User:JelloPiranha/Sandbox2

I'll try to keep the format as close to the following as possible.

Question number and section Full question, taken directly from the text Relevant quotes, again taken directly from the text My own words on the problem. Can range from a summation of a quote to an explanation of a formula or simply be the answer to the question.

And without further ado...

10.
'''10a. Why does a column of mercury in a tube that is inverted in a dish of mercury have a height of about 760mm at sea level?''' From experiments like Torrecelli's, it is known that at sea level at 0°C, the average pressure of the atmosphere can support a 760mm column of mercury. The atmospheric pressure exerts force on the mercury in the dish and pushes it upwards into the tube.

'''10b. What height would be maintained by a column of water inverted in a dish of water at sea level?''' ''Torrecelli wondered why water pumps could raise water to a maximum height of only about 34 feet. . . He reasoned that liquid mercury, which is about 14 times as dense as water, could be raised only 1/14 as high as water.'' Knowing that a standard atmosphere can support about 760mm of mercury, and knowing that water can be raised 14 times as high as mercury, we can set up the following equation: $$760mmHg\frac{14mmH_2O}{1mmHg}=10640mmH_2O$$ And upon converting from millimeters to feet (There are about 0.0032 ft in a mm.), $$10640mmH_2O\frac{0.0032ftH_2O}{1mmH_2O}=34.028ftH_2O$$ we find that our column of water would achieve a height of roughly 34 feet, just as Torrecelli observed.

'''10c. What accounts for the difference in the heights of the mercury and water columns?''' He reasoned that liquid mercury, which is about 14 times as dense as water, could be raised only 1/14 as high as water. Mercury has a much greater density than water, which means that given the same amount of space, mercury has much more mass than water does, making it heavier and thus harder for the atmospheric pressure to push upward.

13.
'''13a. What is the Celsius equivalent of absolute zero?''' The temperature -273.15°C is referred to as absolute zero... The Celsius equivalent of the absolute zero temperature is -273.15°.

13 b. What is the significance of this temperature? ''The temperature -273.15°C is referred to as absolute zero and is given a value of zero in the Kelvin scale. This fact gives the following relationship between the two temperature scales. K=273.15+°C'' Absolute zero, since it is measurable in both Celsius and Kelvin, creates a conversion scale between the two: degrees Celsius plus 273.15 equals Kelvins.

'''13c. What is the relationship between Kelvin temperature and the average kinetic energy of gas molecules?''' Gas volume and Kelvin temperature are directly proportional to each other. The kinetic energy of the gas molecules is what allows the gas to expand infinitely: as long as the molecules move fast enough, they can overcome intermolecular forces and avoid being close enough together to condense into a liquid. So it could be said that a gas's volume is directly proportional to the average kinetic energy of its molecules. And since we have just set Kelvin temperature, gas volume, and kinetic energy proportional to each other, we can safely say that kinetic energy is directly proportional to Kelvin temperature as well.

14.
'''14a. Explain what is meant by the partial pressure of each gas within a mixture of gas.''' ''The pressure of each gas in a mixture is called the partial pressure of that gas. Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.'' When two or more gases are present in the same container, they will each exert their own pressure. The total pressure exerted by the gas mixture is the sum of all the partial pressures of its individual gases, as per Dalton's law.

'''14b. How do the partial pressures of gases in a mixture affect each other?''' The pressure that each gas exerts in the mixture is independent of that exerted by other gases present. The partial pressures of a gas mixture do not affect each other in any way.

16.
Convert each of the following into a pressure reading expressed in torr.

'''16a. 1.25 atm''' We now know that one atmosphere of pressure (atm) is equal to 760 mm Hg, and 1 mm Hg is equal to 1 torr as well. Thus: $$1.25atm\frac{760mmHg}{1atm}\frac{1torr}{1mmHg}=950torr$$ 1.25 atm are equal to 950 torr.

'''16b. 2.48×10-3 atm''' You may have noticed in the last equation that converting from mm Hg to torr did not change our numerical values at all. As such, I will be converting directly from atm to torr in future. First, convert the scientific notation into a decimal to yield 0.00248 atm. $$0.00248atm\frac{760torr}{1atm}=1.8848torr$$ Compensating for significant figures, 2.48×10-3 atm are equal to 1.88 torr.

'''16c. 4.75×104 atm''' Again, we should convert to decimal for our calculations: 475,000 atm. $$475000atm\frac{760torr}{1atm}=361000000torr$$ The proper number of significant figures are present, so 4.75×104 atm are equal to 361,000,000 torr.

'''16d. 7.60×106 atm''' The decimal equivalent is 7,600,000 atm. $$7600000atm\frac{760torr}{1atm}=5776000000torr$$ Correcting significant figures leaves us with 7.60×106 atm equals 5,780,000,000 torr.

17.
Convert each of the following into the unit specified.

'''17a. 125 mm Hg into atm''' $$125mmHg\frac{1atm}{760mmHg}=0.16447atm$$ Thus, 125 mm Hg is equal to about 0.164 atm.

'''17b. 3.20 atm into Pa''' For this next equation, is it necessary to know that 1 atm is equal to 1.01325×105 Pa. It would be wise to convert that right now: 101325 Pa. $$3.20atm\frac{101325Pa}{1atm}=324240Pa$$ Knock off a few digits to satisfy the significant figures, and you're left with 3.20 atm equaling 324,000 Pa.

'''17c. 5.38 kPa into torr''' This one is a bit tricky as there's not a direct relation between kPa and torr, but fortunately we can use atm to get where we need. $$5.38kPa\frac{1atm}{101.325kPa}\frac{760torr}{1atm}=40.3533torr$$ Let's deal with the significant figures. 5.38 kPa is equal to 40.4 torr.