KINETIC THEORY OF GASES

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1 LECTURE 8 KINETIC THEORY OF GASES Text Sections 0.4, 0.5, 0.6, 0.7 Sample Problems 0.4 Suggested Questions Suggested Problems Summary None 45P, 55P Molecular model for pressure Root mean square (RMS) speed Average kinetic energy Mean free path of molecules Distribution of molecular speeds Specific objectives Explain the molecular origin of pressure Understand and use Equations 0-17 and 0-18 Explain the concept of mean free path Describe the distribution of molecular speeds in a gas

2 LECTURE 8 KINETIC THEORY OF GASES Molecular model for pressure In the last two lectures we discussed a molecular model based on the assumptions: the gas is composed of a very large number of identical molecules each of which is an aggregate of one or more atoms. the molecules move continuously in accordance with Newton s Laws of Motion. the pressure of the gas is due to collisions of the molecules with one another and with the walls. Assume further that: the collisions are elastic and short range, i.e. they occur only when the molecules are close to one another. the number of molecules is very large so that statistical averages are meaningful. Consider n moles of an ideal gas in a cubical box with side length L (see Fig 0-3). The walls of the box are at temperature T. When a typical molecule collides with the right-hand wall it exerts an impulsive force and its momentum changes by p = ( m v x ) ( mv x ) = mv x where v x is the component of velocity perpendicular to the wall. Momentum m v x is transferred to the wall. If the molecule does not collide with another molecule (a dubious assumption except in a vacuum, as will be seen later) it will collide again with the right-hand wall after a time t = L / v x.. The time averaged force exerted by the molecule is equal to the rate at which momentum is transferred to the wall by the molecule: F x = mv x t = mv x L The time averaged pressure exerted by the molecule on the wall is p molecule = F x L = m v x L 3

3 The total pressure due to all molecules is p = m v x m = n N A L v 3 x L 3 where a bar over a quantity denotes the average of that quantity for all molecules. For a typical molecule v = v x + v y + v z By symmetry v x = v y = v z = 1 3 v and since V = L 3 p = n N A mv = nmv where M = N A m is the mass of a mole of the gas. This equation relates the macroscopic quantity p to the underlying molecular quantity v and to other macroscopic quantities, n, M and V. Root mean square speed *** The square root of v is called the root mean square speed v RMS.. The above equation for p and the Ideal Gas Law give p = nmv = nrt V or v RMS = v = 3RT M or v RMS = 3R T M. For a given ideal gas the root mean square speed of the molecules is proportional to the square root of the temperature. Some values of v RMS for gases at room temperature are given in Table 0-1. The values decrease as the masses of the gas molecules increase.

4 They are larger than the corresponding speeds of sound in the same gases at the same temperature. (In fact v SOUND = v RMS γ - no proof here.) Average translational kinetic energy The average translational kinetic energy of the molecules is K = 1 mv = 1 mv RMS = 3 m R M T = 3 k T. This is the same result as the one obtained much less rigorously using the Equipartition of Energy Theorem. Remember that there are 3 translational degrees of freedom. The equation relates the macroscopic quantity T to the molecular quantity K. Mean free path of molecules *** A typical molecule in a gas follows a zig-zag path as a result of collisions with other molecules (see Fig 0-4). The mean free path, λ, is the average distance travelled by a molecule in a gas between such collisions. λ is inversely proportional to the number of gas molecules per unit volume. For example, λ for air at sea level is about 10-7 m. A typical molecule undergoes more than 10 9 collisions each second. At higher altitudes the number of molecules per unit volume is much less, there are fewer collisions, and λ is much longer. In the Earth s ozone layer at a typical altitude of 5 km, λ is about 100 times larger. Because of these very short mean free paths and high collision rates a typical molecule takes much longer than its high speed would suggest to move a significant distance from its starting point. For example, if no convective motion occurs, carbon dioxide molecules on average take about 70 hours to travel 1 metre in the air at sea level! This process is called diffusion. Diffusion is an efficient mechanism for moving matter only over very short distances, e.g. within cells in living organisms.

5 Distribution of molecular speeds The molecules in a gas in equilibrium at temperature T have a wide range of speeds. The probability of finding a molecule with any specified speed is given by the Maxwell speed distribution (see Fig 0-7). For any given temperature this distribution peaks at a speed slightly less than v RMS. The molecules in a liquid also have a range of speeds (but not the Maxwell distribution). They are much closer together than gas molecules and hence move in an attractive potential well created by the surrounding molecules. Those with high speeds have sufficient energy to escape from the potential well and break through the liquid surface, i.e. to evaporate.

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