1. PV = nRT

2. Molar Volume Avogadro’s Principle -
equal volumes of gases at same T and P contain equal numbers of particles (equal number of moles)

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. PV = nRT P = pressure (atm or kPa) V = volume (L or dm3)
n = number of moles
R = universal gas constant
T = temperature (K)

10. 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

11. 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?

12. 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

13. 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

14. 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

15. 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?

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

17. 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

18. 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

19. 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 ?

20. N 2 + 3H2  2NH3 5 g H2 (1 mol H2)(2 mol NH3) 2 g 3 mol H2

21. 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