The Kinetic Theory of Gases.ppt - Chapter 4 The Kinetic...

This preview shows page 1 out of 44 pages.

Unformatted text preview: Chapter 4 The Kinetic Theory of Gases Heat. Joule's experiment. Equivalence of heat and work CΔT=mgh J = 427 kgf·m/kcal. 4.27 j--------> 1 cal The description of the behavior of a gas in terms of the macroscopic state variables P, V, and T can be related to simple averages of microscopic quantities such as the mass and speed of the molecules in the gas. The resulting theory is called the Kinetic Theory of Gases. From the point of view of kinetic theory, a gas consist of a large number of molecules making elastic collisions with each other and with the walls of container. In the absence of external forces (we may neglect gravity), there no preferred position of the molecule in the container, and no preferred directions for it’s velocity vector. The molecules are separated, on the average, by distances that are large compared with their diameters, and they exert no forces on each other except when they collide. This final assumption is equivalent to assuming a very low gas density, which is the same as assuming that the gas is an ideal gas. Because momentum is conserved, the collisions the molecules make with each other have no effect on the total momentum in any directions – thus such collisions may be neglected. Molecular Model of an Ideal Gas • The model shows that the pressure that a gas exerts on the walls of its container is a consequence of the collisions of the gas molecules with the walls • It is consistent with the macroscopic description developed earlier Assumptions for Ideal Gas Theory • The gas consist of a very large number of identical molecules, each with mass m but with negligible size (this assumption is approximately true when the distance between the molecules is large, compared to the size). The consequence → for negligible size molecules we can neglect the intermolecular collisions. • The molecules don’t exert any action-at-distance forces on each other. This means there are no potential energy changes to be considered, so each molecules kinetic energy remains unchanged. This assumption is fundamental to the nature of an ideal gas. Assumptions for Ideal Gas Theory • The molecules are moving in random directions with a distribution of speeds that is independent of direction. • Collisions with the container walls are elastic, conserving the molecule’s energy and momentum. Pressure and Kinetic Energy • Assume a container is a cube Edges are length d • Look at the motion of the molecule in terms of its velocity components • Look at its momentum and the average force Pressure and Kinetic Energy • Since the collision is elastic, the y component of molecule’s velocity remains unchanged, while the x component reverses sign. Thus the molecule undergoes the momentum change of magnitude 2mvx. • After colliding with the right hand wall, the x - component of molecule’s velocity will not change until it hits the left-hand wall and its x - velocity will again reverses. Δt = 2d / vx Pressure and Kinetic Energy • The average force, due to the each molecule on the wall: 2m vx mv x2 p Fi t (2d / v x ) d • To get the total force on the wall, we sum over all N molecules. Dividing by the wall area A then gives the force per unit area, or pressure: mv x2 2 2 F m v m v F i x x d P A A A Ad V Pressure and Kinetic Energy P Since, 2 v x N m v V 2 x mN V v 2 x N is just the average of the squares of x – components of velocities: mN 2 P vx V Pressure and Kinetic Energy Since, the molecules are moving in random directions the 2 2 2 v v vx , average quantities y , and z must be equal and the average of the molecular speed 2 2 x 2 y v v v v and v2 = 3vx2, pressure: or 2 z vx2 = v2/3 . Then the expression for mN 2 P v 3V Pressure and Kinetic Energy mN 2 • The relationship P v can be written: 3V 2 N P 3V ___ 2 1 mv 2 • This tells us that pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules Pressure and Kinetic Energy • This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed • One way to increase the pressure is to increase the number of molecules per unit volume • The pressure can also be increased by increasing the speed (kinetic energy) of the molecules A 2.00-mol sample of oxygen gas is confined to a 5.00-L vessel at a pressure of 8.00 atm. Find the average translational kinetic energy of an oxygen molecule under these conditions. A 2.00-mol sample of oxygen gas is confined to a 5.00-L vessel at a pressure of 8.00 atm. Find the average translational kinetic energy of an oxygen molecule under these conditions. 2 2N mv P 3V 2 Kav Kav mv2 3PV where N nNA 2NA 2 2N 3PV 2 2NA 3 8.00 atm 1.013105 Pa atm 5.00 10 3 m3 2 2 mol 6.021023 molecules mol Kav 5.0510 21 J molecule Molecular Interpretation of Temperature • We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation of state for an ideal gas 2 N 1 2 Nk B T P m v 3 V 2 V • Therefore, the temperature is a direct measure of the average molecular kinetic energy Molecular Interpretation of Temperature • Simplifying the equation relating temperature and kinetic energy gives ___ 2 1 3 m v k BT 2 2 • This can be applied to each direction, 1 ___2 1 m v x k BT 2 2 with similar expressions for vy and vz A Microscopic Description of Temperature • Each translational degree of freedom contributes an equal amount to the energy of the gas • A generalization of this result is called the theorem of equipartition of energy Theorem of Equipartition of Energy • Each degree of freedom contributes ½kBT to the energy of a system, where possible degrees of freedom in addition to those associated with translation arise from rotation and vibration of molecules Total Kinetic Energy • The total kinetic energy is just N times the kinetic energy of each molecule K tot trans 1 ___2 3 3 N m v Nk BT nRT 2 2 2 • If we have a gas with only translational energy, this is the internal energy of the gas • This tells us that the internal energy of an ideal gas depends only on the temperature Molar Specific Heat • Several processes can change the temperature of an ideal gas • Since ΔT is the same for each process, ΔEint is also the same • The heat is different for the different paths • The heat associated with a particular change in temperature is not unique Molar Specific Heat • We define specific heats for two processes that frequently occur: – Changes with constant volume – Changes with constant pressure • Using the number of moles, n, we can define molar specific heats for these processes Molar Specific Heat • Molar specific heats: Q = nCV ΔT for constant-volume processes Q = nCP ΔT for constant-pressure processes • Q (under constant pressure) must account for both the increase in internal energy and the transfer of energy out of the system by work • Q (under constant volume) account just for change the internal energy • Qconstant P > Qconstant V for given values of n and ΔT Ideal Monatomic Gas • A monatomic gas contains one atom per molecule • When energy is added to a monatomic gas in a container with a fixed volume, all of the energy goes into increasing the translational kinetic energy of the gas – There is no other way to store energy in such a gas Ideal Monatomic Gas • Therefore, E int K tot 3 3 Nk B T nRT trans 2 2 • ΔEint is a function of T only • At constant volume, Q = ΔEint = nCV ΔT This applies to all ideal gases, not just monatomic ones Monatomic Gases • Solving E in t 3 nC V T nR T 2 for CV gives CV = 3/2 R = 12.5 J/mol . K for all monatomic gases • This is in good agreement with experimental results for monatomic gases Monatomic Gases • In a constant-pressure process, ΔEint = Q + W and E in t Q W n C P T P V • Change in internal energy depends only on temperature for an ideal gas and therefore are the same for the constant volume process and for constant pressure process nCV T nC P T nRT CP – C V = R Monatomic Gases CP – CV = R • This also applies to any ideal gas CP = 5/2 R = 20.8 J/mol . K Ratio of Molar Specific Heats • We can also define C P 5R / 2 1.67 CV 3 R / 2 • Theoretical values of CV , CP , and are in excellent agreement for monatomic gases • But they are in serious disagreement with the values for more complex molecules – Not surprising since the analysis was for monatomic gases Sample Values of Molar Specific Heats • A 1.00-mol sample of air (a diatomic ideal gas) at 300 K, confined in a cylinder under a heavy piston, occupies a volume of 5.00 L. Determine the final volume of the gas after 4.40 kJ of energy is transferred to the air by heat. • A 1.00-mol sample of air (a diatomic ideal gas) at 300 K, confined in a cylinder under a heavy piston, occupies a volume of 5.00 L. Determine the final volume of the gas after 4.40 kJ of energy is transferred to the air by heat. nRT The piston moves to keep pressure constant: V P Q nCP T n CV R T Q Q Q 2Q T nCP n(CV R ) 3R 5nR n R 2 nR nR 2Q 2Q 2 QV V T P P 5nR 5 P 5 nRT • A 1.00-mol sample of air (a diatomic ideal gas) at 300 K, confined in a cylinder under a heavy piston, occupies a volume of 5.00 L. Determine the final volume of the gas after 4.40 kJ of energy is transferred to the air by heat. nR nR 2Q 2Q 2 QV V T P P 5nR 5 P 5 nRT 2 V 5 (4.40 103 J )(5L) 3.53L J (1mol ) 8.314 (300 K ) mol K V f Vi V 5.00 L 3.53L 8.53L Molar Specific Heats of Other Materials • The internal energy of more complex gases must include contributions from the rotational and vibrational motions of the molecules • In the cases of solids and liquids heated at constant pressure, very little work is done since the thermal expansion is small and CP and CV are approximately equal Adiabatic Processes for an Ideal Gas • Assume an ideal gas is in an equilibrium state and so PV = nRT is valid • The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV = constant • = CP / CV is assumed to be constant during the process • All three variables in the ideal gas law ( P, V, T ) can change during an adiabatic process Equipartition of Energy • With complex molecules, other contributions to internal energy must be taken into account • One possible way to energy change is the translational motion of the center of mass Equipartition of Energy • Rotational motion about the various axes also contributes We can neglect the rotation around the y axis since it is negligible compared to the x and z axes Equipartition of Energy • The molecule can also vibrate • There is kinetic energy and potential energy associated with the vibrations Equipartition of Energy • The translational motion adds three degrees of freedom • The rotational motion adds two degrees of freedom • The vibrational motion adds two more degrees of freedom • Therefore, Eint = 7/2 nRT and CV = 7/2 R Agreement with Experiment • Molar specific heat is a function of temperature • At low temperatures, a diatomic gas acts like a monatomic gas CV = 3/2 R Agreement with Experiment • At about room temperature, the value increases to CV = 5/2 R This is consistent with adding rotational energy but not vibrational energy • At high temperatures, the value increases to CV = 7/2 R This includes vibrational energy as well as rotational and translational Complex Molecules • For molecules with more than two atoms, the vibrations are more complex • The number of degrees of freedom is larger • The more degrees of freedom available to a molecule, the more “ways” there are to store energy This results in a higher molar specific heat ...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture