Collection of Solved Problems in Physics

Ideal Gas Versus Van der Waals

Task number: 1284

• Numerical values

  V = 0.53 m3	the internal volume of the container
  n = 1 kmol = 103 mol	the amount of substance of carbon dioxide
  p = 5.07 MPa = 5.07·106 Pa	the pressure of carbon dioxide
  ΔT = ?	the difference between the temperatures calculated by each of the models

  ---

  From The Handbook of Chemistry and Physics:

  R = 8.31 Jmol−1K−1	the molar gas constant
  a = 0.365 Jm3mol−2	the "van der Waals" constant for carbon dioxide
  b = 4.28·10−5 m3mol−1	the "van der Waals" constant for carbon dioxide

• Analysis

  We will evaluate the thermodynamic temperature from both the ideal gas equation and the Van der Waals equation and then we will calculate their difference.

• Solution

  First we evaluate the thermodynamic temperature T_i from the ideal gas equation

  \[pV=nRT_{i},\]

  where p is the pressure, V is the volume, n is the amount of substance of carbon dioxide and R is the molar gas constant.

  \[T_{i}=\frac{pV}{nR}.\]

  We also evaluate the thermodynamic temperature T_v from the van der Waals equation

  \[\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)=nRT_{v},\]

  where a and b are the van der Waals constants for carbon dioxide, and thus we obtain

  \[T_{v}=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)}{nR}.\]

  Finally, we determine the temperature difference ΔT:

  \[\mathrm{\Delta}T=T_{v}-T_{i}=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)-pV}{nR}.\]

• Numerical insertion

  \[\mathrm{\Delta}T=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)-pV}{nR}\] \[\mathrm{\Delta}T=\frac{\left(5.07\cdot{10^6}+\frac{10^6\cdot{0.365}}{0.53^{2}}\right)(0.53-10^3\cdot{4.28}\cdot{10^{-5}})-5.07\cdot{10^6}\cdot{0.53}}{8.31\cdot{10^3}}\,\mathrm{K}\] \[\mathrm{\Delta}T\dot{=}50\,\mathrm{K}\]

• Answer

  The temperature calculated by the van der Waals equation is approximately 50 K higher than the temperature calculated by the ideal gas equation.