Required Practical: Investigating Gas Laws (AQA A Level Physics)

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Required Practical: Investigating Gas Laws

Investigating Boyle's Law

  • The overall aim of this experiment is to investigate Boyle's Law

    • This is the effect of pressure on volume at a constant temperature

  • This is just one example of how this required practical might be tackled

Variables

  • Independent variable = Mass, m (kg)

  • Dependent variable = Volume, V (m3)

  • Control variables:

    • Temperature

    • Cross-sectional area of the syringe

Equipment List

Boyles Law Equipment Table, downloadable AS & A Level Physics revision notes
  • Resolution of measuring equipment:

    • Pressure gauge = 0.02 × 105 Pa

    • Volume = 0.2 cm3

    • Vernier Caliper = 0.02 mm

Method

Boyles Law Apparatus, downloadable AS & A Level Physics revision notes

Apparatus setup for Boyle’s Law

  1. Before setting up the apparatus as shown in the diagram the inside diameter, d of the syringe needs to be measured using a vernier calliper after removing the plunger. Remember to take at least 3 repeat readings and find an average

  2. Determine the lowest volume of air visible by pushing the syringe upwards to remove as much air as possible

  3. At this lowest volume, the rubber tubing should be fit over the nozzle and clamped with a pinch clip as close to the nozzle as possible (this is to stop air escaping)

  4. Set up the apparatus as shown in the diagram and ensure the temperature of the room remains constant throughout the experiment

  5. Record the volume shown on the syringe before adding the masses and the mass holder

  6. Add the 100 g mass holder with a 100 g mass on it to the loop of string at the bottom of the plunger. Wait a few seconds before reading volume to ensure the temperature is kept constant (since work is done against the plunger when the volume increases)

  7. Record the value of the new volume from the syringe scale

  8. Repeat the experiment by adding two 100 g masses at a time up to 8-10 readings. This is so a significant change in volume can be seen each time

  9. Record the mass and volume

  • An example table of results might look like this:

Boyles Example Table of Results 1, downloadable AS & A Level Physics revision notes
Boyles Example Table for Diameter, downloadable AS & A Level Physics revision notes

Analysing the Results

  • Boyle’s Law can be represented by the equation:

pV = constant

  • This means the pressure must be calculated from the experiment

  • The exerted pressure of the masses is calculated by:

Pressure Equation
  • Where:

    • F = weight of the masses, mg (N)

    • A = cross-sectional area of the syringe (m2)

  • The cross-sectional area is found from the equation for the area of a circle:

Cross-sectional Area Equation
  • To calculate the pressure of the gas:

    Pressure of the gas = Atmospheric pressure – Exerted pressure from the masses

  • Where:

    • Atmospheric pressure = 101 kPa

  • The table of results may need to be modified to fit these extra calculations. Here is an example of how this might look:

Boyles Example Table of Results 2, downloadable AS & A Level Physics revision notes
  • Once these values are calculated:

  1. Plot a graph of against 1 / V and draw a line of best fit

  2. If this plot is a straight line graph, this means that the pressure is proportional to the inverse of the volume, hence confirming Boyle's Law (pV = constant)

Boyles Law Example Graph, downloadable AS & A Level Physics revision notes

Evaluating the Experiment

Systematic Errors:

  • There may be friction in the syringe which causes a systematic error

    • Use a syringe that has very little friction or lubricate it, so the only force applied is from the masses pulling the syringe downward

Random Errors:

  • The reading of the volume should be taken a few seconds after the mass has been added to the holder

    • Otherwise, a reading will be taken when the temperature is not constant

  • Due to the nature of this experiment setup, it is prone to many random errors

    • Take repeat readings to reduce their effect

Safety Considerations

  • A counterweight or G-clamp must be used to avoid the stand toppling over and causing injury, especially if the surface is not completely flat

Investigating Charles's Law

  • The overall aim of this experiment is to investigate Charles’s law, which is the effect of temperature on volume at constant pressure

  • This is just one example of how this required practical might be tackled

Variables

  • Independent variable = Temperature, T (°C)

  • Dependent variable = Height of the gas, h (cm)

  • Control variables:

    • Pressure

Equipment List

Charles Law Equipment Table, downloadable AS & A Level Physics revision notes
  • Resolution of measuring equipment:

    • 30 cm ruler = 1 mm

    • 2 litre beaker = 50 ml

Method

Charles Law Apparatus, downloadable AS & A Level Physics revision notes

Apparatus setup for Charles’s Law

  1. The capillary tube should have one open end at the top and a closed end at the bottom. This is to keep the pressure constant at atmospheric pressure. Assume the temperature of the water is the same as the temperature of the gas in the tube

  2. Set up the apparatus as shown in the diagram. Add a drop of sulfuric acid halfway up the tube (the gas below this drop is being studied) to ensure no gas escapes

  3. Boil some water in a kettle and pour it into the beaker for the full 2 litres. Make sure the waterline is higher than the drop of sulfuric acid, therefore surrounding all the gas, and stir well

  4. Allow the temperature to drop down to 95 °C, then read the height of the gas (this is up to the bottom of the sulfuric acid) using the ruler

  5. Record the height of the gas as the temperature decreases in increments of 5 °C. Make sure you have at least 8 readings

  • An example table of results might look like:

Charles Law Example Table, downloadable AS & A Level Physics revision notes

Analysing the Results

  1. Plot a graph of the height of the gas in the capillary tube in cm and the temperature of the water in °C

  2. Draw a line of best fit

  3. Calculate the gradient

    Charles Law Example Graph, downloadable AS & A Level Physics revision notes
  4. If this is a straight-line graph, then this means the temperature is proportional to the height. Since the height is proportional to the volume (V = πr2h) then this means Charles’s law is confirmed, and the temperature is proportional to the volume too

  • To find a value of absolute zero T0, the equation of the graph can be written as

h = mT + c

  • Comparing this to the equation of a straight line: y = mx + c

    • y = h

    • x = T

    • gradient = m

    • y-intercept = c

  • c cannot be determined directly, as temperatures this low would cause the water to freeze

    • The straight line equation can be used for two sets of values to determine absolute zero, T0

  • Write the equation with c as the subject at T0 and any temperature T1 and height h1 from the experimental data:

c = h0 − mT0

c = h1 − mT1

h0 - mT0 = h1 - mT1

  • At absolute zero, h0 = 0

- mT0 = h1 - mT1

T subscript 0 space equals space T subscript 1 space minus space h subscript 1 over m

  • Picking any co-ordinate of h and T from the line of best fit, and substituting into the equation will give a value of absolute zero

  • If this value is close to the accepted value of –273°C then the experiment shows that volume and temperature are directly proportional at constant pressure

Evaluating the Experiment

Systematic Errors:

  • Make sure the capillary tube is secured close to the ruler and properly aligned to get an accurate value for the height of the gas

    • Otherwise, the reading taken will be slightly out each time

Random Errors:

  • Take temperature and height readings at eye level to avoid a parallax error

  • Stir the water well so it and the gas are the same temperature throughout the beaker

Safety Considerations

  • Do not spill boiling water onto your skin or electrical equipment 

  • Protect the workbench from the boiling water by using a heat proof mat 

Worked Example

A student investigates the relationship between the temperature and volume of a column of air. They obtain the following results:

Worked Example Table Question (1), downloadable AS & A Level Physics revision notes
Worked Example Table Question (2), downloadable AS & A Level Physics revision notes

Calculate the value of absolute zero from these results and its relative percentage error with the accepted value of –273.15 °C

Answer:

Step 1: Plot a graph of temperature T against volume V

Worked Example Step 1 Graph, downloadable AS & A Level Physics revision notes
  • Make sure:

    • The axes are properly labelled with values, quantities and units

    • The line of best fit is drawn with a ruler so there are equal numbers of points above and below

Step 2: Calculate the gradient of the graph

Worked Example Step 2 Gradient of Graph, downloadable AS & A Level Physics revision notes
  • The gradient is calculated by:

gradient equals fraction numerator 11.6 minus 9.6 over denominator 90 minus 30 end fraction equals 0.033 space cm space divided by degree straight C

Step 3: Calculate the value of absolute zero

  • Write the line equation of the graph

h = mT + c

  • Rearrange to make c (the y intercept) the subject and equate values at absolute zero (T0) and a set of values from the data (T1, h1)

c = h0 − mT0

c = h1 − mT1

h0 - mT0 = h1 - mT1

  • At absolute zero, a gas has no volume so h0 = 0 m3

T subscript 0 space equals space T subscript 1 space minus space h subscript 1 over m

  • Where T0 is absolute zero and (T1, h1) is any co-ordinate on the line of best fit

  • Using the coordinates (60, 10.6) and gradient calculated (0.033)

straight T subscript 0 equals 60 minus fraction numerator 10.6 over denominator 0.033 end fraction equals negative 2.61.2121 equals negative 261 space degree straight C

Step 4: Calculate its relative percentage error with the accepted value of –273.15 °C

Relative space percentage space error equals fraction numerator Measured space value minus Accepted space Value over denominator Accepted space Value end fraction cross times 100 percent sign
Relative space percentage space error equals fraction numerator negative 261.2121 minus open parentheses negative 273.15 close parentheses over denominator negative 273.15 end fraction cross times 100 percent sign equals negative 4.37 percent sign

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.