Gases & Absolute Temperature (Cambridge International IGCSE Physics)

Absolute temperature

Spec point — 2.1.5 · Convert temperatures between kelvin and degrees Celsius; recall and use the equation T (in K) = θ (in °C) + 273

What is absolute temperature?

• Temperature measured in kelvin is called absolute temperature

• The kelvin temperature scale begins at absolute zero

  • **0 K is equal to -273 °C **
  • An increase of 1 K is the same change as an increase of 1 °C

• It is not possible to have a temperature** lower** than 0 K

• This means a temperature in kelvin will never have a negative value

• To convert between a temperature θ in degrees Celsius and a temperature T in kelvin, use the following conversion:

$T(\mathrm{in}K)=θ(\mathrm{in}^{\circ}C)+273$

$θ(\mathrm{in}^{\circ}C)=T(\mathrm{in}K)-273$

Conversion chart relating the temperature on the Kelvin and Celsius scales

Worked Example Convert the following values between the Kelvin (absolute) and Celsius scales of temperature.

a) 0 K = _______ °C [1]

b) 0 °C = _______ K [1]

c) 20 °C = _______ K [1]

Answer:

Part (a)

Step 1: Choose whether to add or subtract 273 to the value

• The question is in kelvin therefore subtract 273 to convert to Celsius

Step 2: Do the calculation

• $0-273=-273$

Step 3: Write the answer with units

• 0 K = −273 °C [1 mark]

Part (b)

Step 1: Choose whether to add or subtract 273 to the value

• The question is in Celsius therefore add 273 to convert to kelvin

Step 2: Do the calculation

• $0+273=273$

Step 3: Write the answer with units

• 0 °C = 273 K [1 mark]

Part (c)

Step 1: Choose whether to add or subtract 273 to the value

• The question is in Celsius therefore add 273 to convert to kelvin

Step 2: Do the calculation

• $20+273=293$

Step 3: Write the answer with units

• 20 °C = 293 K [1 mark]

The gas laws

Spec point — 2.1.5 · Describe qualitatively, in terms of particles, the effect on the pressure of a fixed mass of gas of: (a) a change of temperature at constant volume (b) a change of volume at constant temperature

The gas laws

Pressure and volume (constant temperature)

• For a fixed mass of gas, if the temperature of a gas remains constant, the pressure of the gas changes when it is:

  • Compressed – decreases the volume which increases the pressure
  • Expanded – increases the volume which decreases the pressure

At constant temperature, changing the volume changes the pressure

• Similarly, a change in pressure can cause a change in volume

• A vacuum pump can be used to remove the air from a sealed container

• The diagram below shows the change in volume to a tied-up balloon when the pressure of the air around it decreases:

  • The balloon is tied, so there are a fixed number of air particles within it
  • At normal pressure, the air pressure outside the balloon is greater than the air pressure inside the balloon
  • When air is removed from the bell jar, the air pressure outside the balloon is less than the air pressure inside the balloon
  • The space between the air particles in the balloon increases, so the volume of the balloon increases

At constant temperature, changing the pressure changes the volume

• When a gas is compressed, the molecules will hit the walls of the container more frequently

  • Each wall is struck more often, and so the total force on each wall increases, which increases the pressure

Pressure and temperature (constant volume)

• For a fixed mass of gas, increasing temperature increases the pressure of a gas which is kept at a constant volume

• The average speed of molecules increases when the temperature increases (and vice versa)

• As the gas heats up, the molecules will travel at a higher speed

  • They collide with the walls more often and with greater force, increasing the pressure

• Therefore, at a constant volume, an increase in temperature increases the pressure of a gas and vice versa

  • Diagram A shows that molecules in the same volume collide with the walls of the container more frequently, and with greater force, as the temperature increases

  • Diagram B shows that the pressure increases steadily as the temperature rises

    • If the temperature is measured in kelvin, pressure is directly proportional to temperature and the line passes through the origin

At constant volume, an increase in the temperature of the gas increases the pressure due to more collisions on the container walls

Boyle's law

Spec point — 2.1.5 · Recall and use the equation pV = constant for a fixed mass of gas at constant temperature, including a graphical representation of this relationship

Extended Tier Only

• For a fixed mass of gas at constant temperature, Boyle’s law is given by:

$p\propto\frac{1}{V}$

• This means the pressure is inversely proportional to the volume of a gas
• This can also be written as:

$pV=\mathrm{constant}$

• The relationship between the pressure and volume for a fixed mass of gas at constant temperature can also be written as:

$p_{1}V_{1}=p_{2}V_{2}$

• Where:

  • p₁ = initial pressure (Pa)
  • p₂ = final pressure (Pa)
  • V₁* *= initial volume (m³)
  • V₂ = final volume (m³)

• Notice that pressure and volume are measured in Pa and m³ respectively

  • Exam questions often give the pressure and volume in units other than Pa and m³, such as kPa and cm³. You can work in those units, as long as the initial and final units match and you give your answer in the same units

Boyle's law graph: Pressure is inversely proportional to volume

Worked Example A fixed mass of gas has a volume of 35 cm³ at a pressure of 120 kPa. The volume is slowly decreased by 20 cm³ at constant temperature. Calculate the change in pressure.

[3]

Answer:

Step 1: List the known quantities

• Initial pressure, $p_{1}=120\mathrm{kPa}$
• Initial volume, $V_{1}=35\mathrm{cm}^{3}$

Step 2: Write down the equation for Boyle's law

$p_{1}V_{1}=p_{2}V_{2}$

Step 3: Calculate the final volume

$V_{2}=35-20=15\mathrm{cm}^{3}$ [1 mark]

Step 4: Calculate the final pressure using Boyle's law

$p_{2}=\frac{p_{1}V_{1}}{V_{2}}$

$p_{2}=\frac{120\times35}{15}$

$p_{2}=280\mathrm{kPa}$ [1 mark]

Step 5: Calculate the change in pressure

$\Delta p=p_{2}-p_{1}$

$\Delta p=280-120=160\mathrm{kPa}$ [1 mark]

Exam Hint It is an easy mistake to make to think that an inversely proportional graph will be a straight line sloping downwards. After all, a directly proportional graph is a straight line (through the origin) which slopes upwards.

An inversely proportional graph is a curve that tends towards zero — it gets closer and closer to each axis but never touches it, as the graph below shows.