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Question 18M.2.HL.TZ2.2

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Date May 2018 Marks available [Maximum mark: 7] Reference code 18M.2.HL.TZ2.2
Level HL Paper 2 Time zone TZ2
Command term Calculate, Explain, State Question number 2 Adapted from N/A
2.
[Maximum mark: 7]
18M.2.HL.TZ2.2

An ideal monatomic gas is kept in a container of volume 2.1 × 10–4 m3, temperature 310 K and pressure 5.3 × 105 Pa.

(a.i)

State what is meant by an ideal gas.

[1]

Markscheme

a gas in which there are no intermolecular forces

OR

a gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures

OR

molecules have zero PE/only KE

 

Accept atoms/particles.

[1 mark]

(a.ii)

Calculate the number of atoms in the gas.

[1]

Markscheme

N = « p V k T = 5.3 × 10 5 × 2.1 × 10 − 4 1.38 × 10 − 23 × 310 » 2.6 × 1022

[1 mark]

(a.iii)

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

[2]

Markscheme

«For one atom U =  3 2 kT»  3 2  × 1.38 × 10–23 × 310 / 6.4 × 10–21 «J»

U = «2.6 × 1022 ×  3 2  × 1.38 × 10–23 × 310» 170 «J»

 

 Allow ECF from (a)(ii)

Award [2] for a bald correct answer

Allow use of U = 3 2 pV

[2 marks]

The volume of the gas in (a) is increased to 6.8 × 10–4 m3 at constant temperature.

(b.i)

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

[1]

Markscheme

p2 = «5.3 × 105 ×  2.1 × 10 − 4 6.8 × 10 − 4 » 1.6 × 105 «Pa»

[1 mark]

(b.ii)

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

[2]

Markscheme

«volume has increased and» average velocity/KE remains unchanged

«so» molecules collide with the walls less frequently/longer time between collisions with the walls

«hence» rate of change of momentum at wall has decreased

«and so pressure has decreased»

 

The idea of average must be included

Decrease in number of collisions is not sufficient for MP2. Time must be included.

Accept atoms/particles.

[2 marks]

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