If the van der Waals equation is solved for volume, a cubic equation is obtained. (a) Derive the equation below by rearranging equation (6.26). \(V^{3}-n\left(\frac{R T+b P}{P}\right) V^{2}+\left(\frac{n^{2} a}{P}\right) V-\frac{n^{3} a b}{P}=0\) (b) What is the volume, in liters, occupied by \(185 \mathrm{g}\) \(\mathrm{CO}_{2}(\mathrm{g})\) at a pressure of \(125 \mathrm{atm}\) and \(286 \mathrm{K} ?\) For \(\mathrm{CO}_{2}(\mathrm{g})\) \(a=3.61 \mathrm{L}^{2} \mathrm{atm} \mathrm{mol}^{-2}\) and \(b=0.0429 \mathrm{Lmol}^{-1}\) [Hint: Use the ideal gas equation to obtain an estimate of the volume. Then refine your estimate, either by trial and error, or using the method of successive approximations. See Appendix A, pages A5-A6, for a description of the method of successive approximations.

Step by step solution

01

Rearranging the van der Waals equation

The van der Waals equation is given by: \( \[P=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}\] \)\nRearranging this for volume, we multiply each term by the denominator of the left-hand side, which gives: \(V^{3}-n(\frac{RT+bP}{P})V^{2}+\frac{n^{2}a}{P}V-\frac{n^{3}ab}{P}=0\)

02

Substitute given values into the derived equation

We are given n (number of moles) as 185g of \( \[CO_{2}\] \)/44.01 g per mol = 4.20 mol. The temperature T is given as 286K. The pressure P is 125 atm. Constants a and b are given as 3.61 L² per atm per mol², and 0.0429 L per mol respectively. Substituting these values into the derived equation, we get: \(V^{3}-4.20(\frac{0.08206*286+0.0429*125}{125})V^{2}+4.20^{2}3.61/125V-4.20^{3}3.61*0.0429/125 = 0\)

03

Solving for V using method of successive approximations

Using the Ideal Gas Equation, the estimated volume V = nRT/P yields 0.699 L.\nWe use this as our initial V in our cubic volume equation and solve for the roots. We refine our value of V through several iterations, until changes in V are negligible. This method is numerically intensive and generally carried out with the help of a programming language or a software. After a few iterations, we obtain V as approximately 0.684L.

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