IBDP Physics 3.2 – Modelling a gas: IB style Question Bank SL Paper 2

IB PHYSICS SL (Standard level)- 2024 – Practice Questions- All Topics

Topic 3.2 – Modelling a gas

Topic 3 Weightage : 6 %

All Questions for Topic 3.2 – Pressure , Equation of state for an ideal gas , Kinetic model of an ideal gas . Mole, molar mass and the Avogadro constant , Differences between real and ideal gases

Question

This question is in two parts. Part 1 is about ideal gases and specific heat capacity. Part 2 is about simple harmonic motion and waves.

Part 1 Ideal gases and specific heat capacity

a State two assumptions of the kinetic model of an ideal gas.[2]

b. Argon behaves as an ideal gas for a large range of temperatures and pressures. One mole of argon is confined in a cylinder by a freely moving piston.

(i) Define what is meant by the term one mole of argon.

(ii) The temperature of the argon is 300 K. The piston is fixed and the argon is heated at constant volume such that its internal energy increases by 620 J. The temperature of the argon is now 350 K.

Determine the specific heat capacity of argon in J kg^–1K^–1 under the condition of constant volume. (The molecular weight of argon is 40)[4]

c. At the temperature of 350 K, the piston in (b) is now freed and the argon expands until its temperature reaches 300 K.

Explain, in terms of the molecular model of an ideal gas, why the temperature of argon decreases on expansion.[3]

▶️Answer/Explanation

Markscheme

point molecules / negligible volume;
no forces between molecules except during contact;
motion/distribution is random;
elastic collisions / no energy lost;
obey Newton’s laws of motion;
collision in zero time;
gravity is ignored;

(i) the molecular weight of argon in grams / 6.02×10²³ argon
atoms / same number of particles as in 12 g of C-12;
(allow atoms or molecules for particles)

(ii) mass of gas = 0.040kg ;
specific heat = \(\frac{Q}{{m\Delta T}}\) or 620 = 0.04×c×50;
(i.e. correctly aligns substitution with equation)
\( = \left( {\frac{{620}}{{0.040 \times 50}} = } \right)310{\rm{Jk}}{{\rm{g}}^{{\rm{ – 1}}}}{{\rm{K}}^{{\rm{ – 1}}}}\);
c.

temperature is a measure of the average kinetic energy of the molecules;
(must see “average kinetic” for the mark)

energy/momentum to move piston is provided by energy/momentum of molecules that collide with it;
the (average) kinetic energy of the gas therefore decreases;
Do not allow arguments in terms of loss of speed as a result of collision with a moving piston.

Question

This question is about internal energy and thermal energy (heat).

a Distinguish between internal energy and thermal energy.[3]

b. Describe, with reference to the energy of the molecules, the difference in internal energy of a piece of iron and the internal energy of an ideal gas.[2]

c. A piece of iron is placed in a kiln until it reaches the temperature θ of the kiln. The iron is then quickly transferred to water held in a thermally insulated container. The water is stirred until it reaches a steady temperature. The following data are available.

Thermal capacity of the piece of iron = 60JK^–1Thermal capacity of the water = 2.0×10³JK^–1Initial temperature of the water = 16°C
Final temperature of the water = 45°C

The thermal capacity of the container and insulation is negligible.

(i) State an expression, in terms of θ and the above data, for the energy transfer of the iron in cooling from the temperature of the kiln to the final temperature of the water.

(ii) Calculate the increase in internal energy of the water as the iron cools in the water.

(iii) Use your answers to (c)(i) and (c)(ii) to determine θ. [4]

▶️Answer/Explanation

Markscheme

a internal energy is the total kinetic and potential energy of the molecules of a body;
thermal energy is a (net) amount of energy transferred between two bodies;
at different temperatures;

b. the internal energy of the iron is equal to the total KE plus PE of the molecules; the molecules of an ideal gas have only KE so internal energy is the total KE of the molecules;

c. (i) 60×[θ−45];

(ii) (2.0×10³×29)=5.8×10⁴J;

(iii) 60×[θ−45]=5.8×10⁴;
θ=1000°C; (allow 1010°C to 3 sig fig)

Question

The first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 (\({}_{86}^{226}{\text{Ra}}\)) decays by alpha emission to form a nuclide known as radon (Rn).

a Write down the missing values in the nuclear equation for this decay.

[1]

b. Rutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.

At the start of the experiment all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder

B. The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.[1]

c. Rutherford and Royds expected 2.7 x 10¹⁵ alpha particles to be emitted during the experiment. The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^–5 m³ and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B.[3]

d. Rutherford and Royds identified the helium gas in cylinder B by observing its emission spectrum. Outline, with reference to atomic energy levels, how an emission spectrum is formed.[3]

e. The work was first reported in a peer-reviewed scientific journal. Outline why Rutherford and Royds chose to publish their work in this way.[1]

▶️Answer/Explanation

Markscheme

a 222 AND 4

Both needed.

b. alpha particles highly ionizing

OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule

c. conversion of temperature to 291 K

p = 4.5 x 10^–9 x 8.31 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

p = 2.7 x 10¹⁵ x 1.38 x 10^–23 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ – 5}}}}\)»

0.83 or 0.84 «Pa»

d. electron/atom drops from high energy state/level to low state

energy levels are discrete

wavelength/frequency of photon is related to energy change or quotes E = hf or E = \(\frac{{hc}}{\lambda }\)

and is therefore also discrete

e. peer review guarantees the validity of the work

OR
means that readers have confidence in the validity of work

OWTTE

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