PV = nRT Molar Volume H Avogadros Principle - equal volumes of gases at same T and P contain equal numbers of particles (equal number of moles)

PV = nRTPV = nRT

Molar VolumeMolar VolumeAvogadro’s Principle - equal volumes of gases at same

T and P contain equal numbers of particles (equal number of moles)

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

H2CO2 CH4

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

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

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

N 2 + 3H2 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

Remember Combined Gas lawRemember Combined Gas law

P1V1 = P2V2

T1 T2Since number of molecules is directly proportional

to volume, it too goes on bottom of equation

P1V1 = P2V2

T1n1 T2n2

P1V1 = constantT1n1

Plug in standard temperature,pressure, volume, & # of moles to find constant

(1atm)(22.4L) = constant (273K)(1mol)

.0821 atm L = constant K . Mol

PV = nRTPV = nRTP = pressure (atm or kPa)V = volume (L or dm3)n = number of molesR = universal gas constantT = temperature (K)

Universal Gas ConstantUniversal Gas ConstantR = 0.0821 L-atm

mole-KR = 8.31 L-kPa

mole-K (1 dm3 = 1 L)

Units in equation must all match

Sample ProblemSample Problem

A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of 200.0 atm at 27oC. How many moles of gas does the cylinder hold?

PV = nRTPV = nRT

V = 20.0 L N2

P = 200.0 atmT = 27oC +273 = 300 K

R=

n = ?

0.0821 L-atm Or 8.315 L-kPa mole-K mole-K

PV= nRTPV= 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) K-mol

n = 162 mol N2

How many grams NHow many grams N2 2 is 162 mol?is 162 mol?

Periodic table mass of nitrogen = 14 g N2 = 28 g

162 mol ( 28 g ) 1 mol N2

= 4,536 g N2

Sample Problem 2Sample 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?

V = 2.24 x 106 L CH4

P = 15.0 atmT = 42oC +273 = 315 KR = .0821 atmxL/molxK? grams

PV = nRT or PV = nPV = nRT or PV = n RT 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

16g1 mol CH4

= 2.08 x 107 g

A really, really difficult one now!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 2 + 3H2 2NH3

PV = nRT PV = nRT

At STP volume of NH3 produced?

P = 1.00 atmV = ?R = .0821 atm-L/K-molT = 273 Kn = 5g H2 ___mol NH3

?

N N 2 2 + 3H+ 3H22 2NH 2NH33

5 g H2 (1 mol H2)(2 mol NH3) 2 g 3 mol H2

1.67 mol NH3

P = 1.00 atmP = 1.00 atm, , V = ?V = ? n = 1.67 mol NHn = 1.67 mol NH33

R = .0821 atm-L/K-mol R = .0821 atm-L/K-mol T = 273 KT = 273 K

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

V = 37.4 L