Chapter 14 Notes Part I Boyle’s, Charles’ and Gay- Lussac’s Laws Combined Gas Laws.
1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’...
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Transcript of 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’...
![Page 1: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/1.jpg)
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Chapter 6 Gases
6.6 The Combined Gas Law
![Page 2: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/2.jpg)
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The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant).
P1 V1 = P2 V2
T1 T2
Combined Gas Law
![Page 3: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/3.jpg)
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A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n is constant)?1. Set up Data TableConditions 1 Conditions 2
P1 = 0.800 atm P2 = 3.20 atm
V1 = 0.180 L (180 mL) V2 = 90.0 mL
T1 = 29°C + 273 = 302 K T2 = ??
Combined Gas Law Calculation
![Page 4: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/4.jpg)
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Combined Gas Law Calculation (continued)
2. Solve for T2 P1 V1 = P2 V2
T1 T2
T2 = T1 x P2 x V2
P1 V1
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
![Page 5: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/5.jpg)
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A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at -95°C and a pressure of 802 mm Hg (n constant)?
Learning Check
![Page 6: 1 Chapter 6 Gases 6.6 The Combined Gas Law. 2 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2.](https://reader035.fdocuments.us/reader035/viewer/2022081420/56649d555503460f94a32483/html5/thumbnails/6.jpg)
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Solution
Data TableConditions 1 Conditions 2
T1 = 308 K T2 = -95°C + 273 = 178K
V1 = 675 mL V2 = ???
P1 = 646 mm Hg P2 = 802 mm Hg
Solve for V2
V2 = V1 x P1 x T2
P2 T1
V2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K