Question 19M.2.SL.TZ1.b.i
| Date | May 2019 | Marks available | [Maximum mark: 2] | Reference code | 19M.2.SL.TZ1.b.i |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Show that | Question number | b.i | 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.
Show that (po + pm) × 0.190 = where
po = atmospheric pressure
pm = pressure due to the mercury column
T = temperature of the trapped gas
n = number of moles of the trapped gas
A = cross-sectional area of the tube.
[2]
use of PV = nRT and V = Area × (0.190) seen ✔
substitution of P = po + pm «re-arrangement to give answer»✔
This question was generally well answered. Most candidates took the time to show the set up and substitutions they used to derive the given expression. A small number of candidates attempted to “show that” by making unit substitutions - this is not acceptable for a question like this.
