Home / iGCSE Physics (0625) 2.2.1 Thermal expansion of solids, liquids and gases Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 2.2.1 Thermal expansion of solids, liquids and gases Paper 3 -Exam Style Questions- New Syllabus

Question

A tight-fitting lid keeps air inside a metal can. An airtight rubber bung holds a liquid-in-glass thermometer that is inserted through a hole in the lid, as shown in Fig. 4.1.
(a)(i) State what happens to the liquid in the thermometer when the air temperature rises.
(a)(ii) The temperature of the air in the can is 18°C. Calculate the temperature of the air in kelvin.
(b) The can is placed in a refrigerator. The temperature of the air inside the can decreases. State and explain what happens to the pressure exerted by the air in the can. Use your ideas about gas particles.
(c) The air in another can exerts a pressure of 102,000 N/m² on the lid. The area of the can lid is 0.0082 m². Calculate the force on the lid due to the air in the can.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 2.2.1 — Thermal expansion of solids, liquids and gases (Part (a)(i))

• Topic 2.1.3 — Gases and the absolute scale of temperature (Parts (a)(ii), (b))

• Topic 1.8 — Pressure (Part (c))

▶️ Answer/Explanation

(a)(i)
For the correct answer: The liquid rises/expands.
When the temperature increases, the liquid particles gain kinetic energy and move more vigorously. This leads to an increase in the average separation between the particles, causing the liquid to expand and rise up the narrow bore of the thermometer tube.

(a)(ii)
For the correct answer: $291\text{ K}$
To convert a temperature from degrees Celsius to the absolute Kelvin scale, add $273$ to the Celsius value. Following the formula $T\text{ (in K)} = \theta\text{ (in °C)} + 273$, the calculation is $18 + 273 = 291\text{ K}$.

(b)
For the correct answer: Pressure decreases.
As the temperature drops, gas particles lose kinetic energy and move more slowly. This results in less frequent and less forceful collisions between the particles and the internal walls of the metal can, thereby reducing the outward pressure exerted by the air.

(c)
For the correct answer: $836.4\text{ N}$
Pressure is defined as the force acting per unit area, expressed by the equation $p = F / A$. By rearranging this to solve for force ($F = p \times A$), we multiply the pressure of $102,000\text{ N/m}^2$ by the area of $0.0082\text{ m}^2$ to find the total force of $836.4\text{ N}$.

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