Cap 5 Chang

CHAPTER 5 GASES
5.13
562 mmHg × 1 atm = 0.739 atm 760 mmHg

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:
? atm = 606 mmHg × ? kPa = 0.797 atm × 1 atm = 0.797 atm 760 mmHg

101.325 kPa = 80.8 kPa 1 atm

5.17

(a) (b)

If the finaltemperature 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). 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). Recall that V ∝
1. As the pressure is tripled, the volume will decrease to P assuming constant n and T. The correct choice is (b).
1

5.18

(1)

3

of its original volume,

(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 constantmoles of gas, the density of the gas will decrease. This decrease in gas density is indicated by the lighter shading. Recall that V ∝ n. Starting with n moles of gas, adding another n moles of gas (2n 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.
1 . Halving the temperature would decrease the volume to1 2 its original P volume. However, reducing the pressure to 1 4 its original value would increase the volume by a factor

(3)

(4)

Recall that V ∝ T and V ∝

of 4. Combining the two changes, we have 1 ×4 = 2 2 The volume will double. The correct choice is (a).

114

CHAPTER 5: GASES

5.19

P1 = 0.970 atm V1 = 725 mL

P2 = 0.541 atm V2 = ? P1V1 = P2V2
V2 = PV1 (0.970 atm)(725mL) 1 = = 1.30 × 103 mL 0.541 atm P2

5.20

Temperature and amount of gas do not change in this problem (T1 = T2 and n1 = n2). Pressure and volume change; it is a Boyle's law problem.
PV1 PV 1 = 2 2 n1T1 n2T2

P1V1 = P2V2 V2 = 0.10 V1
P2 = P2 = PV1 1 V2 (5.3 atm)V1 = 53 atm 0.10V1

5.21

P1 = 1.00 atm = 760 mmHg V1 = 5.80 L P1V1 = P2V2
P2 =

P2 = ? V2 = 9.65 L

PV1 (760mmHg)(5.80 L) 1 = = 457 mmHg 9.65 L V2

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.
PV1 PV 1 = 2 2 n1T1 n2T2

Because n1 = n2 and T1 = T2, P1V1 = P2V2 which is Boyle's Law. The given information is tabulated below.Initial conditions P1 = 1.2 atm V1 = 3.8 L Final Conditions P2 = 6.6 atm V2 = ?

CHAPTER 5: GASES

115

The final volume is given by:
V2 = V2 = PV1 1 P2 (1.2 atm)(3.8 L) = 0.69 L (6.6 atm)

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 finalvolume calculated is less than the initial volume, so the answer seems reasonable. (b) 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 pressure? Solution: You should also come up with the equation P1V1 = P2V2 for this problem. The given information is tabulated below. Initial conditionsP1 = 1.2 atm V1 = 3.8 L The final pressure is given by:
P2 = P2 = PV1 1 V2 (1.2 atm)(3.8 L) = 61 atm (0.075 L)

Final Conditions P2 = ? V2 = 0.075 L

Check: To decrease the volume of the gas fairly dramatically from 3.8 L to 0.075 L, the pressure must be increased substantially. A final pressure of 61 atm seems reasonable. 5.23 T1 = 25° + 273° = 298 K V1 = 36.4 L T2 = 88° + 273° = 361 K V2...