Question 19M.2.SL.TZ1.4b.ii
| Date | May 2019 | Marks available | [Maximum mark: 4] | Reference code | 19M.2.SL.TZ1.4b.ii |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Determine | Question number | b.ii | Adapted from | N/A |
A tube of constant circular cross-section, sealed at one end, contains an ideal gas trapped by a cylinder of mercury of length 0.035 m. The whole arrangement is in the Earth’s atmosphere. The density of mercury is 1.36 × 104 kg m–3.
When the mercury is above the gas column the length of the gas column is 0.190 m.
The tube is slowly rotated until the gas column is above the mercury.
The length of the gas column is now 0.208 m. The temperature of the trapped gas does not change during the process.
Determine the atmospheric pressure. Give a suitable unit for your answer.
[4]
recognition that is constant OR 190po + 190pm = 208po − 208pm
OR po = pm ✔
pressure due to mercury pm = 0.035 × 1.36 × 104 × 9.81(= 4.67 × 103 Pa) ✔
1.03 × 105 ✔
Pa OR Nm-2 OR kgm-1s-2 ✔
Do not award for a bald correct answer. Working must be shown to award MP3.
Award MP4 for any correct unit of pressure (eg “mm of mercury / Hg”).
This question was left blank by many candidates. Of those who attempted a solution, few appreciated the difference made to the derived equation from 4bi when the tube was rotated. It should be noted that this was also the “unit question” on this exam, and a candidate could have been awarded a mark for clearly writing any correct unit of pressure without doing any calculations. Candidates should be reminded to keep an eye out for this opportunity and to at least write a unit even if the question seems unapproachable.
