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PV = nRT

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**Molar Volume Avogadro’s Principle -**

equal volumes of gases at same T and P contain equal numbers of particles (equal number of moles)

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**one mole of any gas at STP has a volume of 22.4 L**

1 mole H2 = 22.4 L at STP 1 mole CO2 = 22.4 L at STP 1 mole CH4 = 22.4 L at STP CO2 CH4 H2

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**No!!!!!!! Molar mass is equal to mass on periodic table.**

Would the mass of this 22.4 L of gas be the same for all gases? No!!!!!!! Molar mass is equal to mass on periodic table. 22.4 L = 1 mole = mass (g) on periodic table

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**one mole of any gas at STP has a volume of 22.4 L**

1 mole H2 = 22.4L at STP 1 mole CO2 = 22.4 L at STP 1 mole CH4 = 22.4 L at STP H2 CO2 CH4 2 g 44 g 16 g

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**Volume Ratios = Molar ratios in balanced equation**

N H2 2NH3 1 mole nitrogen reacts with 3 moles hydrogen to produce 2 moles ammonia Or 1 liter nitrogen reacts with 3 liters hydrogen to produce 2 liters ammonia

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**Remember Combined Gas law**

P1V1 = P2V2 T T2 Since number of molecules is directly proportional to volume, it too goes on bottom of equation P1V1 = P2V2 T1n1 T2n2

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**P1V1 = constant T1n1 (1atm)(22.4L) = constant (273K)(1mol)**

Plug in standard temperature,pressure, volume, & # of moles to find constant (1atm)(22.4L) = constant (273K)(1mol) .0821 atm . L = constant K . Mol

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**PV = nRT P = pressure (atm or kPa) V = volume (L or dm3)**

n = number of moles R = universal gas constant T = temperature (K)

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**Universal Gas Constant**

R = L-atm mole-K R = 8.31 L-kPa mole-K (1 dm3 = 1 L) Units in equation must all match

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Sample Problem A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of atm at 27oC. How many moles of gas does the cylinder hold?

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**PV = nRT V = 20.0 L N2 P = 200.0 atm T = 27oC +273 = 300 K R= n = ?**

L-atm Or L-kPa mole-K mole-K

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**n = 162 mol N2 PV= nRT (200 atm)(20 L N2) = n (.0821 atm-L)(300 K)**

K-mol (200 atm)(20 L N2) = n (.0821 atm-L)(300 K) n = 162 mol N2

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**How many grams N2 is 162 mol? 162 mol ( 28 g ) 1 mol N2 = 4,536 g N2**

Periodic table mass of nitrogen = 14 g N2 = 28 g 162 mol ( 28 g ) 1 mol N2 = 4,536 g N2

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Sample Problem 2 A deep underground cavern contains 2.24 x 106 L of methane gas (CH4) at a pressure of 15.0 atm and a temperature of 42oC. How many grams of methane does this natural gas deposit contain?

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V = 2.24 x 106 L CH4 P = 15.0 atm T = 42oC = 315 K R = atmxL/molxK ? grams

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**PV = nRT or PV = n RT (15.0 atm)(2.24 x 106 L CH4) = n**

(315K)(.0821 atm-L/K-mol) n = 1.30 x 106 mol CH4 CH4 = 16g 16g 1 mol CH4 = 2.08 x 107 g

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**A really, really difficult one now!**

How many liters of ammonia (NH3) gas would be produced at STP if 5.00 g of hydrogen reacts with excess nitrogen? N H2 2NH3

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**PV = nRT P = 1.00 atm V = ? R = .0821 atm-L/K-mol T = 273 K**

At STP volume of NH3 produced? P = 1.00 atm V = ? R = atm-L/K-mol T = 273 K n = 5g H2 ___mol NH3 ?

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**N 2 + 3H2 2NH3 5 g H2 (1 mol H2)(2 mol NH3) 2 g 3 mol H2**

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**P = 1.00 atm, V = ? n = 1.67 mol NH3 R = .0821 atm-L/K-mol T = 273 K**

V = nRT P (1.67 mol)(.0821 atm-L/K-mol )(273K) (1.00 atm) V = 37.4 L

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Chapter 14 The Ideal Gas Law and Its Applications

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