Van der Waals Equation for Hydrogen Sulfide

In the ideal-gas Law, the number of moles per volume \(n/V\) is equal to \(p/RT\). But for the van der Waals equation, solving for \(n/V\) in terms of pressure \(p\) and temperature \(T\) is hard.

(a) Show the van der Waals equation can be written as

(b) The van der Waals parameter for Hydrogen Sulfide gas (H₂ S) are \(a = 0.448~\mathrm{J}\cdot\mathrm{m}^{3}/\mathrm{mol}^{2}\) and \(b = 4.29\times10^{-5}\mathrm{m}^{3}/\mathrm{mol}\). Determine the number of moles per volume H₂ S gas at \(127^{\circ}~\mathrm{C}\) and an absolute pressure of \(9.80\times 10^{5}~\mathrm{Pa}\) as follows:

1. Calculate as a first approximation \(n/V = p/RT\)
2. Plug this approximation into the right-hand side of the Equation from part (a) to get an improved approximation for \(n/V\).
3. Repeat step 2 with "the current approximation" used in the right-hand side and "the next approximation" being equal to the result. Iterate until the approximation is stable to three significant figures.

(c) Compare results from (b) to the result \(p/RT\) obtained using the ideal-gas Law. Which is bigger? Why?

Identify: We need to use the van der Waals equation and the ideal-gas laws.

Set up: We recall the ideal gas Law readily enough:

The van der Waals equation modifies the left-hand side to take into account finite volume occupied by the gas \(nb\) and the surface tension affecting pressure \(a n^{2}/V^{2}\). We recall it to be:

We are given

and

We work at tempterature \(T = 127^{\circ}~\mathrm{C}\) (which must be converted to Kelvin) and pressure \(p = 9.80\times 10^{5}~\mathrm{Pa}\).

Set up: (a) This is just algebra, divide through by \(V\) and rearrange terms gives the result.

(b) We can write code to compute this

(let* ((temperature (+ 273.15 127)) ; kelvin
       (pressure 9.8e5) ;; pascal
       (a 0.448)
       (b 4.29e-5)
       (gas-constant 8.31446261815324)
       (rhs (lambda (guess)
              (* (/ (+ pressure (* a guess guess))
                    gas-constant temperature)
                 (- 1
                    (* b guess)))))
       (initial-guess (/ pressure gas-constant temperature)))
  (list
   :initial-guess initial-guess
   :iter-1 (funcall rhs initial-guess)
   :iter-2 (funcall rhs (funcall rhs initial-guess))))

We see it is stable to 3 digits after 2 iterations.

(c) We see that this van der Waals approach boils down to

The \(bn/V\sim 10^{-2}\) so the right factor \(1 - bn/V\) is very nearly 1. The left factor is on the order of 100, so an initial guess for \(n/V\) would be on the order of 100. Since \(a\sim 10^{-1/2}\), we see that \(\Delta p\sim 10^{3/2}\) or larger.

Hence the problem boils down to comparing \((10 + \pi)(1 - 10^{-2})\) to 10 (the order of ideal-gas answer), which is not "immediately obvious" (to me, at least). But doing the computation, we find \((10 + \pi)(1 - 10^{-2})\approx 13\) which is bigger than the ideal-gas's order-of-magnitude.

Evaluate: We see the order-of-magnitude makes sense, since \(T\sim400~\mathrm{K}\) and \(p\sim 10^{6}~\mathrm{Pa}\) gives us \(n/V\sim 10^{6}/3325\approx 300\) moles per meter cube. This should be slightly greater than the answer we obtained because we rounded up for pressure, down for temperature.

We can also observe that \(bn/V\ll 1\), so we end up with a quadratic equation

This has its solution be:

We find the two values numerically are:

and

These give us upper-bounds on what we expect the solution should be. It also gives us a sense of the correction contribution of the \(bn/V\) term: small (about 5 moles per cubic meter) but noticeable.