IB Physics HL B 3 Gas Laws Questions

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

Exam points

Use P=F/A to calculate pressure or force on doors, walls and stoppers, with the force normal to the surface.
Convert gas mass to moles or particles using n=m/M and N=nN_A, including isotope samples.
Apply pV/T=constant to compare fixed-mass gas states and interpret pressure–volume or temperature graphs.
Use PV=nRT or PV=NkBT to calculate gas states, masses or particle numbers from data.
Explain gas pressure through molecular momentum transfer at walls; use P=⅓ρv² and Newton’s third law.
Calculate monatomic-gas internal energy with U=3/2NkBT=3/2nRT and connect it to total kinetic energy.
Judge ideal-gas limits from low pressure, high temperature and small-volume cases.

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.

(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 ]

$N=\frac{pV}{kT}=\frac{5.3\times10^{5}\times2.1\times10^{-4}}{1.38\times10^{-23}\times310}=2.6\times10^{22}$

(iii)

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

[ 2 ]

For one atom $U=\frac{3}{2}kT$ $U=2.6\times10^{22}\times\frac{3}{2}\times1.38\times10^{-23}\times310=170\ \mathrm{J}$

(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 ]

$p_2=5.3\times10^{5}\times\frac{2.1\times10^{-4}}{6.8\times10^{-4}}=1.6\times10^{5}\ \mathrm{Pa}$

(ii)

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

[ 2 ]

[Maximum number: 1]

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

State what is meant by an ideal gas.

[Maximum number: 2]

The equipment shown in the diagram was used by a student to investigate the variation with volume, of the pressure p of air, at constant temperature. The air was trapped in a tube of constant cross-sectional area above a column of oil.

The pump forces oil to move up the tube decreasing the volume of the trapped air.

Outline how the results of this experiment are consistent with the ideal gas law at constant temperature.

$V\propto H$ , thus the ideal gas law gives $p\propto\frac{1}{H}$ .

So the graph should be a straight line through the origin, as observed.

[Maximum number: 2]

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

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

[Maximum number: 2]

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

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

For one atom $U=\frac{3}{2}kT$ $U=2.6\times10^{22}\times\frac{3}{2}\times1.38\times10^{-23}\times310=170\ \mathrm{J}$

[Maximum number: 2]

The pump forces oil to move up the tube decreasing the volume of the trapped air.

The equation of the line of best fit is:

p=a+\frac{b}{H}

The equation above may be used to predict the pressure of the air at extremely large values of $\frac{1}{H}$ . Suggest why this will be an unreliable estimate of the pressure.

very large $\frac{1}{H}$ means very small volumes / very high pressures at very small volumes the ideal gas does not apply

OR at very small volumes some of the assumptions of the kinetic theory of gases do not hold

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question 2

Question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8