According to the CRC Handbook of Chemistry and Physics (83rd ed.), the molar volume of \(\mathrm{O}_{2}(\mathrm{g})\) is \(0.2168 \mathrm{Lmol}^{-1}\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). (Note: \(1 \mathrm{MPa}=\) \(\left.1 \times 10^{6} \mathrm{Pa} .\right)\)(a) Use the van der Waals equation to calculate the pressure of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) if the volume is 0.2168 L. What is the \% error in the calculated pressure? The van der Waals constants are \(a=1.382 \mathrm{L}^{2}\) bar \(\mathrm{mol}^{-2}\) and \(b=0.0319 \mathrm{L} \mathrm{mol}^{-1}\) (b) Use the ideal gas equation to calculate the volume of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). What is the \% error in the calculated volume?

Short Answer

Expert verified
The calculated pressure using the Van der Waals equation is 10.3 MPa with a 3% error. The calculated volume using the ideal gas equation is 0.232 L with a 7.04% error.

Step by step solution

01

Solve using the Van der Waals Equation

The Van der Waals equation can be used to calculate the pressure of one mole of \(O_{2}(g)\) at 280 K if the volume is 0.2168 L. Inserting the given values, we find: \[P = \frac{(1)(8.3145)(280)}{(0.2168 - (1)(0.0319))} - \frac{(1^{2})(1.382)}{(0.2168^{2})}\]
02

Calculate the Pressure and % error

After calculation, the pressure is found to be 10.3 MPa. The % error in the calculated pressure can then be found by \(\frac{|10 - 10.3|}{10} \times 100% = 3%\).
03

Solve using the Ideal Gas Equation

The ideal gas equation can be used to calculate the volume of one mole of \(O_{2}(g)\) at 280 K and 10 MPa. Inserting the given values, we find: \[V = \frac{(1)(8.3145)(280)}{10 \times 10^{6}}\]
04

Calculate the Volume and % error

After calculation, the volume is found to be 0.232 L. The % error in the calculated volume can then be found by \(\frac{|0.2168 - 0.232|}{0.2168} \times 100% = 7.04%\).

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Most popular questions from this chapter

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.

Calculate the average kinetic energy, \(\bar{e}_{k},\) for \(\mathrm{O}_{2}(\mathrm{g})\) at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\)

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) has a volume of \(26.7 \mathrm{L}\) at 762 Torr. What is the new volume if, with the temperature and amount of gas held constant, the pressure is (a) lowered to 385 Torr; (b) increased to 3.68 atm?

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