IB Physics HL B.3 Gas Laws Question Bank

Practise IB Physics HL B.3 by combining ideal-gas calculations with molecular momentum, internal-energy and model-limit analysis.

Exam points

Solve multi-state ideal-gas problems using PV=nRT, absolute temperature and significant figures.
Link pressure to molecular momentum transfer and relate temperature to mean kinetic energy.
Evaluate ideal-gas assumptions against density, pressure and temperature conditions in data.

Covered syllabus points

B.3.1—Pressure B.3.2—Amount of substance B.3.3—Ideal gas model B.3.4—Empirical gas laws B.3.5—Ideal gas equations B.3.6—Molecular pressure model B.3.7—Ideal gas internal energy B.3.8—Ideal gas approximation limits

[Maximum number: 5]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

Figure for Question B.3 Gas laws question 1 — IB Physics HL

(a)

When fully inflated, an unloaded airbag has a diameter of 1.80 m and a length of 24.0 m . At a temperature of $15^{\circ} \mathrm{C}$ , an airbag can hold 4200 mol of gas.

[ 5 ]

(i)

Show that the pressure in an airbag is about 0.2 MPa .

When the boat is placed on the airbags, the airbags are compressed so that the effective contact area is as shown in the diagram.

In this arrangement, the maximum safe pressure before an airbag bursts is four times the value calculated in (a)(i). The boat is supported by airbags on a slope that is at an angle of $4.0^{\circ}$ to the horizontal. There are fifteen airbags supporting the boat at all times.

[ 2 ]

$\begin{aligned} V=\pi r^{2} h=\pi 0.9^{2} \times 24=61.07 \mathrm{~m}^{3} P=\frac{(4200)(8.31)(273+15)}{61.07}=0.16 \mathrm{MPa} \end{aligned}$

Allow MP1 to be seen in MP2.

(ii)

Estimate the maximum safe mass that this arrangement can hold.

[ 3 ]

$\begin{aligned} \mathrm{F}=4 P A=4\left(1.6 \times 10^{5}\right)(15 \times 24 \times 1.8)=4.14 \times 10^{8} \mathrm{~N} \mathrm{F}=m g \cos \theta=\mathrm{m}(9.8)(\cos 4)=9.78 \mathrm{~m} \mathrm{~N} m=\frac{4 P A}{g \cos \theta}=\frac{4.14 \times 10^{8}}{9.78}=4.3 \times 10^{7} \mathrm{~kg} \end{aligned}$

Do not award MP2 if the cos θterm is omitted.

Allow ECF for MP3.

[Maximum number: 2]

Magnesium-27 nuclei $\left({ }_{12}^{27} \mathrm{Mg}\right)$ decay by beta-minus $\left(\beta^{-}\right)$ decay to form nuclei of aluminium-27 (Al).

Small amounts of magnesium in a material can be detected by firing neutrons at magnesium-26 nuclei. This process is known as irradiation.

Magnesium-27 is formed because of irradiation. The products of the beta-particle emission are observed as the magnesium-27 decays to aluminium-27.

The smallest mass of magnesium that can be detected with this technique is $1.1 \times 10^{-8} \mathrm{~kg}$ .

Show that the smallest number of magnesium atoms that can be detected with this technique is about $10^{17}$ .

So $1.1 \times 10^{-8} \mathrm{~kg} \equiv \frac{1.1}{0.027} \times 10^{-8}$ «mol»

OR

Mass of atom $=27 \times 1.66 \times 10^{-27}$ «kg»

2.4-2.5 × $10^{17}$ atoms

[Maximum number: 7]

(a)

An ideal monatomic gas is kept in a container of volume $2.1 \times 10^{-4} \mathrm{~m}^{3}$ , temperature 310 K and pressure $5.3 \times 10^{5} \mathrm{~Pa}$ .

[ 4 ]

(i)

State what is meant by an ideal gas.

[ 1 ]

(ii)

Calculate the number of atoms in the gas.

[ 1 ]

$N=\ll \frac{p V}{k T}=\frac{5.3 \times 10^{5} \times 2.1 \times 10^{-4}}{1.38 \times 10^{-23} \times 310}-2.6 \times 10^{22}$

(iii)

Calculate, in J , the internal energy of the gas.

[ 2 ]

«For one atom $U=\frac{3}{2} k T$ » $\underset{2}{\stackrel{3}{\text { 豊 }} \cdot 1^{2}} 1.38^{\cdot} 10^{-23} \cdot 310 / 6.4^{\cdot} 10^{-21}$ «J»
$U=\ll 2.6 \times 10^{22} \times \frac{3}{2} \times 1.38 \times 10^{-23} \times 310 \gg 170<\mathrm{J} \gg q$

Marking guidance:

Allow ECF from (a)(ii) Award [2] for a bald correct answer

Allow use of $U=\frac{3}{2} p V$

(b)

The volume of the gas in (a) is increased to $6.8 \times 10^{-4} \mathrm{~m}^{3}$ at constant temperature.

[ 3 ]

(i)

Calculate, in Pa , the new pressure of the gas.

[ 1 ]

$p_{2}=\ll 5.3 \times 10^{5} \times \frac{2.1 \times 10^{-4}}{6.8 \times 10^{-4}}>1.6 \times 10^{5}<\mathrm{Pa} \gg$

(ii)

Explain, in terms of molecular motion, this change in pressure.

[ 2 ]

All question bank results loaded

Covered syllabus points

B.3 Gas laws question 1

Question (a)

Question (i)

Question (ii)

B.3 Gas laws question 2

B.3 Gas laws question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (i)

Question (ii)