in moles is N divided by the Avogadro constant.

II.; R equals N_A multiplied by k_B.

III.; Each molecule carries, on average, kinetic energy k_BT.

How many of the statements are correct?

A.; none B.; one C.; two D.; three

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I is the definition of amount of substance. II follows from comparing PV=nRT with PV=Nk_BT: nR=Nk_B, so R=(N/n)k_B=N_Ak_B. III is missing the factor: the mean translational kinetic energy is \tfrac32k_BT. So two are correct; D is the answer for students who forget the \tfrac32.

Final answer

Question A2

Type A2 · Mass and particle count / P1A[1 mark]

Four samples are prepared: 4.0\,\mathrm{g} of helium (He), 16\,\mathrm{g} of oxygen gas (O_2), 20\,\mathrm{g} of neon (Ne) and 3.0\,\mathrm{g} of hydrogen gas (H_2). Molar masses: He 4.0, O_2 32, Ne 20, H_2 2.0\,\mathrm{g\,mol^{-1}}. Which two samples contain the same number of atoms but different numbers of molecules?

A.; helium and neon B.; helium and oxygen

C.; neon and hydrogen D.; oxygen and hydrogen

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Amounts: He 1.0\,\mathrm{mol} of atoms; O_2 0.50\,\mathrm{mol} of molecules, so 1.0\,\mathrm{mol} of atoms; Ne 1.0\,\mathrm{mol}; H_2 1.5\,\mathrm{mol} of molecules, 3.0\,\mathrm{mol} of atoms. He and O_2 both hold 1.0\,\mathrm{mol} of atoms, but 1.0 against 0.50\,\mathrm{mol} of molecules. A has equal atoms and equal molecules (both gases are monatomic).

Final answer

Question A3

Type A3 · Model assumptions / P1A[1 mark]

A student lists four statements about the molecules of an ideal gas. Which one is not part of the kinetic model?

A.; Together the molecules fill only a tiny fraction of the container.

B.; Molecules push or pull on one another only while they collide.

C.; Every molecule has the same speed.

D.; No kinetic energy is lost when molecules hit each other or the walls.

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The model has molecules moving randomly with a spread of speeds; that spread is why \overline{v^2} and the rms speed appear in P=\tfrac13\rho\overline{v^2}. A, B and D are all assumptions.

Final answer

Question A4

Type A4 · When a real gas is ideal / P1A[1 mark]

A sample of nitrogen, which boils at 77\,\mathrm{K}, can be held at any of the states below. In which state does it come closest to obeying PV=nRT?

A.; 2\times10^{7}\,\mathrm{Pa}, 120\,\mathrm{K} B.; 2\times10^{7}\,\mathrm{Pa}, 600\,\mathrm{K}

C.; 1\times10^{4}\,\mathrm{Pa}, 600\,\mathrm{K} D.; 1\times10^{4}\,\mathrm{Pa}, 120\,\mathrm{K}

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Low pressure means low density: the molecules are far apart, so their own volume is a tiny fraction of the gas volume and they rarely come close enough to attract. High temperature makes the molecules fast, so the weak attractions hardly change their motion. A, high pressure close to the boiling point, is the worst choice; B and D each fix only one of the two conditions.

Final answer

Question B1

Type B1 · Two states, fixed amount / P1A[1 mark]

A fixed amount of gas occupies 450\,\mathrm{cm^3} at 87\,\mathrm{^\circ C}. It is cooled at constant pressure to 27\,\mathrm{^\circ C}. What is its new volume?

A.; 140\,\mathrm{cm^3} B.; 375\,\mathrm{cm^3} C.; 450\,\mathrm{cm^3} D.; 540\,\mathrm{cm^3}

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At constant pressure V/T is constant, with T in kelvin: V=450\times300/360=375\,\mathrm{cm^3}. A uses the Celsius ratio 27/87; D inverts the ratio; C ignores the change.

Final answer

Question E3

Type E3 · What U means / P1A[1 mark]

What does the internal energy of a real gas consist of?

A.; random kinetic energy of the molecules, nothing else

B.; random kinetic energy of the molecules together with the potential energy stored in the forces between them

C.; kinetic energy of the container moving as a whole

D.; potential energy between the molecules, nothing else

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Internal energy is the sum of the random kinetic energies and the intermolecular potential energies. An ideal gas has no intermolecular forces, so its potential energy term is zero and U is all kinetic; a real gas keeps both terms (A describes the ideal case). Kinetic energy of the container as a whole is not internal energy (C).

Final answer

Question D4

Type D4 · Pressure and density / P1A[1 mark]

Container X holds helium and container Y holds argon. The gases have the same density and the same temperature. An argon atom is 10 times as massive as a helium atom. Find \dfrac{\text{pressure in X}}{\text{pressure in Y}}.

A.; 0.1 B.; 1 C.; \sqrt{10} D.; 10

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P=Nk_BT/V and \rho=Nm/V give P=\rho k_BT/m. With \rho and T equal, P\propto1/m, so helium's pressure is 10 times larger: at equal density there are ten times as many helium atoms per cubic metre. B assumes equal density means equal pressure; A puts the heavier atoms in front.

Final answer

Question E1

Type E1 · KE and speed comparisons / P1A[1 mark]

Helium and argon are at the same temperature. An argon atom is 10 times as massive as a helium atom. Which option is correct?

Options give (mean kinetic energy of He)/(mean kinetic energy of Ar); (rms speed of He)/(rms speed of Ar).

A.; 1;; \sqrt{10} B.; 1;; 10 C.; 10;; \sqrt{10} D.; 0.1;; 1

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Mean kinetic energy depends only on temperature, \tfrac32k_BT, so the ratio is 1. Then \tfrac12m\overline{v^2} is equal, so v_{\text{rms}}\propto1/\sqrt m: helium is \sqrt{10}=3.2 times faster. B forgets the square root; C and D give heavier atoms more energy or less.

Final answer

Question E2

Type E2 · KE and speed comparisons / P1A[1 mark]

An ideal gas is heated from 27\,\mathrm{^\circ C} to 327\,\mathrm{^\circ C}. By what factor does the rms speed of its molecules increase?

A.; 1.41 B.; 2.00 C.; 3.48 D.; 12.1

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v_{\text{rms}}=\sqrt{3k_BT/m}\propto\sqrt T. The kelvin temperature doubles, 300\,\mathrm{K}\to600\,\mathrm{K}, so the speed rises by \sqrt2=1.41. B is the energy factor; C and D use the Celsius ratio 327/27.

Final answer

Question D1

Type D1 · Isothermal change, molecules / P1A[1 mark]

A fixed amount of ideal gas expands slowly at constant temperature. What happens to the mean speed of the molecules and to the number of collisions per second on each square metre of the wall?

A.; mean speed unchanged; collision rate decreases

B.; mean speed decreases; collision rate decreases

C.; mean speed unchanged; collision rate unchanged

D.; mean speed decreases; collision rate unchanged

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Constant temperature fixes the mean kinetic energy, so the speeds do not change. The same molecules now spread through a larger volume and reach each square metre of wall less often, which is why the pressure falls. B is the classic slip: an isothermal expansion does not slow the molecules.

Final answer

Tutorial practice

Question B3

Type B3 · Two containers: ratios / P1A[1 mark]

Container X holds 2.0\,\mathrm{L} of an ideal gas at 3.0\times10^{5}\,\mathrm{Pa} and 27\,\mathrm{^\circ C}. Container Y holds 6.0\,\mathrm{L} of an ideal gas at 1.0\times10^{5}\,\mathrm{Pa} and 127\,\mathrm{^\circ C}. What is \dfrac{\text{number of molecules in X}}{\text{number of molecules in Y}}?

A.; 0.75 B.; 1.0 C.; 1.3 D.; 4.7

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N=PV/(k_BT) for each, so N_X/N_Y=\dfrac{(3.0)(2.0)/300}{(1.0)(6.0)/400}=1.33. D uses Celsius temperatures, A is the ratio upside down, B ignores the temperatures.

Final answer

Question B4

Type B4 · Gas added or removed / P1A[1 mark]

A rigid cylinder holds gas at 8.0\times10^{5}\,\mathrm{Pa} and 290\,\mathrm{K}. One quarter of the gas is let out, and the gas left is then warmed to 319\,\mathrm{K}. What is the final pressure?

A.; 2.2\times10^{5}\,\mathrm{Pa} B.; 6.0\times10^{5}\,\mathrm{Pa} C.; 6.6\times10^{5}\,\mathrm{Pa} D.; 8.8\times10^{5}\,\mathrm{Pa}

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At fixed volume P\propto NT. Three quarters of the molecules remain: P=8.0\times10^{5}\times0.75\times(319/290)=6.6\times10^{5}\,\mathrm{Pa}. A keeps the quarter that left; B ignores the warming; D ignores the loss.

Final answer

Question B2

Type B2 · One state: find P, V, T or n; Force from pressure / P2[5 marks]

A horizontal cylinder is closed by a frictionless piston of area 2.5\times10^{-3}\,\mathrm{m^2}. It contains 0.040\,\mathrm{mol} of nitrogen at 20\,\mathrm{^\circ C}, occupying 1.2\times10^{-3}\,\mathrm{m^3}. The air outside is at 1.0\times10^{5}\,\mathrm{Pa}.

(a) Calculate the pressure of the nitrogen. [2]

(b) Calculate the force that the nitrogen exerts on the piston. [1]

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(a) P=nRT/V=0.040(8.31)(293)/1.2\times10^{-3} (M1) =8.12\times10^{4}\,\mathrm{Pa} (A1); the temperature must be 293\,\mathrm{K}. (b) F=PA=8.12\times10^{4}\times2.5\times10^{-3}=203\,\mathrm{N}, outwards (A1). (c) The air pushes inwards with 1.0\times10^{5}\times2.5\times10^{-3}=250\,\mathrm{N} (M1). The forces differ by 250-203=47\,\mathrm{N}, so a 47\,\mathrm{N} pull outwards is needed (A1). Using the full 250\,\mathrm{N} as the answer is the usual slip: only the pressure difference is unbalanced.

Final answer

(a) 8.12\times10^{4}\,\mathrm{Pa}. (b) 203\,\mathrm{N}. (c) 47\,\mathrm{N}, pulling the piston outwards.

Question C1

Type C1 · Gradient of a gas graph / P1A[1 mark]

The volume V of n moles of an ideal gas at constant pressure is plotted against its Celsius temperature \theta. The graph is a straight line of gradient s. What is the pressure of the gas?

A.; \dfrac{nR}{s} B.; \dfrac{nR}{273\,s} C.; \dfrac{s}{nR} D.; nRs

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V=(nR/P)T and T=\theta+273, so V=(nR/P)\theta+273nR/P. A one-degree step is a one-kelvin step: the gradient is still nR/P, and P=nR/s. Moving to the Celsius axis shifts the intercept, not the slope (B). C inverts the gradient.

Final answer

Question E4

Type E4 · Total U of two gases / P1A[1 mark]

Equal masses of helium (molar mass 4.0\,\mathrm{g\,mol^{-1}}) and neon (molar mass 20\,\mathrm{g\,mol^{-1}}) are both monatomic ideal gases. The helium is at 300\,\mathrm{K} and the neon at 600\,\mathrm{K}. The gases are then placed in thermal contact. Which row is correct?

Options give U_{\text{He}}/U_{\text{Ne}}; direction of thermal energy flow.

A.; 2.5; helium to neon B.; 2.5; neon to helium

C.; 0.5; neon to helium D.; 0.5; helium to neon

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Equal masses: N\propto1/M, so helium has 5 times as many atoms. U=\tfrac32Nk_BT gives U_{\text{He}}/U_{\text{Ne}}=5\times(300/600)=2.5. Energy flows from the higher temperature, neon, even though helium has more internal energy. A follows the larger U; C and D treat equal masses as equal numbers of atoms.

Final answer

Question C3

Type C3 · U–T graph / P1A[1 mark]

For a sample of a monatomic ideal gas, a plot of internal energy U against kelvin temperature T passes through the origin and has a slope of 25\,\mathrm{J\,K^{-1}}. How many moles of gas are there?

A.; 2.0\,\mathrm{mol} B.; 3.0\,\mathrm{mol} C.; 4.5\,\mathrm{mol} D.; 1.2\times10^{24}\,\mathrm{mol}

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U=\tfrac32nRT, so the gradient is \tfrac32nR and n=2(25)/(3\times8.31)=2.0\,\mathrm{mol}. B drops the \tfrac32, C uses \tfrac32 upside down, D uses k_B and so gives the number of atoms, not moles.

Final answer

Question C4

Type C4 · U from P –(1/V ) slope / P1A[1 mark]

Neon is squeezed slowly, its temperature held fixed. Plotting pressure P against 1/V gives a line through the origin whose slope is 360\,\mathrm{Pa\,m^3}. What is the internal energy of the neon?

A.; 4.2\times10^{-3}\,\mathrm{J} B.; 240\,\mathrm{J} C.; 360\,\mathrm{J} D.; 540\,\mathrm{J}

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P=nRT(1/V), so the gradient is nRT=PV, in joules. For a monatomic gas U=\tfrac32nRT=1.5\times360=540\,\mathrm{J}. C forgets the \tfrac32, B inverts it, A treats the gradient as 1/(nRT).

Final answer

Question D5

Type D5 · Pressure from mean KE / P1A[1 mark]

A 2.0\times10^{-3}\,\mathrm{m^3} flask holds 4.8\times10^{22} gas molecules with a mean translational kinetic energy of 6.2\times10^{-21}\,\mathrm{J}. What is the pressure of the gas?

A.; 5.0\times10^{4}\,\mathrm{Pa} B.; 9.9\times10^{4}\,\mathrm{Pa} C.; 1.5\times10^{5}\,\mathrm{Pa} D.; 2.2\times10^{5}\,\mathrm{Pa}

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PV=Nk_BT and \overline{E}_k=\tfrac32k_BT, so k_BT=\tfrac23\overline{E}_k and P=\dfrac{2N\overline{E}_k}{3V}=\dfrac{2(4.8\times10^{22})(6.2\times10^{-21})}{3(2.0\times10^{-3})}=9.9\times10^{4}\,\mathrm{Pa}. D uses \tfrac32 instead of \tfrac23, C drops the factor, A uses \tfrac13. Check: \,\mathrm{J\,m^{-3}}=\,\mathrm{N\,m^{-2}}.

Final answer

Question D2

Type D2 · Explain pressure (molecules); Two states, fixed amount / P2[4 marks]

A rigid aerosol can is left in a hot car. The gas in it warms from 20\,\mathrm{^\circ C} to 60\,\mathrm{^\circ C}.

(a) Explain, using the kinetic model and Newton's laws, why the pressure of the gas increases. [3]

(b) The pressure was 3.0\times10^{5}\,\mathrm{Pa} at 20\,\mathrm{^\circ C}. Calculate the pressure at 60\,\mathrm{^\circ C}. [1]

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(a) Higher temperature means a higher mean kinetic energy, so the molecules move faster (M1). Each collision reverses a larger momentum, so the wall exerts a larger force on the molecule and, by Newton's third law, the molecule exerts a larger force on the wall (A1). The volume is fixed and the molecules are faster, so they hit each square metre of wall more often; force per unit area, the pressure, rises (A1). Stating only “more collisions'' loses the momentum mark. (b) P=3.0\times10^{5}\times333/293=3.41\times10^{5}\,\mathrm{Pa} (A1).

Final answer

(a) Faster molecules: larger momentum change per collision and more collisions per second, so a larger force per unit area. (b) 3.41\times10^{5}\,\mathrm{Pa}.

Question D3

Type D3 · Fast compression heats; Isothermal change, molecules / P2[4 marks]

Air in a bicycle pump is compressed quickly by pushing the piston in, and the barrel becomes warm.

(a) Use the motion of the molecules to account for the rise in temperature. [2]

(b) The same compression is now done very slowly. State what happens to the temperature of the air and explain why its pressure still rises. [2]

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(a) The piston moves towards the molecules, so each molecule rebounds from it with a larger speed than it arrived with: the piston does work on the gas (M1). The mean kinetic energy of the molecules rises, and temperature is a measure of mean kinetic energy (A1). (b) Slowly, energy has time to leave through the barrel walls, so the temperature stays at room temperature (A1). The molecules keep the same mean speed but the same number now occupy a smaller volume, so they hit each square metre of wall more often and the pressure rises (A1).

Final answer

(a) Molecules rebound faster from the moving piston; their mean kinetic energy, and so the temperature, rises. (b) Temperature stays the same; the molecules hit the walls more often in the smaller volume.

Advanced practice

Question A5

Type A5 · Particle-volume test; One state: find P, V, T or n / P2[5 marks]

A 1.0\,\mathrm{L} flask holds argon at 1.0\times10^{5}\,\mathrm{Pa} and 300\,\mathrm{K}. Treat each argon atom as a sphere of radius 1.9\times10^{-10}\,\mathrm{m}.

(a) Show that the flask contains about 2.4\times10^{22} atoms. [2]

(b) Estimate the fraction of the flask's volume taken up by the atoms themselves. [2]

(c) The argon is compressed at constant temperature to 2.0\times10^{7}\,\mathrm{Pa}. Use your answer to (b) to explain why the ideal-gas model is now less reliable. [1]

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(a) N=PV/(k_BT)=1.0\times10^{5}(1.0\times10^{-3})/(1.38\times10^{-23}\times300) (M1) =2.42\times10^{22} (A1). (b) One atom: \tfrac43\pi(1.9\times10^{-10})^3=2.87\times10^{-29}\,\mathrm{m^3} (M1); all atoms 6.94\times10^{-7}\,\mathrm{m^3}, fraction 6.94\times10^{-7}/1.0\times10^{-3}=6.9\times10^{-4} (A1). (c) At 200 times the pressure the gas volume is 1/200 as large, so the fraction is 0.139, about 14\%: the assumption of negligible molecular volume fails (A1). A small fraction supports only the volume assumption; it says nothing about the attractions.

Final answer

(a) 2.42\times10^{22}. (b) 6.9\times10^{-4}. (c) The atoms now fill about 14\% of the volume, which is no longer negligible.

Question B5

Type B5 · Open the tap / P2[5 marks]

Flask P (3.0\,\mathrm{L}) holds helium at 2.4\times10^{5}\,\mathrm{Pa}. Flask Q (1.0\,\mathrm{L}) holds helium at 0.80\times10^{5}\,\mathrm{Pa}. They are joined by a short tube with a closed tap, and both are at 300\,\mathrm{K}.

(a) Calculate the total amount of helium. [2]

(b) The tap is opened and the temperature stays at 300\,\mathrm{K}. Determine the final pressure. [2]

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(a) n_P=PV/RT=2.4\times10^{5}(3.0\times10^{-3})/(8.31\times300)=0.289\,\mathrm{mol}, n_Q=0.0321\,\mathrm{mol} (M1); total 0.321\,\mathrm{mol} (A1). (b) Opening the tap keeps the amount and spreads it over 4.0\,\mathrm{L} (M1): P=nRT/V=0.321(8.31)(300)/4.0\times10^{-3}=2.0\times10^{5}\,\mathrm{Pa} (A1). Equivalently (P_PV_P+P_QV_Q)/(V_P+V_Q). It must lie between the two starting pressures; the plain average 1.6\times10^{5}\,\mathrm{Pa} ignores the volumes. (c) Fixed amount and volume: P=2.0\times10^{5}\times360/300=2.4\times10^{5}\,\mathrm{Pa} (A1).

Final answer

(a) 0.321\,\mathrm{mol}. (b) 2.0\times10^{5}\,\mathrm{Pa}. (c) 2.4\times10^{5}\,\mathrm{Pa}.

Question B6

Type B6 · Molar mass from density / P2; PREDICTED TYPE[3 marks]

A sample of an unknown gas has density 1.99\,\mathrm{kg\,m^{-3}} at 1.20\times10^{5}\,\mathrm{Pa} and 320\,\mathrm{K}.

(a) Starting from PV=nRT, show that M=\rho RT/P. [1]

(b) Calculate M and identify the gas from: helium 4, nitrogen 28, argon 40, carbon dioxide 44\,\mathrm{g\,mol^{-1}}. [2]

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(a) n=m_{\text{gas}}/M in PV=nRT gives PV=(m_{\text{gas}}/M)RT; divide by V and use \rho=m_{\text{gas}}/V: P=\rho RT/M, so M=\rho RT/P (A1). (b) M=1.99(8.31)(320)/1.20\times10^{5} (M1) =0.0441\,\mathrm{kg\,mol^{-1}}=44.1\,\mathrm{g\,mol^{-1}}, carbon dioxide (A1). Leave M in \,\mathrm{kg\,mol^{-1}} until the comparison.

Final answer

(a) Proof. (b) 0.0441\,\mathrm{kg\,mol^{-1}}: carbon dioxide.

Question C2

Type C2 · Test a P –V isotherm; Gradient of a gas graph / P2[6 marks]

A student traps air in a syringe joined by a short tube to a pressure sensor and pushes the plunger in slowly at 290\,\mathrm{K}. She reads the volume V from the syringe scale:

\begin{array}{lcccc}V/\mathrm{cm^3} & 48.0 & 38.0 & 23.0 & 18.0\\ P/\mathrm{kPa} & 100 & 125 & 200 & 250\end{array}

(a) Use the first and last readings to show that PV is not constant. [2]

(b) She plots V against 1/P. Determine the gradient and the intercept on the V axis. [2]

(d) Calculate the amount of air trapped. [1]

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(a) 100\times48.0=4800 and 250\times18.0=4500\,\mathrm{kPa\,cm^3} (M1): they differ by about 6\%, more than reading error, so PV read this way is not constant (A1). (b) Points (0.0100, 48.0) and (0.0040, 18.0): gradient =30.0/0.0060=5.0\times10^{3}\,\mathrm{kPa\,cm^3} (A1); intercept =48.0-5000(0.0100)=-2.0\,\mathrm{cm^3} (A1). (c) The true volume is V+2.0\,\mathrm{cm^3}: the air in the tube is always there but never read. With it, P(V+2.0)=5000\,\mathrm{kPa\,cm^3} for every row, so Boyle's law holds (A1). (d) PV=5000\,\mathrm{kPa\,cm^3}=5.0\,\mathrm{J}, n=5.0/(8.31\times290)=2.07\times10^{-3}\,\mathrm{mol} (A1).

Final answer

(a) 4800 and 4500\,\mathrm{kPa\,cm^3}. (b) gradient 5.0\times10^{3}\,\mathrm{kPa\,cm^3} (5.0\,\mathrm{J}); intercept -2.0\,\mathrm{cm^3}. (c) 2.0\,\mathrm{cm^3} of air in the tube and sensor is not on the scale: a systematic error. (d) 2.07\times10^{-3}\,\mathrm{mol}.

Question D6

Type D6 · Pressure from density and rms / P2; PREDICTED TYPE[5 marks]

Nitrogen in a sealed rigid tank has density 1.40\,\mathrm{kg\,m^{-3}} and rms molecular speed 520\,\mathrm{m\,s^{-1}}. The mass of a nitrogen molecule is 4.65\times10^{-26}\,\mathrm{kg}.

(a) Calculate the pressure of the nitrogen. [2]

(b) Determine the temperature of the nitrogen. [2]

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(a) P=\tfrac13\rho\overline{v^2}=\tfrac13(1.40)(520)^2 (M1) =1.26\times10^{5}\,\mathrm{Pa} (A1). Forgetting to square the speed gives 243\,\mathrm{Pa}, an absurd value. (b) \tfrac12m\overline{v^2}=\tfrac32k_BT, so T=m\overline{v^2}/(3k_B)=4.65\times10^{-26}(520)^2/(3\times1.38\times10^{-23}) (M1) =304\,\mathrm{K} (A1). (c) The tank is rigid and sealed, so \rho is fixed and P\propto\overline{v^2}: factor 1.10^2=1.21 (A1). The temperature also rises by 21\%.

Final answer

(a) 1.26\times10^{5}\,\mathrm{Pa}. (b) 304\,\mathrm{K}. (c) 1.21.

Question E5

Type E5 · Calculate U; One state: find P, V, T or n / P2[6 marks]

A rigid 0.025\,\mathrm{m^3} tank holds argon, a monatomic gas, at 1.6\times10^{5}\,\mathrm{Pa} and 280\,\mathrm{K}.

(a) Calculate the number of argon atoms in the tank. [2]

(b) Calculate the internal energy of the argon. [2]

(c) The tank stands in sunshine and the argon warms to 315\,\mathrm{K}. Calculate the change in internal energy and state where this energy came from. [2]

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(a) N=PV/(k_BT)=1.6\times10^{5}(0.025)/(1.38\times10^{-23}\times280) (M1) =1.04\times10^{24} (A1). (b) U=\tfrac32Nk_BT=\tfrac32PV (M1) =1.5\times1.6\times10^{5}\times0.025=6.0\times10^{3}\,\mathrm{J} (A1). Using \tfrac32k_BT alone gives the energy of one atom, not the gas. (c) U\propto T at fixed N: \Delta U=6000\times35/280=750\,\mathrm{J} (A1). The tank is rigid, so no work is done on the gas: all of it arrives as thermal energy from the hotter tank walls, warmed by the sunlight (A1).

Final answer

(a) 1.04\times10^{24}. (b) 6.0\times10^{3}\,\mathrm{J}. (c) 750\,\mathrm{J}, transferred thermally from the surroundings (no work is done: the volume is fixed).

Connected questions

Question X1

Type X1 · B.3 + A.2 / Connected / Paper 2[9 marks]

A hot-air balloon has an envelope of volume 2800\,\mathrm{m^3}, open at the bottom. The outside air is at 10\,\mathrm{^\circ C} and 1.0\times10^{5}\,\mathrm{Pa}; the molar mass of air is 0.029\,\mathrm{kg\,mol^{-1}}. Ignore the volume of the basket and fabric.

(a) Show that the density of the outside air is about 1.23\,\mathrm{kg\,m^{-3}}. [2]

(b) Calculate the buoyant force on the envelope. [1]

(c) The air inside is heated to 100\,\mathrm{^\circ C} at the outside pressure, and some escapes through the opening. Calculate the mass of air now inside. [2]

(d) The fabric, basket and passengers have a total mass of 650\,\mathrm{kg}. Determine the initial acceleration of the balloon when it is released. [3]

(e) Explain, in terms of molecules, why fewer molecules per cubic metre are needed inside to give the same pressure. [1]

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(a) \rho=PM/(RT)=1.0\times10^{5}(0.029)/(8.31\times283) (M1) =1.23\,\mathrm{kg\,m^{-3}} (A1). (b) A.2: F_b=\rho Vg=1.233(2800)(9.8)=3.38\times10^{4}\,\mathrm{N} (A1). (c) Inside, \rho=1.0\times10^{5}(0.029)/(8.31\times373)=0.936\,\mathrm{kg\,m^{-3}} (M1); mass =0.936\times2800=2.62\times10^{3}\,\mathrm{kg} (A1). (d) Weight of everything =(2620+650)(9.8)=3.20\times10^{4}\,\mathrm{N} (M1); resultant =3.38\times10^{4}-3.20\times10^{4}=1.8\times10^{3}\,\mathrm{N} upwards (M1); a=1.79\times10^{3}/3270=0.55\,\mathrm{m\,s^{-2}} (A1). The hot air inside is part of the accelerated mass. (e) Hotter molecules are faster, so each one gives a larger momentum change per collision and collides more often; fewer of them per cubic metre give the same pressure (A1).

Final answer

(a) 1.23\,\mathrm{kg\,m^{-3}}. (b) 3.38\times10^{4}\,\mathrm{N}. (c) 2.62\times10^{3}\,\mathrm{kg}. (d) 0.55\,\mathrm{m\,s^{-2}} upwards. (e) Faster molecules transfer more momentum per collision and hit more often.

Question X2

Type X2 · B.3 + B.1 + B.5 / Connected / Paper 2[8 marks]

A rigid 5.0\,\mathrm{L} flask of helium, a monatomic gas of molar mass 4.0\times10^{-3}\,\mathrm{kg\,mol^{-1}}, is at 1.0\times10^{5}\,\mathrm{Pa} and 290\,\mathrm{K}. A small heater inside the flask, connected to a 12\,\mathrm{V} supply, carries 0.50\,\mathrm{A} for 20\,\mathrm{s}. Assume all the electrical energy goes to the helium.

(a) Calculate the amount of helium. [2]

(b) Show that the internal energy of the helium is 750\,\mathrm{J} before heating. [1]

(d) Calculate the specific heat capacity of helium at constant volume. [2]

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(a) n=PV/(RT)=1.0\times10^{5}(5.0\times10^{-3})/(8.31\times290) (M1) =0.207\,\mathrm{mol} (A1). (b) U=\tfrac32nRT=\tfrac32PV=1.5(1.0\times10^{5})(5.0\times10^{-3})=750\,\mathrm{J} (A1). (c) B.5: energy supplied =VIt=12(0.50)(20)=120\,\mathrm{J} (M1). \Delta T=2\Delta U/(3nR)=46.4\,\mathrm{K}, so T=336\,\mathrm{K} (A1). Fixed volume: P=1.0\times10^{5}\times336.4/290=1.16\times10^{5}\,\mathrm{Pa} (A1). (d) B.1: mass =nM=8.30\times10^{-4}\,\mathrm{kg} (M1); c=Q/(m\Delta T)=120/(8.30\times10^{-4}\times46.4)=3.1\times10^{3}\,\mathrm{J\,kg^{-1}\,K^{-1}} (A1). The same value follows from c=3R/(2M), with no measurement at all.

Final answer

(a) 0.207\,\mathrm{mol}. (b) 750\,\mathrm{J}. (c) 336\,\mathrm{K}; 1.16\times10^{5}\,\mathrm{Pa}. (d) 3.1\times10^{3}\,\mathrm{J\,kg^{-1}\,K^{-1}}.

Question X3

Type X3 · B.3 + D.1 / Connected / Paper 2[7 marks]

The Moon has mass 7.35\times10^{22}\,\mathrm{kg} and radius 1.74\times10^{6}\,\mathrm{m}. Its sunlit surface reaches 390\,\mathrm{K}. The mass of a nitrogen molecule is 4.65\times10^{-26}\,\mathrm{kg}.

(a) Calculate the rms speed of nitrogen molecules at 390\,\mathrm{K}. [2]

(b) Find g on the Moon's surface. [1]

(c-SL) Instead of (c): find the temperature at which the rms speed of nitrogen is 600\,\mathrm{m\,s^{-1}}. [2]

(d) The escape speed is 2.37\times10^3\,\mathrm{m\,s^{-1}}. The rms speed is well below this, yet the Moon has lost any nitrogen atmosphere it had. Suggest why. [2]

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(a) v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3(1.38\times10^{-23})(390)/4.65\times10^{-26}} (M1) =589\,\mathrm{m\,s^{-1}} (A1). (b) D.1: g=GM/r^2=6.67\times10^{-11}(7.35\times10^{22})/(1.74\times10^{6})^2=1.62\,\mathrm{N\,kg^{-1}} (A1). (c) v_{\text{esc}}=\sqrt{2GM/r} (M1) =2.37\times10^{3}\,\mathrm{m\,s^{-1}}, about 4.0 times v_{\text{rms}} (A1). (d) The rms speed is only a typical value: collisions keep giving a small fraction of molecules speeds several times larger, above the escape speed, and those at the top of the atmosphere leave (M1). Collisions refill the fast end of the spread, so over billions of years the gas drains away; lighter molecules, faster at the same temperature, go first (A1). Alternative (c-SL): T=mv_{\rm rms}^2/(3k_B) (M1) =4.65\times10^{-26}(600)^2/(3\times1.38\times10^{-23})=404\,\mathrm{K} (A1).

Final answer

(a) 589\,\mathrm{m\,s^{-1}}. (b) 1.62\,\mathrm{N\,kg^{-1}}. (c) 2.37\times10^{3}\,\mathrm{m\,s^{-1}}. (d) Molecules have a spread of speeds; the fastest few exceed the escape speed and the rest keep redistributing energy over billions of years. Alternative (c-SL): 404\,\mathrm{K}.

Question X4

Type X4 · B.3 + A.2 + A.3 / Connected / Paper 2[9 marks]

A vertical cylinder of cross-sectional area 4.0\times10^{-3}\,\mathrm{m^2} is closed at the top by a frictionless piston of mass 8.0\,\mathrm{kg}. Air outside is at 1.0\times10^{5}\,\mathrm{Pa}. The cylinder holds helium in a column 0.15\,\mathrm{m} high at 290\,\mathrm{K}.

(a) Draw a free-body diagram of the piston. [1]

(b) Show that the pressure of the helium is about 1.2\times10^{5}\,\mathrm{Pa}. [2]

(d) Calculate the work done by the helium on the piston. [2]

(e) Calculate the change in internal energy of the helium, and explain why the heater must supply more energy than this. [2]

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(a) Three forces: weight F_g down, push of outside air P_{\text{atm}}A down, push of helium PA up (A1). (b) A.2 equilibrium: PA=P_{\text{atm}}A+mg, so P=1.0\times10^{5}+8.0(9.8)/4.0\times10^{-3} (M1) =1.0\times10^{5}+1.96\times10^{4}=1.196\times10^{5}\,\mathrm{Pa} (A1). (c) The piston's balance fixes P, so V/T is constant: height =0.15\times348/290=0.180\,\mathrm{m} (M1); rise 0.030\,\mathrm{m} (A1). (d) A.3: W=Fs=PA\,\Delta h=1.196\times10^{5}(4.0\times10^{-3})(0.030) (M1) =14.4\,\mathrm{J} (A1). (e) \Delta U=\tfrac32nR\Delta T=\tfrac32P\Delta V=1.5(1.196\times10^{5})(4.0\times10^{-3})(0.030)=21.5\,\mathrm{J} (A1). Energy is conserved: the heater must supply the rise in internal energy and the work done, 21.5+14.4\approx36\,\mathrm{J} (A1).

Final answer

(a) F_g down, force of outside air down, force of helium up. (b) 1.196\times10^{5}\,\mathrm{Pa}. (c) 0.030\,\mathrm{m}. (d) 14.4\,\mathrm{J}. (e) 21.5\,\mathrm{J}; the helium also does 14.4\,\mathrm{J} of work lifting the piston and pushing back the air, so about 36\,\mathrm{J} is needed.

From the book

Photon Physics: Study Guide for the IB Diploma (SL & HL)

Gas laws from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book