{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chpt 5 Solutions

# Chpt 5 Solutions - CHAPTER 5 GASES Questions 20 Molecules...

This preview shows pages 1–3. Sign up to view the full content.

137 CHAPTER 5 GASES Questions 20. Molecules in the condensed phases (liquids and solids) are very close together. Molecules in the gaseous phase are very far apart. A sample of gas is mostly empty space. Therefore, one would expect 1 mol of H 2 O(g) to occupy a huge volume as compared to 1 mol of H 2 O(l). 21. The column of water would have to be 13.6 times taller than a column of mercury. When the pressure of the column of liquid standing on the surface of the liquid is equal to the pressure of air on the rest of the surface of the liquid, then the height of the column of liquid is a measure of atmospheric pressure. Because water is 13.6 times less dense than mercury, the column of water must be 13.6 times longer than that of mercury to match the force exerted by the columns of liquid standing on the surface. 22. A bag of potato chips is a constant-pressure container. The volume of the bag increases or decreases in order to keep the internal pressure equal to the external (atmospheric) pressure. The volume of the bag increased because the external pressure decreased. This seems reasonable as atmospheric pressure is lower at higher altitudes than at sea level. We ignored n (moles) as a possibility because the question said to concentrate on external conditions. It is possible that a chemical reaction occurred that would increase the number of gas molecules inside the bag. This would result in a larger volume for the bag of potato chips. The last factor to consider is temperature. During ski season, one would expect the temperature of Lake Tahoe to be colder than Los Angeles. A decrease in T would result in a decrease in the volume of the potato chip bag. This is the exact opposite of what actually happened, so apparently the temperature effect is not dominant. 23. The P versus 1/V plot is incorrect. The plot should be linear with positive slope and a y - intercept of zero. PV = k, so P = k(1/V). This is in the form of the straight-line equation y = mx + b . The y -axis is pressure, and the x -axis is 1/V. 24. The decrease in temperature causes the balloon to contract (V and T are directly related). Because weather balloons do expand, the effect of the decrease in pressure must be dominant. 25. d = (molar mass)P/RT; density is directly proportional to the molar mass of a gas. Helium, with the smallest molar mass of all the noble gases, will have the smallest density. 26. Rigid container: As temperature is increased, the gas molecules move with a faster average velocity. This results in more frequent and more forceful collisions, resulting in an increase in pressure. Density = mass/volume; the moles of gas are constant, and the volume of the container is constant, so density must be temperature-independent (density is constant).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 5 GASES 138 Flexible container: The flexible container is a constant-pressure container. Therefore, the internal pressure will be unaffected by an increase in temperature. The density of the gas, however, will be affected because the container volume is affected. As T increases, there is an immediate increase in P inside the container. The container expands its volume to reduce
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern