TUTORIAL INSTRUCTIONS: Carefully go through the tutorial, step by step. You may return to any section, as necessary.Once you are satisfied that you understand the concepts and procedures, click CLOSE to get out of the tutorial mode. You will then be taken to the post-quiz.

PLEASE LISTEN CAREFULLY

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if n mol gas contain N particles,

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then 2n mol gas contain 2N particles,

and n/2 mol gas contain N/2 particles

P = F/A

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T of gas

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average kinetic energy of particles

average velocity of gas particles

Vgas Vcontainer

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IDEAL GAS LAW

PV = nRT

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R = 0.082053 L atm mol-1 K-1 = 62360.3 mL torr mol-1 K-1 = 8.3144 pa m3 mol-1 K-1

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R = PV nT

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P 1/V

(Boyle’s Law )

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n 1/T (in Kelvin)

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n P

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For a gas sample composed of CO(g) at 2 atm, Ar(g) at 3 atm and O2(g) at 1 atm:Ptotal = PCO (g) +PAr(g) + PO2(g)

= 2 atm + 3 atm + 1 atm = 6 atm

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Pmixture of gases = Pgas 1 +Pgas 2 + Pgas 3 + …

Dalton’s Law of Partial Pressures:

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V T (in Kelvin)

(Charles’ Law )

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n V

(Avogadro’s Law )

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P T (in Kelvin)

(Gay-Lussac’s Law)

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P1V1 = P2V2

T1 T2

COMBINED GAS LAW:

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Example 1: A student prepares chlorine gas in the chemistry lab where the temperature is 20 oC, and collects it in a gas jar over water. On that day the atmospheric pressure is 0.997 atm. At 20 oC water has a vapor pressure of 17.5 torrs. What was the actual press- ure of the chlorine collected?

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Given properties:

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temperature = 20 0C atmospheric pressure = 0.997 atm water vapor pressure at 200 = 17.5 torrs

Property to be found:

(ANALYSIS OF THE PROBLEM)

pressure of chlorine gas

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Ptot = P1 + P2 + P3 +…

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Ptot = Patm = Pchlorine + Pwater vapor

0.997 atm

= Pchlorine + 17.5 torr

Pchlorine = 0.997 atm - 17.5 torr

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Ptot = P1 + P2 + P3 +…

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Ptot = Patm = Pchlorine + Pwater vapor

Pchlorine = Patm - Pwater vapor

= 0.997 atm - 17.5 torr

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 0.997 atm x 760 torrs

1 atm

Pchlorine = 758 torrs – 17.5 torrs

= 758 torrs

= 740 torrs

Example 2: A sample of neon which occupies 40 mL at 800 torrs and 30 oC is transferred to another container in which it attains S.T.P. conditions. What is the volume of this new container?

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V1 = 40 mLV2 = x

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P1 = 800 torrs P2 = 760 torrs (standard

pressure)T1 = 30 oC T2 = 25 oC (standard

temperature)

P1V1 = P2V2

T1 T2 CLICK TO CONTINUE

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V1: mL V2: mLP1: torrs; P2: torrsT1 = (30 + 273) K = 303 K T2 = (25 + 273) K = 298 K

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800 torrs x 40 mL = 760 torrs x V2

300 K 298 KV2 = 800 torrs x 40 mL x 298 K 300 K x 760 torrs

P1V1 = P2V2 T1 T2

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V2 = P1V1T2

T1P2

P1V1 = P2V2 T1 T2

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= 800 torrs x 40 mL x 298 K

300 K x 760 torrs

V2 = 41.8 mL

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Example 3: A pure sample of CO2 was collected and stored in a

2.5-liter flask on a day when the atmospheric pressure was one atmosphere. The next day when the temperature in the storage area was 27 oC the gas was at a pressure of 776 mm Hg. At what temperature was the gas originally collected?

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V1 = 2.5 L V2 = 2.5 L

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P1 = 1.0 atm P2 = 776 mm Hg

T1 = x T2 = 27 oC

P1 = P2

T1 T2

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P1 = 1.0 atm = 760 mm Hg T2 = 27 oC = (27 + 273) K = 300 K

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EITHER

760 mm Hg = 776 mm Hg T1 300 K

T1 = 760 mm Hg x 300 K 776 mm Hg

OR

T1 = P1T2 P2

T1 = 760 mm Hg x 300 K 776 mm Hg

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T1 = 294 K

T1 = (294 – 273) oC = 21 oC

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Example 4: A pure sample of CO2 was collected at 299 K

and stored in a 2.5-liter flask on a day when the atmospheric pressure was 1.04 atmosphere. How much carbon dioxide was collected? CLICK TO CONTINUE

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T = 299 K

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P = 1.04 atmsn = xGas Law: PV = nRT (Ideal gas

law)

V = 2.5 L

R = 0.082053 L atm mol-1 K-1

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EITHER

1.04 atm x 2.5 L = n x 0.082053 atm L x 299 K mol K

n = 1.04 atm x 2.5 L 0.082053 atm L mol-1 K-1 x 299 K OR

n = PV RT

n = 1.04 atm x 2.5 L

0.082053 atm L mol-1 K-1 x 299 K

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n = 0.106 mol

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mass CO2

= n x molar mass = 0.106 mol x 44.010 g/mol = 4.67 g

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volume CO2 = n x molar volume at S.T.P.= 0.106 mol CO2 x 22.414 L/mol= 2.34 L CO2 at S.T.P.

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END OF TUTORIAL

SUMMARY OF GAS LAWSGas Law Equation

Avagadro’s n/V1 = n/V2

Boyle’s P1V1 = P2V2

Charles’ V1/T1 = V2/T2

Combined P1V1/T1 = P2V2 /T2

Dalton’s Ptot = P1 + P2 + P3 +…

Gay-Lussac’s P1/T1 = P2/T2

Ideal PV = nRTCLICK TO CONTINUE

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