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Question 19N.2.SL.TZ0.2

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Date November 2019 Marks available [Maximum mark: 8] Reference code 19N.2.SL.TZ0.2
Level SL Paper 2 Time zone TZ0
Command term Calculate, Comment, Determine, Show that Question number 2 Adapted from N/A
2.
[Maximum mark: 8]
19N.2.SL.TZ0.2

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.

(a)

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]

Markscheme

N=pVkT OR N=1.0×105×0.361.38×10-23×295 ✔

N=8.8×1024 ✔

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.

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.

(b(i))

Determine the pressure of the air inside the refrigerator.

[2]

Markscheme

use of pT = constant  OR  p=nRTV  OR  NkTV ✔
p=9.4×104« Pa »✔

NOTE: Allow ECF from (a)
Award [2] for bald correct answer

(b(ii))

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]

Markscheme

F=A×∆p ✔

F=0.72×1.0-0.94×105 OR 4.3 × 103 « N »✔

NOTE: Allow ECF from (b)(i)
Allow ECF from MP1

(b(iii))

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

[2]

Markscheme

force is «very» large ✔

there must be a mechanism that makes this force smaller
OR
assumption used to calculate the force/pressure is unrealistic ✔

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