L25 - "Kinetic Theory of Gases" MQ Chapter 27...

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“Kinetic Theory of Gases” MQ Chapter 27 Prof. GianluigiVeglia
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Let’s recall the definition of an ideal gas: The gas molecules which have mass m and diameter d are in constant motion The molecules are small compared to the distances they travel between collisions The molecules do not interact except when they collide and then the collisions are perfectly elastic.
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Physical interpretation of gas pressure In this vessel, the velocity of a molecule of mass m has components: u 1x , u 1y, u 1z . The particle moves along the direction of the vector and collides with the surface of the box. Assuming an elastic collision: The time elapsed between the collision is (the molecule needs to travel a distance 2a) x x x x mu mu mu mu 1 1 1 1 2 ) ( ) ( = = The momentum is conserved x u a t 1 / 2 =
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Applying Newton’s law: Where F is the force that the molecule exerts on the wall. From the force it is possible to calculate the pressure: Where V=abc is the volume of the box. The total pressure is given by The average value of u 2 is F a mu u a mu t mu x x x x = = = 2 1 1 1 1 / 2 2 ) ( V mu abc mu bc F P x x 2 1 2 1 1 = = = = = = = = = N j jx N j x N j j u V m V mu P P 1 2 1 2 1 1 = = N j jx x u N u 1 2 2 1
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If we substitute this into the expression for the pressure: Since the motion is isotropic and the total speed is: Then . This implies that each of the component contributes equally to the total velocity: The so-called equipartition of the kinetic energy. If we substitute this in the expression for the pressure: Now we have linked macroscopic properties to the molecular properties. 2 x u Nm PV = 2 2 2 z y x u u u = = 2 2 2 2 z y x u u u u + + = 2 2 2 2 z y x u u u u + + = 2 2 2 2 3 1 z y x u u u u = = =
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On page 700 of MQ, using partition functions we learned that the average translational energy per molecule is 3/2 k B T or per mole of gas 3/2RT (multiplying by Avogadro’s number). Since N A m=M (molar mass of the gas): We can use the expression to determine the average speed of a gas: the units here are m 2 s -2 , to have m s -1 we need to take the root square: This is called root-mean-square-speed. T k u m B 2 3 2 1 2 = RT u m N A 2 3 2 1 2 = RT u M 2 3 2 1 2 = RT u M = 2 3 1 M RT u 3 2 =
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The theoretical treatment of the distribution of molecular speeds was due to Maxwell and refined by Boltzmann.
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This note was uploaded on 10/30/2011 for the course CHEM 3501 taught by Professor Blank during the Fall '08 term at Minnesota.

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L25 - "Kinetic Theory of Gases" MQ Chapter 27...

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