Presentation is loading. Please wait.

Presentation is loading. Please wait.

Gases Chapter 9. What parameters do we use to describe gases? pressure: force/unit area 1 atm = 101 kPa; volume: liters (L) Temperature: K.

Similar presentations


Presentation on theme: "Gases Chapter 9. What parameters do we use to describe gases? pressure: force/unit area 1 atm = 101 kPa; volume: liters (L) Temperature: K."— Presentation transcript:

1 Gases Chapter 9

2 What parameters do we use to describe gases? pressure: force/unit area 1 atm = 101 kPa; volume: liters (L) Temperature: K

3 What is meant by % volume? In principle, the actual molecular volume of a gas is so small in comparison to the volume it will occupy that we treat gases at mathematical points.

4 How do we measure pressure? above 1 atmosphere below 1 atmosphere A column of air 1 m 2 has a mass of 10,300 kg, producing a pressure of 101 kPa due to gravity (14.7 pounds/in 2 ) 1 atm = 76 cm Hg; 101 kPa

5 Boyle’s Law is concerned with the relationship of pressure and volume using a fixed amount of gas ( a fixed number of mols of gas) P*V = constant at constant temperature Avogadro’s Law is concerned with the relationship between the number of molecules or mols (n) and the volume of a gas under conditions of constant pressure and temperature V  n at constant pressure and temperature Charles’ Law is concerned with the relationship of temperature and volume when dealing with a constant amount of gas (mols) V  T when T is expressed in K. The K temperature scale is derived from the behavior of gases if V  T then V = kT where k is a constant at constant pressure

6 Ideal gas law: PV = nRT where R is a constant R = L. atm/K. mol Note that at constant n and T, PV = constant Boyle’s Law Note that at constant P and T V/n = constantAvogadro’s Law Note that at constant P and n, V/T = constant Charles’s Law

7 Standard conditions of pressure and temperature T = 0 °C (273 K) Pressure: 1 atm What volume does a mol of any ideal gas occupy at STP? PV = nRT V = 1mol( L*atm/K*mol)(273 K)/(1 atm) V = 22.4 L This means that equal volumes of gases under identical conditions of temperature and pressure contain equal number of molecules

8 What is the difference between an ideal gas and a real gas?

9 The ideal gas equation was generated from the kinetic theory of gases making the following assumptions 1.The molecules could be treated as points (ie molecular volume = 0) 2.There are no attractive interactions between molecules. 3.Gas particles move around at random 4.Collision of gas molecules with the wall are totally elastic 5.The kinetic energy of the gas particle is  to temperature (K) In general, the ideal gas law works best at low pressures and high temperatures

10 Real Gases: van der Waal’s equation (P + an 2 /V 2 )(V-nb) = n RT an 2 /V 2 corrects for intermolecular attractions nb corrects for the real volume of molecules

11 Dalton’s Law of partial pressures: Total atmospheric pressure = 1 atm; How much of the pressure is contributed by N 2 ?

12 Pressure is a consequence of molecules colliding with each other and the walls of the container *062

13 For air if If P T V = n T RT and n T = (n o2 + n N2 +...) at constant T, P T V = (n o2 + n N2 +...)RT Since the actual volume of the molecules is small in comparison to the volume occupied by the gas, all molecule occupy the same volume V. The contribution to the total pressure is dependent on the number of collision of each gas with the wall and this is dependent on the number of molecule of each gas. Hence: P = (P N2 + P O2 +...) P O2 V = n O2 RT ; P N2 V = n N2 RT...

14 Temperature: a measure of the average kinetic energy of molecules

15 Distribution of molecular speeds as a function of temperature

16 What are some of the consequences associated with the fact that molecules at the same temperature have different speeds? The size of the pinhole needs to be small *07 Diffusion: mixing of gasesEffusion: escape through a small opening

17 Two molecules of different mass at the same temperature effusing through an opening

18 Two molecules of different mass at the same temperature effusing through an opening

19 From the kinetic theory of gases speed of a molecule u = (3RT/M) 1/2 For two gases at the same temperature 1/2m a u a 2 = 1/2m b u b 2 u a = average speed of molecule a u b = average speed of molecule b m a /m b = u b 2 /u a 2 The rate at which molecule a hits the pinhole  u if the comparisons are made at the same concentration and temperature. u b /u a = (m a /m b ) 1/2

20 Solving some problems involving gases 1.A sample of gas at 25 °C and 2 atm pressure in a 5 L vessel was found to have a mass of 18 g. What is its molecular weight? PV = n RT 2 atm*5 L = n* (Latm/K mol)*298 K n = 10/(0.0821*298) mol; n = wt/ mw; = 18g/mw mw = 44 g/mol

21 Suppose the gas at the right exerted a pressure of 15 cm as shown. Would the pressure of the gas be greater or less than 1 atm? How many atm of pressure is the gas exerting? 15.2 cm 1 atm = 76 cm = 60.8

22 Suppose we have a sample of equal amounts of H 2 and D 2 in a vessel and a small opening is introduced. What will be the initial rates of effusion? u H2 /u D2 = m D2 /m H2 = (4/2) 0.5 = 1.42 Will the relative rate change with time?

23 What is the density of natural gas (CH 4 ) at STP? PV = nRT density is g/mL or g/L We know the molar volume of any gas is 22.4 L at STP How many g of methane in a mole? 16g/22.4 L = g/l or 7.14*10 -4 g/mL or PV =(wt/mw)*RT; mw*P/RT = (wt/V)

24 The surface temperature of Venus is about 1050 K and the pressure is about 75 Earth atmospheres. Assuming these conditions represent a “Venusian STP, what is the standard molar volume of a gas on Venus? PV = nRT 75 atmV =1mol*0.0821(Latm/K mol)*1050 K; V = 1.15 L

25 Natural gas is a mixture of a number of substances including methane (mol fraction, 0.94); ethane (mol fraction, 0.04); propane (mol fraction, 0.015). If the total pressure of the gases is 1.5 atm, calculate the actual pressure contributed by each of the gases described. mol fraction = mol A/(mol A + mol B +....) P T = 1.5 = P CH4 + P C2H P x V = n x RT n CH4 /n C2H6 = P CH4 /P C2H6 = 0.94/.04 CH 4 = 0.94*1.5 C 2 H 6 = 0.04*1.5 C 3 H 8 = 0.015*1.5


Download ppt "Gases Chapter 9. What parameters do we use to describe gases? pressure: force/unit area 1 atm = 101 kPa; volume: liters (L) Temperature: K."

Similar presentations


Ads by Google