Ideal gas equation

pV=nRTwhereppressure in Pa=Nm−2 or atm=1.01×105PaVvolume in m3namount in molRmolar gas constant =8.31JK−1 mol−1Tabsolute temperature in K\begin{array}{c} \begin{aligned} pV &= nRT \end{aligned} \\ \hspace{-9em}\textbf{where} \\ \begin{array}{rl} p & \text{pressure in $\text{Pa} = \text{Nm}^{-2} $ or $\text{atm} = 1.01 \times 10^5 \text{Pa}$} \\ V & \text{volume in $\text{m}^3$} \\ n & \text{amount in mol} \\ R & \text{molar gas constant $= 8.31\text{JK}^{-1}\text{ mol}^{-1}$} \\ T & \text{absolute temperature in K} \end{array} \end{array}

Assumptions

  • Gas particles occupy negligible volume as compared to the volume of the container
  • The forces of attraction and repulsion between gas particles are negligible
    • i.e. there are negligible intermolecular forces
  • Gas particles collide elastically
    • i.e. there is no loss in kinetic energy after collision
  • Gas particles are in constant random motion
  • The average kinetic energy of the gas particles is directly proportional to the absolute Kelvin temperature
    • At a particular temperature, all particles possess the same average kinetic energy

Deviations from ideal gas behaviour

  • Real gas particles occupy significant volume

  • Real gas particles experience significant intermolecular forces of attraction between them

  • Pressure

    • Low Pressure
      • Few gaseous particles exist in a unit volume; the volume of particles is negligible compared to the container volume
      • Molecules are far apart, making intermolecular forces insignificant
    • High Pressure
      • Many gaseous particles exist in a unit volume; the volume of particles cannot be considered negligible
      • Molecules are close together, making intermolecular forces significant
  • Temperature

    • High Temperature
      • Gaseous particles have great speed and average kinetic energy
      • Particles can easily overcome intermolecular forces, which are thus insignificant
    • Low Temperature
      • Gaseous particles have less speed and average kinetic energy
      • Particles are less able to overcome intermolecular forces, which are thus significant
  • Conditions for Ideal and Non-Ideal Behaviour

    • Real gases behave most ideally at low pressures and high temperatures
    • Real gases deviate most from ideal behaviour at high pressures and/or low temperatures
  • Gases that Deviate Significantly from Ideal Behaviour

    • Large molecules (e.g., \ceCO2\ce{CO2}): Have a large electron cloud and thus stronger intermolecular forces between molecules
    • Polar molecules (e.g., \ceHCl\ce{HCl} and \ceNH3\ce{NH3}): Have stronger intermolecular forces of attraction such as permanent dipole-permanent dipole interactions or hydrogen bonding
  • Deviation from ideal gas is compared with graph of PvRT\frac{Pv}{RT} against P

    • The graphs begins at 1 and dips before rising above the line of ideal gases
    • dip
      • cause by intermolecular forces of attraction which pull molecules together, causing the gas to compress more easily than an ideal gas
      • The actual volume of the gas is less than the expected ideal volume, making pVpV smaller and causing the graph to dip below 1
    • rise
      • At high pressures, molecules are packed tightly and resist further compression
      • The space occupied by the molecules becomes a significant fraction of the total container volume
      • The actual volume is greater than the expected ideal volume, making pVpV larger and causing the graph to rise above 1
  • For smaller molecules like \ceH2\ce{H2}, they do not dip at all as the dispersion forces between them are too weak

    IMG-20260919180556813
  • When changing temperature, deviation is greatest at low temperatures

    IMG-20260919180644980

Van der Waals equation

[P+a(nv)2](V−nb)=nRT[P+a\left( \frac{n}{v} \right)^2](V-nb)=nRT

  • a (intermolecular forces correction)
    • unit is atm⋅L2⋅mol−2atm\cdot L^2\cdot mol^{-2}
  • b (molecular volume correction)
    • unit is L⋅mol−2L\cdot mol^{-2}
  • 1L=1dm31L = 1dm^3
  • 1bar=100000Pa1\text{bar} = 100000Pa
  • since 1L⋅bar=100J1L\cdot \text{bar} = 100J, molar gas constant can also be in 0.0831 L bar K−1 mol−10.0831\ L\ \text{bar}\ K^{-1}\ \text{mol}^{-1}
  • will provide during exam
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