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Chapter 14Gases

The Combined Gas LawVolume and Moles (Avogadro’s Law)Partial Pressures

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Combined Gas LawP1V1 = P2V2

T1 T2

Rearrange the combined gas law to solve for V2

P1V1T2 = P2V2T1 V2 = P1V1T2

P2T1

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Learning Check C1

Solve the combined gas law for T2.

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Solution C1

Solve the combined gas law for T2. (Hint: cross-multiply first.)P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

T2 = P2V2T1

P1V1

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Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

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Data Table

Set up Data Table

P1 = 0.800 atm V1 = 0.180 L T1 = 302 K

P2 = 3.20 atm V2= 90.0 mL T2 = ????

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Solution

Solve for T2Enter data

T2 = 302 K x atm x mL = K atm mL

T2 = K - 273 = °C

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Calculation

Solve for T2

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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Learning Check C2

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

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Solution G9

T1 = 308 K T2 = ?

V1 = 675 mL V2 = 0.315 L = 315 mL

P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg

T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec

= 178 K - 273 = - 95°C

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Avogadro’s Law

When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas

V1 = V2

n1 n2 initial final

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Daltons’ Law of Partial Pressures

Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container

Dalton's Law of Partial PressuresThe total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.

PT = P1 + P2 + P3 + .....

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Gases in the AirThe % of gases in air Partial pressure (STP)

78.08% N2 593.4 mmHg

20.95% O2 159.2 mmHg

0.94% Ar 7.1 mmHg

0.03% CO2 0.2 mmHg

PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2

Total Pressure 760 mm Hg

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Learning Check C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) 557 2) 9.14 3) 0.109

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Solution C6

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?2) 156

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) 557

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Ideal Gas Law

The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written

PV = nRT

R = ideal gas constant

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Ideal Gases

Behave as described by the ideal gas equation; no real gas is actually ideal

Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less

In real gases, particles attract each other reducing the pressure

Real gases behave more like ideal gases as pressure approaches zero.

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PV = nRT

R is known as the universal gas constant

Using STP conditions P V

R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K)

n T = 0.0821 L-atm

mol-K

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Learning Check G15

What is the value of R when the STP value for P is 760 mmHg?

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Solution G15

What is the value of R when the STP value for P is 760 mmHg?

R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K)

= 62.4 L-mm Hg mol-K

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Learning Check G16

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?

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Solution G16

Set up data for 3 of the 4 gas variables Adjust to match the units of R

V = 20.0 L 20.0 L

T = 23°C + 273 296 K

n = 2.86 mol 2.86 mol

P = ? ?

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Rearrange ideal gas law for unknown P

P = nRT V

Substitute values of n, R, T and V and solve for P

P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol)

= 2.64 x 103 mm Hg

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Learning Check G17

A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

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Solution G17

Solve ideal gas equation for n (moles)n = PV

RT

= (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K)

= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2

1 mol O2

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Density of a Gas

Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.

P = 1.00 atm T = 273 K

Rearrange the ideal gas equation for moles/L PV = nRT PV = nRT P = n RTV RTV RT V

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Substitute (1.00 atm ) mol-K = 0.0446 mol O2/L (0.0821 L-atm) (273 K)

Change moles/L to g/L

0.0446 mol O2 x 32.0 g O2 = 1.43 g/L 1 L 1 mol O2

Therefore the density of O2 gas at STP is 1.43 grams per liter