Download presentation

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

Similar presentations

Presentation is loading. Please wait....

OK

Chapter 12 The Behavior of gases

Chapter 12 The Behavior of gases

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on linear equations in two variables worksheet Ppt on pricing policy objective Ppt on travels and tourism Ppt on aircraft emergencies with atc Ppt on properties of rational numbers class 8 Ppt on campus recruitment system analysis Project ppt on introduction to business finance Ppt on submerged arc welding Ppt on motivational techniques Ppt on beer lambert law graph