### TOPIC 5: THERMAL EXPANSION | PHYSICS FORM 3

**Thermal Energy**

The Concept of Thermal Expansion of Gases

Explain the concept of thermal expansion of gases

expand when heated just like solids and liquids. This is because the

average kinetic energy of the molecules in a gas is directly

proportional to the absolute temperature of the gas. Heating the gas

increases the kinetic energy of its molecules, making them vibrate more

vigorously and occupy more space.

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

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?

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}=30×333/303= 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.

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}PascalsV

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

_{2}=2.0 x 10^{5}PascalsV

_{2}=?p1xV1= p2xV2

V

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

_{2}=1.2×10^{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

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 = 423Kp2 =?

p1/T1= p2/T2

p

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

_{2}=1.0×10^{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:

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?

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

zero is the lowest temperature that can be attained theoretically. It

is not possible to attain this temperature because all gases liquefy

before attaining it. The kelvin scale of temperature is obtained by

shifting the vertical axis to -273 degrees Celsius and renaming it 0 K.

On the scale 0 degrees Celsius becomes 273 K and 100 degrees Celsius

corresponds with 373 K.

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.

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)

standard temperature and pressure (S.T.P) is a set of conditions for

experimental measurements to enable comparisons to be made between sets

of data. The standard temperature is 0 degrees Celsius (273 K) while the

standard pressure is 1 atmosphere (101300 Pa or 760 mm of mercury).

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.

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.

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