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Chapter 5: Gases. Substances that exists as Gases Diatomic Elements – H 2 – N 2 – O 2 – F 2 – Cl 2 Other States – O 3 -Ozone – He – Ne – Ar – Kr – Xe.

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Presentation on theme: "Chapter 5: Gases. Substances that exists as Gases Diatomic Elements – H 2 – N 2 – O 2 – F 2 – Cl 2 Other States – O 3 -Ozone – He – Ne – Ar – Kr – Xe."— Presentation transcript:

1 Chapter 5: Gases

2 Substances that exists as Gases Diatomic Elements – H 2 – N 2 – O 2 – F 2 – Cl 2 Other States – O 3 -Ozone – He – Ne – Ar – Kr – Xe – Rn Compounds – HF – HCl – CO – CH 4 – NH 3 – NO-Nitric Oxide – NO 2 -Nitric Dioxide – N 2 O-Nitrous oxide – SO 2 – H 2 S – HCN

3 Pressure of a Gas Before we can investigate how to solve gas equations, we need to know specific values of the SI measurements. – V=D/t – Acceleration=∆V/ ∆t – F=MA (measured in Newtons=1kg●m/s 2 ) – Pressure=F/area (measured in Pascals= 1 N/m 2 )

4 Atmospheric Pressure Defined as-“the force experienced by an area to the Earths atmosphere is equal to the weight of the column of air above it.” – Measured in multiple units, here are the conversion factors: 1 torr=1mmHg 1 atm=760 mm Hg (standard atmospheric pressure) 1 atm = 101, 325 Pascal (Pa) – 1 atm = 1.01325 x 10 5 Pa – 1 atm = 1.01325 x 10 2 kPa

5 Some Practice Problems Convert 688 mmHg to atmospheres – Pressure = 688 mmHg x 1 atm/760 mmHg Convert 749 mmHg to atm. The atmospheric pressure in San Francisco is 732 mmHg, What is the pressure in kPa?

6 The Gas Laws Boyle’s Law – Presents the inverse relationship between pressure and volume. In other words, when pressure increases, volume decreases. Therefore, P= k 1 x 1/V k = the proportionality constant (this will either be given to you or you will have to calculate it). – k= PV – Before and after equation: P 1 V 1 =P 2 V 2 – Any unit of volume can be used as long as the same unit is used on both sides of the equation.

7 Charles Law (Charles and Gay-Lussac’s Law)

8 The Ideal Gas Equation Defines the relationship between Pressure, volume, and Temperature. Ideal gas- “a hypothetical gas whose pressure- volume-temperature behavior can be completely accounted for by the ideal gas equation” – PV = nRT

9 The Ideal Gas Equation

10 Ideal conditions Standard Temperature and pressure – 0° Celsius=273° Kelvin – 1 atm.

11 Density of a Gas

12 For Example

13 Next Example

14 Daltons Partial Pressure Partial Pressures=the pressures of individual gas components. Simply put, total pressure in a multi-gas system is equal to the sum of all individual gas pressures. – P TOTAL =∑P 1  ∞ – REMEMBER THAT PV=nRT, so the total Pressures = nRT/V

15 But it doesn’t stop there!!!

16 Examples A mixture contains 4.46 moles of Ne,.74 moles of Ar, and 2.15 moles of Xe. Calculate the partial pressures of the gasses if the total pressure is 2.00 atm.(the question does not mention Temperature/volume so ignore it for now). – First calculate mole fractions (X Ne, X Ar, X Xe ) – P T =? – P i =X i* P T

17 Example #2 A mixture of gases contains.31 moles of CH 4,.25 moles of C 2 H 6, and.29 moles of C 3 H 8. The total pressure is 1.50atm. Calculate the partial pressures. – First calculate mole fractions (X CH 4, X C 2 H 6, X C 3 H 8 ) – P T =? – P i =X i* P T

18 Homework. READ READ READ READ READ READ READ READ READ READ READ READ the chapter on Gas Stoichiometry. – Work practice problems on your own. I am not going to assign you homework this week, take it upon YOURSELF to challenge YOURSELF. Elite University professors rarely assign homework unless they are insane. My job is to prepare you for this. – If I do not have your lab report by the end of today it is a zero (unless of course you attach a mole dollar). – We will do this weeks lab in class tomorrow, it is very quick.

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