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Date May 2019 Marks available 2 Reference code 19M.2.SL.TZ2.2
Level Standard level Paper Paper 2 Time zone 2
Command term Estimate Question number 2 Adapted from N/A

Question

A container of volume 3.2 × 10-6 m3 is filled with helium gas at a pressure of 5.1 × 105 Pa and temperature 320 K. Assume that this sample of helium gas behaves as an ideal gas.

A helium atom has a volume of 4.9 × 10-31 m3.

The molar mass of helium is 4.0 g mol-1. Show that the mass of a helium atom is 6.6 × 10-27 kg.

[1]
a.

Estimate the average speed of the helium atoms in the container.

[2]
b.

Show that the number of helium atoms in the container is about 4 × 1020.

[2]
c.

Calculate the ratio   total volume of helium atoms volume of helium gas .

[1]
di.

Explain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.

[2]
dii.

Markscheme

m = 4.0 × 10 − 3 6.02 × 10 23 «kg»

OR

6.64 × 10−27 «kg» ✔

 

a.

1 2 m v 2 = 3 2 k T / v = 3 k T m / 3 × 1.38 × 10 − 23 × 320 6.6 × 10 − 27  ✔

v = 1.4 × 103 «ms−1» ✔ 

b.

N = p V k T / 5.1 × 10 5 × 3.2 × 10 − 6 1.38 × 10 − 23 × 320

OR

N = p V N A R T / 5.1 × 10 5 × 3.2 × 10 − 6 × 6.02 × 10 23 8.31 × 320   ✔

N = 3.7 × 1020 ✔

c.

« 4 × 10 20 × 4.9 × 10 − 31 3.2 × 10 − 6 = » 6 × 10 − 5  ✔

di.

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

dii.

Examiners report

The mark was awarded for a clear substitution or an answer to at least 3sf. Many gained the mark for a clear substitution with a conversion from g to kg somewhere in their response. Fewer gave the answer to the correct number of sf.

a.

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

b.

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by NA to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by NA.

c.

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

di.

In general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

dii.

Syllabus sections

Core » Topic 3: Thermal physics » 3.2 – Modelling a gas
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