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Chapter 6: Problem 35
A 0.418 g sample of gas has a volume of \(115 \mathrm{mL}\) at \(66.3^{\circ} \mathrm{C}\) and \(743 \mathrm{mm} \mathrm{Hg} .\) What is the molar mass of this gas?
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At times, a pressure is stated in units of mass per unit area rather than force per unit area. Express \(P=1 \mathrm{atm}\) in the unit \(\mathrm{kg} / \mathrm{cm}^{2}\) [Hint: How is a mass in kilograms related to a force?]
When working with a mixture of gases, it is sometimes convenient to use an apparent molar mass (a weightedaverage molar mass). Think in terms of replacing the mixture with a hypothetical single gas. What is the apparent molar mass of air, given that air is \(78.08 \% \mathrm{N}_{2}\) \(20.95 \% \mathrm{O}_{2,0.93 \%} \mathrm{Ar}_{\left(\text {and } 0.036 \% \mathrm{CO}_{2},\text { by volume? } \right.}\)
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?
A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]
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.
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