| Date | May 2018 | Marks available | 1 | Reference code | 18M.2.SL.TZ2.2 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate | Question number | 2 | Adapted from | N/A |
Question
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.
The volume of the gas in (a) is increased to 6.8 × 10–4 m3 at constant temperature.
State what is meant by an ideal gas.
Calculate the number of atoms in the gas.
Calculate, in J, the internal energy of the gas.
Calculate, in Pa, the new pressure of the gas.
Explain, in terms of molecular motion, this change in pressure.
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]
N = «\(\frac{{pV}}{{kT}} = \frac{{5.3 \times {{10}^5} \times 2.1 \times {{10}^{ - 4}}}}{{1.38 \times {{10}^{ - 23}} \times 310}}\)» 2.6 × 1022
[1 mark]
«For one atom U = \(\frac{3}{2}\)kT» \(\frac{3}{2}\) × 1.38 × 10–23 × 310 / 6.4 × 10–21 «J»
U = «2.6 × 1022 × \(\frac{3}{2}\) × 1.38 × 10–23 × 310» 170 «J»
Allow ECF from (a)(ii)
Award [2] for a bald correct answer
Allow use of U = \(\frac{3}{2}\)pV
[2 marks]
p2 = «5.3 × 105 × \(\frac{{2.1 \times {{10}^{ - 4}}}}{{6.8 \times {{10}^{ - 4}}}}\)» 1.6 × 105 «Pa»
[1 mark]
«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]
