Chapter 6 Solutions

Chapter 6 Solutions - CHAPTER 6 SOLUTION SET(10th Edition...

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Unformatted text preview: CHAPTER 6 SOLUTION SET (10th Edition) Keep in mind that there will often be multiple ways of approaching a given question; at times, my approach will differ from what you may have done or from how it is done in the book. Understanding how different approaches all fundamentally mean the same thing will go much further towards improving your understanding than memorizing one ʻ best ʼ way. If your approach is VERY different, this may be brilliant or it may signal a Faw in understanding that will come out and bite you later. Ask a tutor if you ʼ re worried. 2. a) Recalling that 1 atm is equal to the pressure of 760 mm Hg: 0.984 atm x 760 mmHg/atm= 748 mm Hg b) Torr is just another way of writing mm Hg, really. So 928 mm Hg . c) This is a comparison of densities. Multiply the height of the water column by the ratio of the densities (1/13.6), or derive it by comparing the two pressures thusly: g x h water x d water = g x h mercury x d mercury (142 feet x 12 inches/foot x 2.54 cm/inch) x 1 g/ml= h mercury x 13.6 g/ml h mercury = 318 cm . 6. Manometers work on the principal that the difference in the height of mercury is equal to the difference in pressure between the two sides. This question is dressed up to appear more complicated than it is, by giving extraneous information. Since the difference in mercury height has been given as h1= 30 mm, this means that the sealed gas must be 30 mmHg more pressurized than the exterior... hence it is: 740 mmHg + 30 mmHg= 770 mmHg 13. Recall your general gas law, used to compare a gas going through some change of state (you add 1875 L to your available volume, which changes pressure): P 1 V 1 /n 1 T 1 =P 2 V 2 /n 2 T 2 Since T and n don ʼ t change, you can eliminate them. You are left with: P 1 (35.8 L)=721mmHg (1875 +35.8 L) Solve for P 1 , then remember at the last minute to change it to atm (divide your answer by 760 mm Hg/atm). You should end up with 50.6 atm . 25. Again, a gas system that is changing. This time, P and V don ʼ t change, so you have: n 1 T 1 =n 2 T 2 You don ʼ t know what n is, and have no way of Fguring it out; it could be any gas. This doesn ʼ t matter, however, because whatever that n is, it ʼ s directly linked to the mass that we have, which has the same relation to n before and after. Mathematically: n= m/M m 1 /M x T 1 = m 2 /M x T 2 m 1 x T 1 = m 2 x T 2 So it comes to the same thing. 12.5 g (294.15 K) = m 2 (483.15 K) m 2 = 7.61 g remain....
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Chapter 6 Solutions - CHAPTER 6 SOLUTION SET(10th Edition...

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