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Chapter 6 Gases

6.6 The Combined Gas Law

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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

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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

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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

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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

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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