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II. Ideal Gas Law Ch. 10 - Gases. A. Ideal Gas Law P 1 V 1 P 2 V 2 T 1 n 1 T 2 n 2 = This is where we ended with the Combined Gas Law: Play video!

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Presentation on theme: "II. Ideal Gas Law Ch. 10 - Gases. A. Ideal Gas Law P 1 V 1 P 2 V 2 T 1 n 1 T 2 n 2 = This is where we ended with the Combined Gas Law: Play video!"— Presentation transcript:

1 II. Ideal Gas Law Ch. 10 - Gases

2 A. Ideal Gas Law P 1 V 1 P 2 V 2 T 1 n 1 T 2 n 2 = This is where we ended with the Combined Gas Law: Play video!

3 PV T VnVn PV nT A. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 L  kPa/mol  K R=62.4 L. mmHg/mol. K = R You don’t need to memorize these values!

4 A. Ideal Gas Law PV=nRT “Piv-nert”

5 GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm B. Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

6 GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 L  kPa/mol  K B. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 L

7 GIVEN: V = 604 cm 3 (ml) n = 0.0851 mol T = ? P = 753.1 torr R = 62.4 L  mm Hg/mol  K C. Ideal Gas Law Problems b At what temperature is a gas of 0.0851 mol if it is contained in a 604 cm 3 vessel at 753.1 torr? PV = nRT (753.1)(0.604)=(.0851) (62.4) (?) torr L mol L  mmHg/mol  K K T = 85.7 K (or -187 °C ) WORK:


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