chem456_hw1key - Chemistry 456 Winter 2010 Bruce H Robinson...

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Chemistry 456 Winter 2010 Bruce H. Robinson Problems: Z5.22, P1.2, PX1, P1.24, PX2,P1.37 (Formerly 1 st Ed: P1.4, P1.5, P1.14, P1.15) Homework 1A key ( due Friday ). Problems denoted by Z mean they come from Zumdahl’s text (Chemical Principles). You may work all problems to a single significant figure, taking care to notice differences (such as in P1.15) and explain them. The PX problems are those that were in the 1 st Ed but are not in the second edtion. Z5.22) Two examples for an open-tube manometer are shown. If the flask (attached on the right but not shown) is open to the atmosphere, the mercury levels will be equal in height (The height difference will be zero.) For each of the following situations, in which a gas is contained in the flask, calculate the pressure in the flask in torr, atmospheres, and Pascals. The pressure of the mercury is : Fm gm g P h dgh AAA h == = = , where d is the density of mercury and h is the height of the column, g is the gravitational constant (9.8S.I.). The density of mercury is about 13.6 g/cc. Because 1 torr = 1mm Hg, we can just read off the pressure in torr. For A: P=760-118 = 660 torr, and B: P=760+215 = 960 torr . Noticed how I rounded the difference and just added and subtracted simple numbers from the reference of 760. So now we can look up conversion tables that tell us 1Atm ~ 1Bar = 5 11 0 Bar Pa = . So the pressure is also 660 960 : ~ .85 1.25 760 760 A BA t m C If the pressure of the atmosphere is 635 torr instead of 760 (but the height differences are the same) calculate the pressures in the flasks in parts a and b (in torr only). For A: P=635-118 = 520 torr, and B: P=635+215 = 850 torr . 1
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Chemistry 456 Winter 2010 Bruce H. Robinson P1.2 ) A compressed cylinder of gas contains 1.50 × 10 3 g of N 2 gas at a pressure of 2.00 × 10 7 Pa and a temperature of 17.1ºC. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 1.80 × 10 5 Pa? Assume ideal behavior and that the gas temperature is unchanged. For I.G. assume the volume of the gas cylinder does not change and there is no temperature change, so from the pressure change we compute the change in the number of moles, which is the number of moles released from the cylinder. The volume of the cylinder is computed from the initial conditions of the cylinder: 3 3 33 7 1.5 10 50 28 2.4 10 50 0.05 210 ii PV n RT nm o Vm l e s m = =⋅= == The number of moles released then is found from the pressure difference. ff i i f if PV nRT nRT PP P nn n RT Pn V = Δ≡ − Δ= Δ i Notice that almost all of the moles are released. Those moles occupy a volume at 1Atm and the same temperature, and that is: air air air air air nR T n R T P VV PP P Δ Δ === This could all be done more simply, by realizing that nearly all the gas is released, so 50 moles of gas at STP occupies . 3 50 22.4 1 1 km ⋅= = AA PX1) A balloon filled with 10.50 L of Ar at 18.0ºC and 1 atm rises to a height in the atmosphere where the pressure is 248 Torr and the temperature is –30.5ºC. What is the final volume of the balloon?
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chem456_hw1key - Chemistry 456 Winter 2010 Bruce H Robinson...

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