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Date	November 2019	Marks available	2	Reference code	19N.2.SL.TZ0.2
Level	Standard level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Calculate	Question number	2	Adapted from	N/A

Question

The air in a kitchen has pressure 1.0 × 10⁵ Pa and temperature 22°C. A refrigerator of internal volume 0.36 m³ is installed in the kitchen.

The refrigerator door is closed. The air in the refrigerator is cooled to 5.0°C and the number of air molecules in the refrigerator stays the same.

With the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator.

[2]

Determine the pressure of the air inside the refrigerator.

[2]

b(i).

The door of the refrigerator has an area of 0.72 m². Show that the minimum force needed to open the refrigerator door is about 4 kN.

[2]

b(ii).

Comment on the magnitude of the force in (b)(ii).

[2]

b(iii).

Markscheme

$N = \frac{p V}{k T}$ OR $N = \frac{1.0 \times 10^{5} \times 0.36}{1.38 \times 10^{- 23} \times 295}$ ✔

$N = 8.8 \times 10^{24}$ ✔

NOTE: Allow [1 max] for substitution with T in Celsius.
Allow [1 max] for a final answer of n = 14.7 or 15
Award [2] for bald correct answer.

use of $\frac{p}{T}$ = constant OR $p = \frac{n R T}{V}$ OR $\frac{N k T}{V}$ ✔
$p = 9.4 \times 10^{4}$ « Pa »✔