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# Chapter 5 Continued for Exam 3 Chem 109

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This 6 page Class Notes was uploaded by mkennedy24 on Monday February 29, 2016. The Class Notes belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 44 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.

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Date Created: 02/29/16

Chapter 5 (continued for exam 3) Section 5.8: Kinetic Molecular Theory Moles for Ideal Gas Behavior Objective: Describe parts and apply where needed. Kinetic Molecular Theory: o Postulate 1: Particle size is so small it can be ignored. Particles themselves occupy no volume but still have a mass Ar(gas) V ArV T . 01% Space taken up by molecules/ o Postulate 2: Average Kinetic Energy atoms is proportional to temperature in Kelvins At any given moment, some particles **Rest of space (.99%) are moving faster than others (there is is space between distribution of velocities) atoms** Relationship to temperature is related to velocity: a hot(greater average kinetic energy) gas moves faster than a cold gas(lesser average kinetic energy) 1 2 KE= m v → where m is the mass and v is the 2 velocity o Postulate 3: Collision are completely elastic No energy loss in a collision only exchanges Elastic Collision Inelastic Collision Boyles Law: With constant number of particles at constant temperature, the volume of the gas is inversely proportional to its pressure (V ↑ , P ↓ or vice versa) **Decreasing volume forces particles to occupy a smaller space. As long as temperature stays the same, the number of collisions increase, resulting in greater pressure. Charles Law: With constant number of particles at constant pressure, the volume of a gas is proportional to temperature (V ↑ , T ↓ ) **As temperature increases 100 200 particles speed increases as ℃ ℃ well. In order for pressure to remain constant, volume must increase. Which makes volume and temperature directly proportional. **Note**: temperatures used Longer arrows indicate faster moving particles. in visuals are hypothetical Avogadro’s Law: With constant temperature and pressure, the volume of gas is proportional to number of particles. **When the number of particles are increased in a gas sample, the number of collisions increase. The Dalton’s Law: P TotalAP +B +C n only way for pressure to remain constant is for volume to increase. Total pressure of a gas mixture is the sum of partial pressures of it components Kinetic Molecular Theory and Ideal Gas Law Kinetic molecular theory is a quantitive model that implies PV=nRT The pressure on a wall of a container occupied by particles in constant motion is the total force on the wall(due to collisions) divided by the area of the wall: FTotal P= A Force imparted for each collision: ∆V ∆T ) F =m¿ If a particle collides elastCollisionth the wall, it bounces off the wall with no loss of energy. For a straight line collision, the change in velocity is 2v (the particles velocity was v before the collision and –v after the collision, there for 2v) The force per collision is: Number of collisions is proportional to number of particles within 2v ∆T ) v∆T : F Collision n is #of particles ) V is volume Volume Density of particles Total Force: Is equal to ( Force )(# of collisions) Collision The pressure on the wall is equal to the total force divided by the surface area of the wall: 2 Average Kinetic Energy (1/2mv ) is proportional to temperature in Contain Avogadro’s LC:ontains Boyles Law: Vn P 1/V Kelvins: mv ∝T By combining the P∝ T×n ∨PV ∝nT above: V Adding the constant R and the equal sign(=), we get PV=nRT These equations above are just showing how the kinetic molecular theory implies the Ideal Gas Law. NONE of the above needs to be memorized except PV=nRT and the kinetic molecular theory. Temperature and Molecular Velocities Particles of different masses have same average kinetic energy at given temperatures 1 2 KE= m2v o The only way for particles of different masses to have the same average kinetic energy is for them to have different velocities In a gas mixture at a given temperature, lighter particles travel faster on average than heavier ones Root Mean Square Velocity (urms of a particle: 2 2 urms√u´ where u´ is the average of the squares of the particles velocities Mean Square Velocity: R=8.314 Product of NAm where NA is avogadro’s number and m is mass results in kg/mol Example: Calculate the root mean square velocity of oxygen molecules at 25 ℃ J= T=25+273=298K 2 2 kg∙2 The 2xponent in m 32.00gO2 1kg s −3 because of theut M= 1molO × 1000g=32.00×10 kgO 2mol radical 2 J 2 3(8.31mol∙K )(298K) 2.32×10 J 2.32×105kg∙m urms −3 s2 32.00×10 kgO2 √ kg kg √ 1mol √ 482 m/s Section 5.9: Mean Free Path, Diffusion, and Effusion of Gases Mean Free Path: Average distance that a molecule travels between collisions o Increases with decreasing pressure Diffusion: The process by which gas molecules spread out in response to a concentration gradient o Root mean square velocity influences rate of diffusion o Heavy molecules diffuse slower than lighter molecules o You are two meters away from your friend who just sprayed perfume. The first molecule that is smelt in the perfume mixture is the lighter one. Effusion: The process by which a gas escapes from a container into a vacuum through a small hole o Rate of effusion is also related to root mean square 1 rate∝ √M velocity o The ratio of effusion rates of two different gases is given by Graham’s Law of Effusion: Example: An unknown gas effuses at a rate that is 0.462 times that of nitrogen gas (at same temperature). Calculate molar mass of the unknown gas in g/mol. ratunknown Given: rate = 0.462 N 2 rateA= M B ratB √ M A ratA= M B→ M = M N 2 → 28.02g/mol→131g/mol ratB √M A unk rate 2 0.4622 ( un) ratN 2 Section 5.10: Real Gases: The effects of Size and Intermole Forces Gases behave ideally when: o (a) The volume of the gas particles is small compared to the space between them o (b) The forces between the gas particles are not significant The effect of finite volume of gas particles o Finite volume actual size o Van der Waals correction: Ideal Behvaior: V=nRT/P Corrected for Volume of Gas Particles: V=nRT/P +(nb) Where n is number of moles Where b is a constant that depends on gas Effect of Intermolecular Forces o Van de Waals correction factor that accounts for intermolecular forces between gas particles: Ideal Behavior: P=nRT/V n Corrected for Intermolecular forces: P=nRT/V –a( V 2 ) Where n is number of moles Where V is Volume Where a is a constant that depends on the gas n **Correction factor increases as V increases because a greater concentration of particles makes it more likely that they will interact with one Rearranging cor2ected equations as: P+a( n ) =nRT V V nRT V−nb= P Van Der Waals Equation: (combo of both corrected) 2 P+a n × [−nb ]nRT [ (V) ]

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