Part 2 More advanced topics on the ideal gas laws, calculations, kinetic particle model–theory (sections 4a to 4b)

The basic particle theory and properties of gases, liquids and solids, state changes & solutions are described on GCSE/IGCSE notes on particle models of gases–liquids–solids, describing and explaining their properties and advanced students should be familiar with ALL of its contents before studying this page ....

More advanced sections on another page: The ideal gas equation PV=nRT * Dalton's Law of partial pressures Graham's Law of diffusion * The deviations of a gases from ideal behaviour and their causes * The Van der Waals equation of state * Compressibility factors * The Critical Point – The Critical Temperature and Critical Pressure

For other calculations see the Calculations Index page including Mole definition and Avogadro Constant

AND molar gas volume and reacting gas volume ratios.

The Maxwell Boltzmann distribution of particle kinetic energies is discussed in the KINETICS pages.

4.(a) Ideal gas behaviour and the gas laws and the Kelvin scale of temperature

Introduction – the kinetic particle model of an ideal gas

• The (advanced) kinetic theory of gases is founded on the following six fundamental postulates:

  1. Gases are composed of minute discrete particles (usually molecules).

  2. The particles are in continuous chaotic motion moving in straight lines between very frequent collisions with each other and the sides of the container (approximately 10⁹/s).

  3. The bombardment of the container walls by the particles causes the phenomenon we call pressure (i.e. force of impacts/unit area). The greater the force of collision and the more frequent the collisions the greater the gas pressure exerted on the container surface.

  4. The collisions are perfectly elastic i.e. no energy loss on collision due to friction.

  5. At relatively low pressures the average distance between particles is large compared to the diameter of the particles and therefore the inter–molecular forces between the particles is negligible.

  6. The average kinetic energy of the particles is directly proportional to their absolute temperature on the Kelvin scale (K) i.e. KE(J) T(K)

     • This means if you heat up a gas the average kinetic energy of the particles increases, therefore the average speed increases too.

     • The Kelvin scale of temperature is explained below.

• When a gas behaves according to this model, the gas laws described in sections 4a to 4e are obeyed.

• However in real gases things are not so simple and this non–ideal behaviour is discussed in section 5.

The Kelvin Scale of Temperature

In the past lots of measurements have been made to investigate how (i) the pressure and volume of a given mass of vary at constant temperature and how (ii) the pressure and volume gas of a fixed mass of gas varies with temperature (see the two graphs left and right). This resulted in the formulation of the laws of gases described in the next section 4a. along with how to use them in calculations and problem solving.

However, before this, if you look at these two graphs of gas behaviour when changing pressure or volume with temperature, one thing becomes clear, when the graph lines are extrapolated back to the x axis they give a value of –273^oC. This gave rise to the idea that there was a minimum possible temperature of –273 ^oC and further experimentation has confirmed this time and time again. At –273^oC all substances are solid and in terms of the kinetic particle theory of matter, at –273 ^oC the particles have virtually no motion i.e. ~no vibration of the atoms in the solid.

Therefore as well as the established Celsius scale (centigrade scale), a new temperature scale was proposed in which the lowest value was 0 (known as absolute zero) rather than –273. This is called the Kelvin scale of temperature or the absolute temperature scale, denoted by the unit K. Incidentally you don't say degrees Kelvin like you say degrees Celsius, you just say Kelvin. The Kelvin temperature scale was also designed so that a 1 K temperature change or interval, exactly equals 1^oC Celsius change or interval. Therefore you can easily convert between the two temperature scale by a simple calculation

K = ^oC + 273   and  ^oC = K – 273   and it is the temperature in K that you must use in gas law calculations (4b.)

Some examples are worked out below and a practice in reading a Celsius thermometer, which is what you use in the school or college laboratory!

–7oC –7 + 273 = 267 K

36oC 36 + 273 = 309 K

77oC 77 + 273 = 350 K

101.5oC 101.5 + 273.0 = 374.5 K

132oC 132 + 273 = 405 K

206oC 206 + 273 = 479 K

It seems a bit weird to say you body has a temperature of 310, which is why it is always important to state the units too!

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4.(b) The particle model of a gas - and gas pressure-volume calculations using Boyle's Law

• All particles have mass and their movement gives them kinetic energy and momentum.

• The particles in a gas are in constant random motion - random direction, variety of velocities and kinetic energies.

• When the fast moving gas particles collide with a surface, their millions of impacts create a force that we measure as gas pressure - the total force of impacts per unit area.

• The particles collide with the container surface completely at random and impact at every angle, BUT, the effect is to create a net force at right angles to the surface - gas pressure!

• The greater the number of collisions per unit area of surface, the greater the pressure, assuming the gas volume and temperature are kept constant.

• If the temperature is kept constant and the volume increased, the impacts are more spread out and less frequent per unit area, so the gas pressure decreases.

  • Conversely, if a gas is compressed into a smaller volume at constant temperature, the number of impacts per unit area increases, so the pressure increases.

  • From measurements of volumes and pressure of gases at constant pressure, a numerical inverse law can be formulated - Boyle's law.

  • pressure x volume = a constant (at constant temperature)

  • pV = constant

  • p = pressure in pascals (Pa), V = volume (m³)

  • You can connect two pressure and two volumes by the simple equation

  • p₁ x V₁ = p₂ x V₂

  • where 1 represent the original conditions, and 2 the final situation if an enforced change of p₁ or V₁ is made.

  • Examples of simple gas calculations

    • (i) 5 m³ volume of a gas at a pressure 101 300 Pa was compressed to a volume of 2.8 m³.

      • Calculate the final pressure.

      • p₁ x V₁ = p₂ x V₂

      • rearranging gives p₂ = (p₁ x V₁) / V₂

      • p₂ = (101 300 x 5) / 2.8 = 180893 Pa

    • (ii) 10m³ of gas at a pressure of 100 000 Pa was compressed to a pressure of 300 000 Pa.

      • Calculate the final volume of the gas.

      • p₁ x V₁ = p₂ x V₂

      • rearranging gives V₂ = (p₁ x V₁) / p₂

      • V₂ = (100 000 x 10) / 300 000 = 3.33 m³

More on gas pressure and volume calculations

Boyle's Law for volume and gas pressure

• The particle theory of gas pressure was explained in Part 1 so this section concentrates on the gas law calculations involving pressure and volume.

• Boyle's Law states that for given mass of gas at a constant temperature (^oC or K), the product of the pressure multiplied by the volume is a constant.

• p x V = constant

• Therefore, for initial values of p₁ and V₁, which change to final values of p₂ and V₂, the following equation applies ...

• p₁ x V₁ = p₂ x V₂ (for fixed amount of gas at constant temperature)

• or p₂ = p₁ x V₁/V₂ or V₂ = p₁ x V₁/p₂

• The graph shows how the pressure and volume vary according to Boyles Law at two different temperatures.

• At lower temperatures the volume and pressure values are lower (see next section).

• You can use any volume or pressure units you like as long as both p's and both V's have the same units.

• Using particle theory and simple arithmetical values to explain Boyles Law.

  • If a gas is compressed to half its original volume the concentration or density of the gas is doubled. Therefore there will be twice as many collisions with the surface causing twice the impact effect i.e. double the pressure.

  • If the volume of a gas is increased by a factor of three, the concentration is reduced by the same factor, so the chance of particle collision with the container walls is similarly reduced, so the pressure decreases by a factor of three.

• Gases e.g. oxygen for hospitals, can be stored under high pressure enabling reasonably efficient storage. Because the internal pressure in the cylinder is so much greater than the external pressure, on fitting a valve, a large volume of gas can be released to flow slowly under controlled conditions for a patients respiration.

• Examples of Boyle's Law calculations (constant temperature assumed)

• Ex. Q4a 1

  • 240cm³ of air at a pressure of 100kPa in a bicycle pump is compressed to a volume of 150cm³.

  • What is the pressure of the compressed air in the pump?

  • p₁ x V₁ = p₂ x V₂ , rearranging to scale up for the new higher pressure

  • p₂ = p₁ x V₁/V₂  = 100 x 240/150 = 160 kPa

• Ex. Q4a 2

  • 10 m³ of butane gas at 1.2 atm was required to be stored at 6 atm pressure. To what volume must the gas be compressed to give the required storage pressure?

  • p₁ x V₁ = p₂ x V₂ , rearranging to scale down for the new lower volume

  • V₂ = p₁ x V₁/p₂ = 1.2 x 10/6 = 2.0 m³

• Ex. Q4a 3

  • A 100 cm³ gas syringe containing 80 cm³ of gas that was compressed to 60 cm³. If atmospheric pressure is 101325 Pa, and the temperature remains constant, what is the pressure of the gas in the syringe after compression.

  • p x V = constant

  • p₁ x V₁ = p₂ x V₂

  • p₂ = p₁ x V₁/V₂

  • p₂ = 101325 x 80/60 = 135100 Pa

• Ex. Q4a 4

  • In hospital the gas pressure in a 100 dm³ cylinder of oxygen is 5.52 atm (5 x atmospheric pressure). What volume of gas can be released slowly to a patient on releasing it to  an atmospheric pressure 1.01?

  • p x V = constant

  • V₂ = p₁ x V₁/p₂

  • V₂ = p₁ x V₁/p₂ = 5.52 x 100/1.01 = 546.5 dm³

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4.(c) Charles's Law/Gay–Lussac's Law for pressure/volume and temperature

• The particle theory of gas pressure was explained in Part 1 so this section concentrates on the gas law calculations involving pressure and volume and their variation with temperature.

• Charles's/Gay–Lussac's Law states that for a fixed mass of gas ...

  • (i) the volume of a gas is directly proportional to the absolute temperature (K) at constant pressure

    • V = constant x T (right graph), or

    • V/T = constant, or

    • V₁/V₂ = T₁/T₂  for conditions changing from 1 (initial) to 2 (final),

    • or V₁/T₁ = V₂/T₂  for constant pressure

    • V₁ x T₂ = V₂ x T₁

    • V₂ = V₁ x T₂/T₁

    • or T₂ = T₁ x V₂/V₁

    • Kinetic particle reasoning - increasing the temperature increases the kinetic energy of the molecules giving more forceful collisions which 'push out' to expand the gas at constant pressure (major factor) and the chance of particle collision with the sides of the container is also increased (minor factor).

    • So both effects contribute to an increase in volume at constant pressure with increase in temperature (above right graph.

    • Note that the graphs extrapolate back to 0K (absolute zero, Kelvin scale) or -273^oC (Celsius scale).

  • OR (ii) the pressure of a gas is directly proportional to the absolute temperature (K) at constant volume,

    • p = constant x T (right graph), or

    • p/T = constant, or

    • p₁/p₂ = T₁/T2  for conditions changing from 1 (initial) to 2 (final),

    • or p₁/T₁ = p₂/T₂  for constant volume

    • p₁ x T₂ = T₁ x p₂

    • p₂ = p₁ x T₂/T₁

    • or T₂ = T₁ x p₂/p₁

    • Kinetic particle reasoning - increasing the temperature increases the kinetic energy of the molecules giving more forceful collisions (major factor) and greater chance of collision (minor factor), both of which contribute to an increase in the pressure if the volume is constrained (kept constant).

    • Note again that the graphs extrapolate back to 0K (absolute zero, Kelvin scale) or -273^oC (Celsius scale).

• In all calculations, the absolute or Kelvin scale of temperature must be used for T (K = ^oC + 273).

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4.(d) The combined gas law equation - more complicated calculations

• If all the laws described in 4b and 4c are combined, you get the following general expression

• p x V/T = a constant (for a given mass of gas).

• This can be expressed in generalised form for calculations based on an initial set of conditions₁ (1) changing to a new and final set of conditions₂ (2) for a given mass of gas, giving the combined pressure–volume–temperature gas calculation equation ...

• p1 x V1		p2 x V2
  ––––––––––––	=	––––––––––––
  T1		T2

• In shorthand': p₁V₁ /T ₁ = p₂V₂ / T₂
• therefore the three permutations for problem solving involving all three variables are:
  • p₂ = p₁V₁T₂ / V₂T₁
  • V₂ = p₁V₁T₂ / p₂T₁
  • T₂ = p₂V₂T₁ / p₁V₁

• If one variable is constant, the permutations for the other two variables are:

  • p₂ = p₁V₁ / V₂   (at constant temperature)
  • V₂ = V₁T₂ / T₁   (at constant pressure)
  • T₂ = p₂T₁ / p₁   (at constant volume)

• Note:

  • If the temperature is constant you get Boyle's Law.

  • If p or V is constant you get Charles's/Gay–Lussac's Law.

  • You can use any volume or pressure units you like as long as both p's or both V's have the same units.

  • The graphs of p or V versus temperature become invalid once the gas has condensed into a liquid BUT when extrapolated back all the lines seem to originate from y = 0 (for p or V), x = –273^oC (for T).

  • This was part of the scientific evidence that led to the belief that –273^oC was the lowest possible temperature, though there is no theoretical upper limit at all.

  • This led to the devising of a new thermodynamic absolute temperature scale or Kelvin scale which starts at OK.

    • e.g. ice melts at 0^oC or 273K and water boils at 100^oC or 373K.

• Examples of P–V–T calculations

• Ex. 4b Q1

  • The pressure exerted by a gas in sealed container is 100kPa at 17^oC. It was found that the container might leak if the internal pressure exceeds 120kPa. Assuming constant volume, at what temperature in ^oC will the container start to leak?

  • 17^oC + 273 = 290K

  • p₁/T₁ =  p₂/T₂

  • rearranging to scale up to the higher temperature

  • T₂ = T₁ x p₂/p₁

  • T₂ = 290 x 120/100 = 348 K or 348 – 273 = 75^oC when the container might leak

• Ex. 4b Q2

  • A cylinder of propane gas at 20^oC exerted a pressure of 8.5 atmospheres. When exposed to sunlight it warmed up to 28^oC. What pressure does the container side now experience?

  • 20^oC = 273 + 20 = 293K, 28^oC = 273 + 28 = 301K

  • p₂ = p₁ x T₂/T₁

  • p₂ = 8.5 x 301/293 = 8.73 atm

• Ex. 4b Q3

  • A student was investigating the speed of reaction between limestone granules and different concentrations of hydrochloric acid. However after doing a whole series of experiments at different acid concentrations, there was no time to do the last planned experiment. The volume of carbon dioxide collected after 5 minutes in a 100cm³ gas syringe was used to determine the rate of reaction. All the experiments were done in one lesson at a temperature of 22^oC except for the last one. This was done in the next lesson, giving a carbon dioxide volume of 47.0 cm³ after 5 minutes, but at a higher temperature of 27^oC (when in Kelvin call this T₁, and the other temperature T₂).

  • To make the data analysis fair, all the gas volumes should be ideally measured at the same temperature, but a correction can be made for the last experiment.

  • (a) Calculate the volume the of 47.0 cm³ of gas at 27^oC,  would occupy at 22^oC.

    • V₁/V₂ = T₁/T₂ so V₂ = V₁ x T₂/T₁

    • V₁ = 47.0 cm³, T₁ = 273 + 27 = 300K, T₂ = 273 + 22 = 295K

    • V₂ = 47.0 x 295/300 = 46.2 cm³

  • (b) If the temperature was ignored, what is the % error in the rate of reaction measurement?

    • Volume error = 47.0 – 46.2 = +0.8 cm³, therefore ....

    • % error = 0.8 x 100/47 = +1.7% (so you would over calculate the reaction rate without this correction)

    • The % error in the volume would be the same as calculated for the rate e.g. in cm³/min.

  • (c) Should the calculated value for 22^oC be used in the rate calculation analysis? and are this still other sources of error?

    • The theoretical–calculated gas volume for 22^oC should be used for calculating the rate, it will improve the accuracy a little, BUT there is another problem!

    • If the reaction was unfortunately carried out at a higher temperature (i.e. 27^oC) there is a second source of error. At a higher temperature the reaction is faster, so you are bound to get a higher volume of gas formed in five minutes. Therefore you will calculate a faster rate of reaction e.g. in cm³ gas/minute at 27^oC, that would have occurred/been measured at 22^oC and so an unfair comparison with all the other results from the previous lesson.

    • So, although you can correct reasonably well the volume error due to an 'expanded' gas volume at the higher temperature, the gas volume will still be too high because of the faster rate of reaction at 27^oC and there isn't much you can do about that error except repeat the experiment at 22^oC, which is the best thing to do anyway!

      • Note that if the temperature of a rates experiment was too low compared to all the other experiments, the 'double error' would occur again, but this time the measured gas volume and the calculated speed/rate of reaction would be lower than expected.

  • (d) Would you need to do any correction for the volume of acid added to the limestone? Explain your decision.

    • No correction needed for this at all. Although liquids expand/contract on heating/cooling, the volume changes are far less compared to gas volume changes for the same temperature change. This is because of the relatively strong intermolecular forces between liquid molecules, which are almost absent in gases.

• Ex. 4b Q4

  • 25 cm³ of a gas at 1.01 atm. at 25^oC was compressed to 15 cm³ at 35^oC.

  • Calculate the final pressure of the gas.

  • p₁ = 1.01 atm, p₂ = ?, V₁ = 25 cm³, V₂ = 15 cm³,

  • T1 = 25 + 273 =  298 K, T2 = 35 + 273 = 308 K

  • (p₁ x V₁)/T₁ = (p₂ x V₂)/T₂

  • p₂ = (p₁ x V₁ x T₂)/(V₂ x T₁)

  • p₂ = (1.01 x 25 x 308)/(15 x 298) = 1.74 atm

• Ex. 4b Q5

  • 12.0 dm³ of gas in a cylinder and piston system is heated from 290 K to 340 K. If the pressure remains constant, calculate the final volume of gas in the cylinder.

    • V/T = constant

    • V₁/V₂ = T₁/T₂

    • V₁ x T₂ = V₂ x T₁

    • V₂ = V₁ x T₂/T₁

    • V₂ = 12 x 340/290 = 14.1 dm³

• Ex. 4b Q6

  • The fuel and air gases in the cylinders of a 1200 cm³ car engine go from 25^oC before combustion and rise to a peak temperature of 2100^oC after combustion. If normal atmospheric pressure is 101kPa, calculate the peak pressure reached after combustion. Although the movement of the piston changes the volume, (i) for the sake of argument assume the volume is constant.

  • T₁ = 25 + 273 = 298 K, T₂ = 2100 + 273 = 2373 K, P1 = 101 KPa

  • p/T = constant

  • p₁/p₂ = T₁/T2

  • p₂ = p₁ x T₂/T₁

  • p₂ = 101 x 2373/298 = 804 kPa

    • (ii) To be more realistic, assume the initial volume of fuel vapour plus air was 400 cm3, now re-calculate the final pressure.

    • You now need to use the full PVT expression.

    • (p₁ x V₁)/T₁ = (p₂ x V₂)/T₂
    • p₂ = (p₁ x V₁ x T₂)/(V₂ x T₁)
    • p₂ = (101 x 400 x 2373)/(1200 x 298) = 268 kPa

• Ex. Q7

  • ?

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The combined gas law equation for an ideal gas (for advanced A level chemistry students only)

There is an equation that combines both Boyle's law and Charles law plus moles of gas involved.

The equation, known as the ideal gas equation, is given as:

pV = nRT

p = pressure in pascals (unit Pa)

V = volume in cubic metres (m³)

n = moles of gas (mol = mass in g / molecular mass of gas M_r)

R = ideal gas constant = 8.314 joules per kelvin per mol (J K^-1 mol^-1)

T = temperature in kelvin (K)

You must convert to these units for a correct calculation using pV = nRT

More on pV = nRT calculations - a more advanced chemistry page

OTHER USEFUL PAGES

See also for gas calculations

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume ratios, Avogadro's Law & Gay–Lussac's Law Calculations

All other calculation pages

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass – the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay–Lussac's Law (ratio of gaseous reactants–products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do volumetric titration calculations e.g. acid–alkali titrations (and diagrams of apparatus)

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws (this page)

17. Radioactivity & half–life calculations including dating materials

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