Question

10 POINTS each question. SHOW DETAILED WORK, CALCUATIONS, FORMULAS, AND UNITS. EEDS T。OF-AND-RI rreN 1. Chlorofluorocarbons with the general formula CF Cln were used for decades in refrigerators and air conditioners until they were found to degrade Earth's ozone layer. For which value of n is the rate of effusion the slowest? Which compound will effuse at the fastest rate? Convert the following pressures into millimeters of mercury: (a) 0.541 atm; (b) 2.8 kPa 2. 3. The volume of a quantity of gas at 1.00 atm is compressed from 3.25 L to 2.24 L What is the final pressure of the gas if there is no change in temperature? 4. The pressure on a sample of an ideal gas is increased from 715 mm Hg to 3.55 atm at constant temperature. If the initial volume of the gas is 485 mL, what is the final volume of the gas? A 22.5 L sample of gas is cooled from 145° C to a temperature at which its volume is 18.3L. What is the new temperature? Assume no change in pressure of the gas 5. 6. Which of the following processes are exothermic, and which are endothermic? a. Ice cubes solidify in the freezer b. Ice cubes in a frost free-freezer slowly lose mass. c. Dew forms on a lawn overnight. What happens to the internal energy of a gas when it expands (with no heat flow)? 7. Instant cold pack chemistry. Breaking the small pouch of water inside a chemical cold pack containing ammonium nitrate activates the pack, which is used by sports trainers for injured athletes. What is the sign of AH for the process taking place in the cold pack? 8. 9. Keeping an athlete cool. During a strenuous workout, an athlete generates 233k of thermal energy. What mass of water would have to evaporate from the athlete's skin to dissipate this energy?

Answer 1

0.541 atm 1 atm = 760.0 mm Hg 0.541 atm = 0.541 times 760.0 mm Hg 0.541 atm = 411.16 mm Hg 2.8 kpa 2.8 kpa = 2.8 times 10^3 Partialdifferential = 2800 Partialdifferential = 2800 times 1/133.322 mm Hg 2.8 kpa = 21.001 mm Hg 1 mm Hg = 133.322 pascals 1 pascal = 1/133.322 mm Hg Given P_1 = 1.00 atm P_2 = x (say) v_1 = 3.25 L v_2 = 2.24 L (P_1, P_2 = Initial & final pressure v_1, v_2 = Initial & final volumes) at constant temperature p infinity 1/v = pv = constant p_1 v_1 = p_2 v_2 1.00 atm times 3.25 L = x times 2.24 L x = 1.00 atm times 3.25/2.24 = > x = 1.4508 atm Therefore Final pressure, P_2 = 1.4508 atm at constant temperature p infinity 1/v = > p_1 v_1 = p_2 v_2 Given p_1 = 715 mm Hg P_2 = 3.55 atm = > p_1 = 715/760