Chapter 5

# Chapter 5 - CHAPTER 5 GASES 5.13 562 mmHg 1 atm = 0.739 atm...

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CHAPTER 5 GASES 5.13 1atm 562 mmHg 760 mmHg ×= 0.739 atm 5.14 Strategy: Because 1 atm = 760 mmHg, the following conversion factor is needed to obtain the pressure in atmospheres. 1 atm 760 mmHg For the second conversion, 1 atm = 101.325 kPa. Solution: 606 mmHg 760 mmHg = ? atm 0.797 atm 101.325 kPa 0.797 atm = ?kPa 80 .8kPa 5.17 (a) If the final temperature of the sample is above the boiling point, it would still be in the gas phase. The diagram that best represents this is choice (d) . (b) If the final temperature of the sample is below its boiling point, it will condense to a liquid. The liquid will have a vapor pressure, so some of the sample will remain in the gas phase. The diagram that best represents this is choice (b) . 5.18 (1) Recall that 1 V P . As the pressure is tripled, the volume will decrease to 1 3 of its original volume, assuming constant n and T . The correct choice is (b) . (2) Recall that V T . As the temperature is doubled, the volume will also double, assuming constant n and P . The correct choice is (a) . The depth of color indicates the density of the gas. As the volume increases at constant moles of gas, the density of the gas will decrease. This decrease in gas density is indicated by the lighter shading. (3) Recall that V n . Starting with n moles of gas, adding another n moles of gas (2 n total) will double the volume. The correct choice is (c) . The density of the gas will remain the same as moles are doubled and volume is doubled. (4) Recall that V T and 1 V P . Halving the temperature would decrease the volume to 1 2 its original volume. However, reducing the pressure to 1 4 its original value would increase the volume by a factor of 4. Combining the two changes, we have 1 42 2 × = The volume will double. The correct choice is (a) .

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CHAPTER 5: GASES 88 5.19 P 1 = 0.970 atm P 2 = 0.541 atm V 1 = 725 mL V 2 = ? P 1 V 1 = P 2 V 2 11 2 (0.970 atm)(725 mL) 0.541 atm == = 3 2 1.30 10 mL PV P V × 5.20 Temperature and amount of gas do not change in this problem ( T 1 = T 2 and n 1 = n 2 ). Pressure and volume change; it is a Boyle's law problem. 2 2 = P V nT n T P 1 V 1 = P 2 V 2 V 2 = 0.10 V 1 2 2 = P V 1 1 (5.3 atm) 0.10 2 53 atm V V P 5.21 P 1 = 1.00 atm = 760 mmHg P 2 = ? V 1 = 5.80 L V 2 = 9.65 L P 1 V 1 = P 2 V 2 2 (760 mmHg)(5.80 L) 9.65 L = 2 457 mmHg V P 5.22 (a) Strategy: The amount of gas and its temperature remain constant, but both the pressure and the volume change. What equation would you use to solve for the final volume? Solution: We start with Equation (5.9) of the text. = Because n 1 = n 2 and T 1 = T 2 , P 1 V 1 = P 2 V 2 which is Boyle's Law. The given information is tabulated below. Initial conditions Final Conditions P 1 = 1.2 atm P 2 = 6.6 atm V 1 = 3.8 L V 2 = ? The final volume is given by: 2 2 = V P
CHAPTER 5: GASES 89 (1.2 atm)(3.8 L) (6.6 atm) == 2 0.69 L V Check: When the pressure applied to the sample of air is increased from 1.2 atm to 6.6 atm, the volume occupied by the sample will decrease. Pressure and volume are inversely proportional. The final volume calculated is less than the initial volume, so the answer seems reasonable.

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Chapter 5 - CHAPTER 5 GASES 5.13 562 mmHg 1 atm = 0.739 atm...

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