Chapter 17 - Phases and Phase Changes1 Ideal GasesAn “ideal gas” is a gas in which intermolecular interactions are negligibly small. Idealgases don’t really exist in nature, but the behavior of real gases can be understoodby studying the ideal case.The equation of state for an ideal gas is of pressure P, volume V , number of moleculesN, and temperature T isP V = NkT (1)where k = 1.38 × 10−23J/K is the Boltzmann constant.In Chapter 16, we saw that the temperature and pressure of a gas are linearly re-lated, with a temperature of absolute zero (0 K) corresponding to a pressure of zero.From the ideal gas equation, we can see the constant of proportionality is Nk/V ,P = (Nk/V )T : for a fixed number of molecules in a fixed volume, the pressure islinearly related to the temperature.Similarly, for a fixed volume and fixed temperature, the pressure depends linearlyon the number of molecules, P = (kT/V )N. (Pumping more air molecules into analready inflated tire increases the pressure.) For a fixed number of molecules and afixed temperature, the pressure depends inversely on the volume, P = (NkT )(1/V ).(You take a balloon that’s pumped up and squeeze it: as the volume decreases, thepressure increases.)Remember pressure is force per unit area and is measured in pascals (Pa).1Pa = 1N/m2= 1kg/(ms)2.A helium-filled balloon at 23◦C has a volume of 0.025 m3and is at a pressure of1.8 × 105Pa. How many helium atoms are in the balloon?L. Whitehead 1 Phys 130223◦C = 296.15 K.P V = NkTN =P VkTN =(1.8 × 105Pa)(0.025m3)(1.38 × 10−23J/K)(296.15K)N = 1.1 × 1024moleculesA mole is a unit used to measure the amount of a substance. It is defined as theamount of a substance that contains as many molecules as there are atoms in 12 gof carbon-12. There are 6.022 × 1023atoms in 12 g of carbon-14. NA= 6.022 × 1023is called “Avagadro’s number.” Thus1mole = 6.022 × 1023moleculesNA= 6.022 × 1023molecules/mole (2)Let n be the number of moles in a gas, then the number of molecules is N = nNA.If we plug this into the ideal gas equation:P V = nNAkTR = NAk = (6.022 × 1023molecules/mole)(1.38 × 10−23J/K) = 8.31J/mol KP V = nRT (3)where R is called the universal gas constant.One mole of any substance has the same number of molecules, but one mole corre-sponds to a different mass for different substances. One mole of helium atoms hasa mass of 4.00260 g, while one mole of copper atoms has a mass of 63.546 g. Theatomic or molecular mass M of a substance is the mass in grams of one mole of thatsubstance. The mass of an individual atom or molecule then is m = M/NA.1.1 × 1024molecules is how many moles?N = nNAn =NNA=1.1 × 1024molecules6.022 × 1023molecules/mole= 1.8molesL. Whitehead 2 Phys 1302Boyle’s Law: The pressure of a gas varies inversely with volume as long as tempera-ture and the number of molecules are head constant:P V = NkT = constantwhich means the product of pressure and volume doesn’t change:PiVi= PfVf(4)Each of the curves in the figure corresponds to a different temperature; thus thesecurves are called “isotherms.”Charles’s Law: The volume of a gas divided by its temperature is constant, as longas the pressure and number of molecules are constant:VT=NkP= constantwhich means volume over temperature doesn’t change:ViTi=VfTf(5)If the temperature of the helium inside the balloon in the example above increasesto 40◦C, but the pressure remains constant, what volume will the gas occupy?The number of molecules isn’t changing, because we are not pumping air into or outL. Whitehead 3 Phys 1302of the balloon. Thus we can use Charles’s Law.40◦C = 313.15 KViTi=VfTfVf=ViTiTfVf=0.025m3296.15K(313.15K)Vf= 0.026m32 Kinetic TheoryThe kinetic theory of gases makes the connection between what’s happening in a gason the microscopic level to what we observe on the macroscopic level. Suppose wehave a gas made up of a collection of molecules moving about inside a container ofvolume V . We make the following assumptions:1. The container holds a very large number N of identical molecules, each of massm. We assume the size of the molecules is negligible compared to the size ofthe container and compared to the distance between the molecules.2. The molecules move in a random manner, and they obey Newton’s law ofmotion.3. When molecules hit the walls of the container or collide with one another, theybounce elastically. Other than these elastic collisions, the molecules have nointeractions.The pressure of a gas is due to the collisions between gas molecules and the wallsof the container. Suppose we have a container that is a cube with side length L.Consider a molecule moving in the negative x direction toward a wall of the container.Its speed is vxand thus its initial momentum is pi= −mvx. When it hits the wall,it bounces off elastically, meaning that its momentum is now the same magnitudebut in the opposite direction: pf= mvx. The change in the momentum is ∆p =mvx− −mvx= 2mvx. The time it takes for the molecule to travel from one wall tothe another wall, bounce off, and get back to the original wall is ∆t = 2L/vx. Theaverage force is thenF =∆p∆t=2mvx2L/vx=mv2xLL. Whitehead 4 Phys 1302and the average pressure isP =FA=mv2xL1L2=mv2xV(6)The molecules in a gas will have different speeds, which are constantly changingas each molecule undergoes collisions with the walls and other moelcules. But thedistribution of speeds remains constant. The Maxwell speed distribution (namedafter James Clerk Maxwell) is the probability that a molecule in a gas will have aparticular speed. The distribution depends on the temperature, shown for O2gas inthe figure. The most probable speed increases with temperature.To make the pressure equation more general, we replace v2xwith the average value(v2x)avg.P =m(v2x)avgVThe above is the pressure exerted by one molecule. To get the total pressure, wemultiply by the number of molecules.P =NVm(v2x)avgThis is only for velocity in the x direction. The total average velocity will comefrom summing the 3 components: (v2)avg= (v2x)avg+ (v2y)avg+ (v2z)avg= 3(v2x)avg, or(v2x)avg= (1/3)(v2)avg. (There’s nothing special about the x direction. We chose itL. Whitehead 5 Phys 1302arbitrarily.) The pressure in terms of the total velocity (instead only one componentof the velocity) is:P =13NVm(v2)avgWe can write it this way so that it’s in terms of the kinetic energy:P =23NV12m(v2)avg=23NVKavg(7)So we see that the pressure of a gas is directly