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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)

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Presentation on theme: "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)"— Presentation transcript:

1

2 PV = nRT

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

4 H one mole of any gas at STP has a volume of 22.4 L 1 mole H 2 = 22.4 L at STP 1 mole CO 2 = 22.4 L at STP 1 mole CH 4 = 22.4 L at STP H2H2 CO 2 CH 4

5 H 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 L = 1 mole = mass (g) on periodic table

6 H one mole of any gas at STP has a volume of 22.4 L 1 mole H 2 = 22.4L at STP 1 mole CO 2 = 22.4 L at STP 1 mole CH 4 = 22.4 L at STP H2H2 CO 2 CH 4 2 g44 g16 g

7 Volume Ratios = Molar ratios in balanced equation H N 2 + 3H 2 2NH 3 H 1 mole nitrogen reacts with 3 moles hydrogen to produce 2 moles ammonia Or H 1 liter nitrogen reacts with 3 liters hydrogen to produce 2 liters ammonia

8 Remember Combined Gas law H P 1 V 1 = P 2 V 2 T 1 T 2 Since number of molecules is directly proportional to volume, it too goes on bottom of equation H P 1 V 1 = P 2 V 2 T 1 n 1 T 2 n 2

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

10 PV = nRT H P = pressure (atm or kPa) H V = volume (L or dm 3 ) H n = number of moles H R = universal gas constant H T = temperature (K)

11 Universal Gas Constant H R = L-atm mole-K H R = 8.31 L-kPa mole-K (1 dm 3 = 1 L) Units in equation must all match

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

13 PV = nRT H V = 20.0 L N 2 H P = atm H T = 27 o C +273 = 300 K H R= H n = ? L-atm Or L-kPa mole-K mole-K

14 PV= nRT H (200 atm)(20 L N 2 ) = n (.0821 atm-L)(300 K) K-mol H (200 atm)(20 L N 2 ) = n (.0821 atm-L)(300 K) K-mol n = 162 mol N 2

15 How many grams N 2 is 162 mol? H Periodic table mass of nitrogen = 14 g H N 2 = 28 g H 162 mol ( 28 g ) 1 mol N 2 = 4,536 g N 2

16 Sample Problem 2 H A deep underground cavern contains 2.24 x 10 6 L of methane gas (CH 4 ) at a pressure of 15.0 atm and a temperature of 42 o C. How many grams of methane does this natural gas deposit contain?

17 Y V = 2.24 x 10 6 L CH 4 Y P = 15.0 atm Y T = 42 o C +273 = 315 K Y R =.0821 atmxL/molxK Y ? grams

18 PV = nRT or PV = n RT H (15.0 atm)(2.24 x 10 6 L CH 4 ) = n (315K)(.0821 atm-L/K-mol) n = 1.30 x 10 6 mol CH 4 CH 4 = 16g 16g 1 mol CH 4 = 2.08 x 10 7 g

19 A really, really difficult one now! H How many liters of ammonia (NH 3 ) gas would be produced at STP if 5.00 g of hydrogen reacts with excess nitrogen? H N 2 + 3H 2 2NH 3

20 PV = nRT H At STP volume of NH 3 produced? H P = 1.00 atm H V = ? H R =.0821 atm-L/K-mol H T = 273 K H n = 5g H 2 ___mol NH 3 ?

21 N 2 + 3H 2 2NH 3 H 5 g H 2 (1 mol H 2 )(2 mol NH 3 ) 2 g 3 mol H 2 H 1.67 mol NH 3

22 P = 1.00 atm, V = ? n = 1.67 mol NH 3 R =.0821 atm-L/K-mol T = 273 K H V = nRT P H (1.67 mol)(.0821 atm-L/K-mol )(273K) (1.00 atm) H V = 37.4 L


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