CH70C7~1

CH70C7~1 - CHAPTER 5 GASES PRESSURE HILL PETRUCCI CHAPTER 5...

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CHAPTER 5 - GASES PRESSURE GASES PRESSURE 14. In the above drawing, the value of h, the height of the liquid in the tube above the liquid in the beaker, would depend on which of the following? X. Density of the liquid Y. Cross sectional area of the tube Z. Atmospheric pressure (a) Z only (b) Y and Z only (c) X and Y only (d) X and Z only (e) X, Y, and Z 1

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This is a barometer. The liquid is held in the tube by atmospheric pressure. If the atmospheric pressure is exactly 1 atmosphere (14.7 lb/in 2 ), then the liquid would have to exert a pressure of 14.7 lb/in 2 . (X) If the liquid was, let’s say, less dense, then it would have to be a greater height to exert a pressure of 14.7 lb/in 2 . If the liquid was more dense then it would have to be a shorter height. (Y) The cross sectional area of the tube is irrelevant, since the pressure (14.7 lb/in 2 ) would be the same regardless of the cross-sectional area. (Z) If the atmospheric pressure decreased, then the height would decrease because it would require less liquid to be equal to a lower pressure. If the atmospheric pressure increased, then the height would need to increase. 2
GAS LAWS 3 Chapter 5 – Gases Gas laws A container of fixed volume contains 22.0 g of CO 2 (g) exerting a pressure of 1.00atm. Some CH 4 (g) is introduced into the container, raising the pressure to 1.50atm. What mass of CH 4 (g) was introduced? A. 8.00g B . 4.00g C. 12.0g D. 16.0g E. 2.00g CO 2 (g) CH 4 (g) 22.0 g = 0.4999 mol ?g 1.00 atm V 1 = V 2 T 1 = T 2 P 2 = 1.50 atm Condition 1 = just CO 2 Condition 2 = Total gases P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 P 1 /n 1 = P 2 /n 2 1.00 atm/0.4999 mol = 1.50 atm/X mol X mol = 0.75 mol = total moles Therefore, 0.75 total mol – 0.50 mol CO 2 = 0.25 mol CH 4 g = 0.25 mol x 16.05g/mol = 4.01 g CH 4 B 3

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GASES GAS LAWS 7. A high vacuum pump can produce a pressure of 1.0 x 10 -6 Torr. How many molecules are in each mL of gas at 25 o C at this low pressure? (a) 3.2 x 10 10 (b) 2.5 x 10 10 (c) 3.9 x 10 11 (d) 2.5 x 10 13 (e) 3.9 x 10 13 Gas 1.0x10 -6 Torr x (1 atm/760 Torr) = 1.316 x 10 -9 atm 0.001L 298K n = PV/RT n = ((1.316 x 10 -9 atm) x 0.001L)/((0.08205Latmdeg -1 mol -1 ) x 298K) = 5.382 x 10 -14 mol 5.382x10 -14 mol x (6.022 x 10 23 molecules/mol) = 3.24 x 10 10 molecules GASES GAS LAWS 1. A bicycle tire is filled with air to a pressure of 610 kPa, at a temperature of 19 o C. Riding the bike on asphalt on a hot day increases the temperature to 58 o C. The volume of the tire increases by 4.0%. What is the new pressure in the tire? (a) 665 kPa (b) 517 kPa (c) 868 kPa (d) 434 kPa (e) 1217 kPa Bike(condition 1) Bike(condition 2) 610 kPa ?kPa = Y kPa 292 K 331 K X L 1.04X L P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 n 1 = n 2 P 1 V 1 /T 1 = P 2 V 2 /T 2 (610kPa x XL)/(292K) = (YkPa x 1.04XL)/(331K) The X’s and units cancel out. (610/292) = (Y x 1.04)/(331)
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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CH70C7~1 - CHAPTER 5 GASES PRESSURE HILL PETRUCCI CHAPTER 5...

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