# FORM THREE PHYSICS STUDY NOTES TOPIC TOPIC 5: THERMAL EXPANSION & TOPIC 6: TRANSFER OF THERMAL ENERGY

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**TOPIC 5: THERMAL EXPANSION**

**Thermal Energy**

The Concept of Thermal Expansion of Gases

Explain the concept of thermal expansion of gases

The Relationship between Volume and Temperature of Fixed Mass of Air at Constant Pressure

Investigate the relationship between volume and temperature of fixed mass of air at constant pressure

Three
properties are important when studying the expansion of gases. These
are; pressure, volume and temperature. Charles law states that the
volume of a fixed mass of gas is directly proportional to the absolute
(Kelvin) temperature provided the pressure remains constant.
Mathematically V

_{1}T_{2}= V_{2}T_{1}.
Example 2

the
volume of gas at the start is recorded as 30 cm3with a temperature of
30°C. The cylinder is heated further till the thermometer records 60°C.
What is the volume of gas?

Solution:

We know,V/T = constant

therefore,

V

_{1}/T_{1}=V_{2}/T_{2}
V

_{1}=30 cm^{3}
T1 =30°C = 30+273 = 303K

*(remember to convert from Celsius to Kelvin)*
T2 =60°C = 60+273 = 333K

V2 =?

V1/T1=V2/T2

V

_{2}=V_{1}xT_{2}/T_{1}
V

_{2}=30x333/303
= 32.97 cm

^{3}
The Relationship between Pressure and Volume of a Fixed Mass of Air at Constant Temperature

Investigate the relationship between pressure and volume of a fixed mass of air at constant temperature

The
relationship obtained when the temperature of a gas is held constant
while the volume and pressure are varied is known as Boyle’s law.
Mathematically, P1V1 = P2V2. Boyle's law states that the volume of a
fixed mass of gas is inversely proportional to its pressure if the
temperature is kept constant.

PressurexVolume = constant

pxV = constant

Example 3

The volume of gas at the start is 50 cm

^{3}with a pressure of 1.2 x 10^{5}Pascals. The piston is pushed slowly into the syringe until the pressure on the gauge reads 2.0 x 10^{5}Pascals. What is the volume of gas?**Solution:**

We know

p x V = constant

therefore,

p

_{1}xV_{1}= p_{2}xV_{2}
p

_{1}=1.2 x 10^{5}Pascals
V

_{1}=50 cm^{3}
p

_{2}=2.0 x 10^{5}Pascals
V

_{2}=?
p1xV1= p2xV2

V

_{2}=p_{1}xV_{1}/p_{2}
V

_{2}=1.2x10^{5}x50/2.0 x 10^{5}
V

_{2}= 30 cm^{3}
The Relationship between Pressure and Temperature of a Fixed Mass of Air at Constant Volume

Investigate the relationship between pressure and temperature of a fixed mass of air at constant volume

To
investigate the relationship between the pressure and the temperature
of a fixed mass, the volume of the gas is kept constant. The pressure is
then measured as the temperature is varied. P1/T1 = P2/T2 ,this is
called pressure law. The pressure law states that the pressure of a
fixed mass of a gas is directly proportional to the absolute temperature
if the volume is kept constant

Example 4

Pressure of gas is recorded as 1.0 x 10

^{5}N/m^{2}at a temperature of 0^{°}C. The cylinder is heated further till the thermometer records 150^{°}C. What is the pressure of the gas?
Solution:

We know,p/T = constant

therefore,

p

_{1}/T_{1}= p_{2}/T_{2}
p

_{1}=1.0 x 10^{5}N/m^{2}
T

_{1}=0^{°}C = 0+273 = 273K*(remember to convert from Celsius to Kelvin)*
T

_{2}=150^{°}C = 150+273 = 423K
p2 =?

p1/T1= p2/T2

p

_{2}=p_{1}xT_{2}/T_{1}
p

_{2}=1.0x10^{5}x423/273
= 1.54 x 10

^{5}N/m^{2}
The General Gas Equation from the Gas Laws

Identify the general gas equation from the gas laws

The three gas laws give the following equations:

- pV =
**constant***(when T is kept constant)* - V/T =
**constant***(when p is kept constant)* - P/T=
**constant***(when V is kept constant)*

These 3 equations are combined to give the ideal gas equation:

*Where,*

*p = the pressure of the gas**V = the volume the gas occupies**T = the gas temperature on the Kelvin scale*

From
this equation we know that if a fix mass of gas has starting values of
p1, V1 and T1, and then some time later has value p2, V2 and T2, the
equation can be written as:

Exercise 1

Sabah
pumps up her front bicycle tyre to 1.7 x 105Pa. The volume of air in
the tyre at this pressure is 300 cm3. She takes her bike for a long ride
during which the temperature of the air in the tyre increases from 20°C
to 30°C. Calculate the new front tyre pressure assuming the tyre had no
leaks and so the volume remained constant?

Absolute Scale of Temperature

Explain absolute scale of temperature

Convertion of Temperature in Degrees Centigrade (Celsius) to Kelvin

Convert temperature in degrees centigrade (celsius) to kelvin

The
Kelvin temperature scale takes its name after Lord Kelvin who developed
it in the mid 1800s. It takes absolute zero as the starting point and
temperature measurements are given the symbol K (which stands for
"Kelvin"). Temperature differences on the Kelvin scale are no different
to those on the Celsius (°C) scale. The two scales differ in their
starting points. Thus, 0°C is 273K.

Converting from Celsius to Kelvin

- Temperature in °C + 273 = Temperature in K

Converting from Kelvin to Celsius

- Temperature in K – 273 = Temperature in °C

Example 5

The temperature of a gas is 65 degrees Celsius. Change it to the kelvin scale.

**Solution**

T(K) = degrees Celsius + 273, T(K) = 65+273

therefore T(K) = 338 K.

Standard Temperature and Pressure (S.T.P)

Explain standard temperature and pressure (S.T.P)

Expansion of Gas in Daily Life

Apply expansion of gas in daily life

Land
and sea breezes are the result of expansion of air caused by unequal
heating and cooling of adjacent land and sea surfaces. The piston engine
and firing bullets from guns work under principles of expansion of
gases.

**TOPIC 6: TRANSFER OF THERMAL ENERGY**

**Conduction**

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