Collection of Solved Problems in Physics

Work of van der Waals Gas

Task number: 3946

• Hint

  Van der Waals gas is described by the equation of state

  \[\left( p+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) = nRT,\]

  where V is the volume of the gas, p is the gas pressure, n is the amount of substance, R is the universal molar gas constant, T is thermodynamic temperature, a and b are the given constants (characterizing the gas).

  How do you determine the work performed by the gas provided that pressure is the function of a volume?

• Analysis

  In this task we need to use integral calculus to determine work performed by the gas since the pressure is the function of volume. From the equation of state for van der Waals gas we can determine pressure using volume and integrate the resulting function with respect to volume. The limits of the integral are initial and final volume.

  To get a numerical solution we will need to determine temperature from the equation for van der Walls gas.

• Given Values

  a = 0.137 J m3 mol−2	constant from a van der Waals gas model
  b = 38.7·10−6 m3 mol−1	constant from a van der Waals gas model
  V1 = 10 l = 10 dm3 = 0.01 m3	initial volume of the gas
  V2 = 5V1 = 0.05 m3	volume after isothermal expansion
  p1 = 300 kPa = 3·105 Pa	initial pressure of the gas
  n = 1 mol	amount of substance of the gas
  W = ?	work performed by the gas during isothermal expansion

  Table values:

  R = 8.31 J K−1 mol−1

  universal gas constant

• Solution

  From the equation of state of van der Waals gas

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

  where V is the volume of the gas, p the gas pressure, n is the amount of substance, R is the universal gas constant, T is thermodynamic temperature, a and b are given constants,

  we first determine gas temperature T. The expansion is isothermal, the temperature T is therefore constant. We determine it using initial volume V₁ and pressure p₁.

  We obtain:

  \[T = \frac{\left( p_1+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) }{nR}.\]

   

  The work performed by the gas during a small change in volume dV is determined by

  \[\mathrm{d}W = p\mathrm{d}V.\]

  The total performed work is then determined by integration between the limits of initial volume V₁ and final volume V₂ = 5V₁. (We need to use integration calculus because the pressure of the gas changes continuously. Therefore, it is not possible to use a simple equation W =p(V₂-V₁) that holds for isobaric process with contant pressure!).

  From the basic form of van der Waals equation we can determine pressure as a volume function. We will need it in the next step of the calculation:

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

  Now we can perform the integration:

  \[W = \int_{V_{1}}^{V_{2}}{p}\, \mathrm{d}V = \int_{V_{1}}^{5V_{1}}{\left( \frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}\right)} \, \mathrm{d}V =\]

  we factor constants out of the integrals

  \[=nRT\int_{V_{1}}^{5V_{1}}{\frac{1}{V-nb}}\, \mathrm{d}V - n^{2}a \int_{V_{1}}^{5V_{1}}{\frac{1}{V^{2}}}\, \mathrm{d}V =\]

  we perform the integration and substitute the limits

  \[=nRT\left[ \ln \left( V-nb\right) \right] _{V_{1}}^{5V_{1}} + n^{2}a\left[ \frac{1}{V} \right] _{V_{1}}^{5V_{1}} =\] \[=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}+n^{2}a\left( \frac{1}{5V_{1}}-\frac{1}{V_{1}}\right) \] \[=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}- \frac{4n^{2}a}{5V_{1}}.\]

• Numerical Solution

  \[T = \frac{\left( p+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) }{nR} = \frac{\left( 3\cdot{ 10^{5}}+\frac{0.137}{10^{-4}}\right) \cdot \left( 0.01-38.7\cdot{ 10^{-6}}\right) }{8.31}\, \mathrm{K}\] \[T \dot{=} 361.3\, \mathrm{K}\]

   

  \[W=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}-\frac{4n^{2}a}{5V_{1}} \] \[W=\left(8.31\cdot{ 361.3}\cdot \ln \frac{0.05-38.7\cdot{ 10^{-6}}}{0.01-38.7\cdot{ 10^{-6}}}- \frac{4\cdot{0.137}}{0.05}\right) \, \mathrm{J}\] \[W \dot{=} 4830.5\, \mathrm{J}\dot{=} 4.8\, \mathrm{kJ}\]

• Answer

  A gas following a van der Waals model performed work of 4.8 kJ during expansion.