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GAS LAWS Boyle’s Charles’ Gay-Lussac’s Combined Gas Ideal Gas Dalton’s Partial Pressure.

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Presentation on theme: "GAS LAWS Boyle’s Charles’ Gay-Lussac’s Combined Gas Ideal Gas Dalton’s Partial Pressure."— Presentation transcript:

1 GAS LAWS Boyle’s Charles’ Gay-Lussac’s Combined Gas Ideal Gas Dalton’s Partial Pressure

2 Boyle’s Law The pressure and volume of a sample of gas at CONSTANT temperature are inversely proportional to each other.

3 Graphing Boyle’s Law (V ↑ as P ↓) and (P ↑as V ↓)

4 Boyle’s Law Equation: P 1 V 1 = P 2 V 2 P 1 V 1 = initial pressure/volume P 2 V 2 = final pressure/volume Must Have Unit Agreement

5 A gas occupies 20.0 L at a pressure of 50.0 mm Hg. What is the volume when the pressure is increased to 80.0 mm Hg? (50.0 mm Hg)(20.0 L) = (80.0 mm Hg) V 2 (80.0 mm Hg) (80.0 mm Hg) V 2 = 12.5 L

6 Charles’ Law At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.

7 Graphing Charles’ Law (V ↑ as T ↑) and (V ↓ as T ↓)

8 Charles’ Law Equation: V 1 T 2 = V 2 T 1 T is measured in K temperature Must Have Unit Agreement

9 A 250. mL sample of neon is collected at 44 o C. Assuming the pressure remains constant, what would be the volume of the neon at standard temperature? (250. mL)(273 K) = V 2 (317K) (317K) (317K) V 2 = 215 mL

10 Gay-Lussac’s Law At constant volume, the pressure of a gas is directly proportional to the temperature (Kelvin).

11 Graphing Gay-Lussac’s Law (P ↑as T ↑ and P ↓ as T ↓)

12 Gay-Lussac’s Equation: P 1 T 2 = P 2 T 1 Must Have Unit Agreement T is measured in K temperature

13 If a gas is cooled from 355 K to 293 K and the volume is kept constant, what final pressure would result if the original pressure were 780.0 mm Hg? (780.0 mm Hg)(293 K) = P 2 (355 K) (355 K) (355 K) P 2 = 644 mm Hg

14 Combined Gas Law The law combines Boyle’s, Charles’s, and Gay- Lussac’s relationship into one equation.

15 Combined Gas Law Equation: P 1 V 1 T 2 = P 2 V 2 T 1 Must Have Unit Agreement T is measured in K temperature

16 A sample of gas occupies a volume of 450.0 mL at 740 mm Hg and 16 o C. Determine the volume of this sample at 760 mm Hg and 37 o C. (740 mm Hg)(450.0 mL)(310K) = (760 mm Hg)(V 2 )(289K) (760 mm Hg) (289K) (760 mm Hg) (289K) V 2 = 470 mL

17 Dalton’s Partial Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the gases

18 Dalton’s Partial Pressure Equation: P total = P A + P B + P C …

19 A mixture of neon and argon gases exerts a total pressure of 2.39 atm. The partial pressure of the neon alone is 1.84 atm, what is the partial pressure of the argon? P total = P A + P B 2.39 atm = 1.84 atm of Ne + ? atm of Ar *Rearrange the equation and subtract to solve for Argon partial pressure 0.55 atm of Ar

20 Ideal Gas Law Relates the pressure, temperature, volume, and moles of gas through the gas constant “R” The value for “R” depends on the pressure and volume R = 0.0821 L atm mol K R = 8.314 L kPa mol K R = 62.4 L mm Hg mol K R = 62.4 L torr mol K

21 Ideal Gas Law Equation: PV = nRT where P = (kPa, atm, mm Hg, torr) V = (Liters) n = # of moles R = Universal Gas Constant T = (Kelvin) Must have unit agreement!

22 How many moles of argon are there in a 725 mL sample of gas at 1875 torr and at standard temperature? P= 1875 torr V= 725 mL* n= ? mol Ar R= 62.4 T= 0 o C* PV=nRT Must convert mL to L Must convert o C to K n = 0.0798 mol Ar (1875 torr)(0.725 L)=n(62.4)(273 K)

23 How many grams of argon are present? Convert moles to grams by using molar mass 0.0798 mol ArX 39.95 g Ar 1 mol Ar = 3.19 g Ar


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