04_ws+3-key

U8 WS#3 Solutions On each of the problems below, start with the given P, V, T, or n; then make a decision as to how a change in P, V, T, or n will affect the starting quantity, and then multiply by the appropriate factor. Draw particle diagrams of the initial and final conditions. 1. A sample of gas occupies 150 mL at 25 ˚C. What is its volume when the temperature is increased to 50˚C? (P and n = constant) V 2 = 150 mL × 323 K 298 K = 163 mL → 160 mL 2. The pressure in a bicycle tire is 105 psi at 25˚C here in Fresno. You take the bicycle up to Huntington, where the temperature is – 5˚C. What is the pressure in the tire? (V and n = constant) P 2 = 724 kPa× 268 K 298 K = 651 kPa 3. What would be the new pressure if 250 cm 3 of gas at standard pressure is compressed to a volume of 150 cm 3 ? ( T, n = constant) P 2 = 1 atm × 250 cm 3 150 cm 3 = 1.7 atm or 1300 mmHg or 170 kPa or 25 psi P T V n Initial -- 298K 150mL -- Final -- 323K ? -- Effect --   -- P T V n Initial 724 kPa 298K -- -- Final ? 268K -- -- Effect   -- -- P T V n Initial 1atm -- 250cm 3 -- Final ? -- 150cm 3 -- Effect  --  --

4. What would be the new volume if 250 cm 3 of gas at 25˚C and 730 mm pressure were changed to standard conditions of temperature and pressure? (__ n __= constant) V 2 = 250 cm 3 × 730 mmHg 760 mmHg × 273 K 298 K = 220 cm 3 5. Sam’s bike tire contains 15 units of air particles and has a volume of 160mL. Under these conditions the pressure reads 13 psi. The tire develops a leak. Now it contains 10 units of air and has contracted to a volume of 150mL). What would the tire pressure be now? P 2 = 13 psi × 160 mL 150 mL × 10 u 15 u = 9.2 psi 6. A closed flask of air (0.250L) contains 5.0 “bobs” of particles. The pressure probe on the flask reads 93 kPa. A student uses a syringe to add an additional 3.0 “bobs” of air through the stopper. Find the new pressure inside the flask. P 2 = 93 kPa× 8 mol 5 mol = 150 kPa 7. A 350 mL sample of gas has a temperature of 30˚C and a pressure of 1.20 atm. What temperature would needed for the same amount of gas to fit into a 250 mL flask at standard pressure? P T V n Initial 730mmHg 298K 250cm 3 -- Final 760mmHg 273K ? -- Effect     -- P T P T V n Initial 13psi -- 160mL 15u Final ? -- 150mL 10u Effect   --   V n P T V n Initial 93kPa -- -- 5mol Final ? -- -- 8mol Effect  -- -- 

8. A 475 cm 3 sample of gas at standard temperature and pressure is allowed to expand until it occupies a volume of 600. cm 3 . What temperature would be needed to return the gas to standard pressure? T 2 = 273 K × 600 cm 3 475 cm 3 = 345 K 9. The diagram below left shows a box containing gas molecules at 25˚C and 1 atm pressure. The piston is free to move. In the box at right, sketch the arrangement of molecules and the position of the piston at standard temperature and pressure. Does the volume decrease significantly? Why or why not? V 2 = V 1 × 273 K 298 K V 2 = 0.92 V 1 The volume will decrease by about 8%. P T V n Initial -- 273K 475cm 3 -- Final -- ? 600cm 3 -- Effect --   --