Physics 231 Lecture 30. Main points of today s lecture: Ideal gas law:

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1 Physics 231 Lecture 30 Main points of today s lecture: Ideal gas law: PV = nrt = Nk BT 2 N 1 2 N 3 3 V 2 3 V 2 2 P = m v = KE ; KE KE = kbt

2 Phases of Matter Slide 12-16

3 Ideal Gas: properties Approximate descrip/on of a gas as a collec/on of atoms or molecules or both that Exert no force upon each other The energy of a system of two atoms/molecules cannot be reduced by bringing them close to each other Atoms themselves take up very li?le of the total volume The volume taken by the atoms/molecules is negligible compared to the volume that the gas occupies.

4 Poten/al Energy R min 0 -E min R Ideal gas: we are neglec/ng the poten/al energy between The atoms/molecules because they only rarely get that close to each other.

5 Important Properties of gases V = volume P = pressure T = temperature in K (Kelvin) N = number of atoms or molecules

6 Amount of substance If there are N particles in a gas then the number of moles: n = N N A N A = 6.02*10 23 Avogadro s number N A is not defined randomly, it is defined such that there is exactly N A atoms of carbon in 12 g of 12 C number of moles = mass mass of one mole

7 Number of atoms and moles N = total number of atoms (or molecules) n = N N A = number of moles or N = n N A so one mole contains N A atoms (or molecules) M = m N = total mass of all atoms (or molecules), each having mass m M = n M molar

8 Molar mass m = mass of one atom (or molecule) N A = M molar = m N A Avagodro's number molar mass for example M molar (carbon) = 12.0 g = kg m(carbon) = M molar N A kg = kg Another way to calculate m(carbon): Approximate mass: m(carbon)=12 Nucleons x atomic mass unit 1 atomic mass unit =1µ = 1.66x10-27 kg. check that it works: 12 x 1.66x10-27 kg =1.99 x10-26 kg

9 Number of electrons Z Name X A molar mass in grams

10 Weight of 1 mol of atoms 1 mol of atoms weighs A grams (A is the molar mass) Examples: 1 mol of Hydrogen weighs 1.0 g 1 mol of Carbon weighs 12.0 g 1 mol of Oxygen weighs 16.0 g 1 mol of Zinc weighs 65.4 g What about molecules? H 2 O 1 mol of water molecules: 2 x 1.0 g (due to Hydrogen) 1 x 16.0 g (due to Oxygen) Total: 18.0 g

11 Example: Silicon cube A cube of Silicon (molar mass 28.1 g) is 250 g. A) M molar =28.1g. How much Silicon atoms are in the cube? Total number of moles n = M / M molar = 250/28.1 = 8.90 N = n N A = (8.9) (6.02x10 23 ) = 5.4x10 24 atoms B) What would be the mass for the same number of gold atoms (molar mass 197 g) a) 197 g b) 394 g c) 875 g M = n M molar = (8.90) (197 g) = 1750 g d) 1750 g

12 Clicker Quiz 1) 1 mol of CO 2 has a larger mass than 1 mol of CH 2 2) 1 mol of CO 2 contains more molecules than 1 mol of CH 2 a) 1) true 2) true b) 1) true 2) false c) 1) false 2) true d) 1) false 2) false

13 Clicker Quiz 1) 1 mol of CO 2 has a larger mass than 1 mol of CH 2 2) 1 mol of CO 2 contains more molecules than 1 mol of CH 2 a) 1) true 2) true 1) 1 mole CO 2 weighs 12g +2*16 g =44 g b) 1) true 2) false 1 mole of CH 2 weighs 12g + 2*1g = 14g c) 1) false 2) true 2) 1 mole of has N A molecules d) 1) false 2) false

14 Boyle s Law At constant temperature and constant number of cons/tuents: P ~ 1/V ½P 0 2V 0 T 0 P 0 V 0 T 0 2P 0 ½V 0 T 0

15 Charles law If you want to maintain a constant pressure with constant number of cons/tuents, the temperature must be increased linearly with the volume V~T P 0 2V 0 2T 0 P 0 V 0 T 0

16 Gay-Lussac s law If, at constant volume, the temperature is increased, the pressure will increase by the same factor P ~ T P 0 V 0 T 0 2P 0 V 0 2T 0

17 All the dependence of pressure P on T, V, and number of moles is contained in the Ideal gas law Ideal gas pressure depends linearly on temperature. P = nr V T K When T is given in Kelvin Ideal gas pressure depends linearly on temperature. Here n=number of moles of the gas. There are N A = 6.02x10 23 molecules or atoms (if the atoms don t combine into molecules) per mole. One mole =.0224 m 3 of gas at T=0 o C and P=1 atm. R=8.31 J/(moleK) Example: A molecular gas is contained in an 8.0-L vessel at a temperature of 20 C and a pressure of 9.0 atm. (a) Determine the number of moles of gas in the vessel. (b) How many molecules are in the vessel? PV ( 5 )( 3) a) n = x10 Pa.008m RT = = 3.0 moles 8.31J / K 293K b) N nn A ( )( ) = 23 = 36.02x10 24 = 1.8x10 molecules

18 Reading Quiz 2. A sample of nitrogen gas is in a sealed container with a constant volume. The temperature of the gas is increase by heating it. The pressure A. increases B. stays the same C. decreases D. can t be determined with the information given P = nrt V Slide 12-8

19 Example Gas is confined in a tank at a pressure of 10.0 atm and a temperature of 15.0 C. If half of the gas is removed and the temperature is raised to 65.0 C, what is the ratio of the final density over the initial density? 1 Nm 0 Nm Nm 0 f ρ 0 = ρ f = 2 1 = 0 V V V = 2 ρ What is the new pressure in the tank? P P P 0 0 B 0 f 0 N = k T V Nf \ kt B f = V\ N0 \ kt V Pf \ B 0 = P Nf T N T = N V k T f f B f f = = = atm = 5.9atm 0.59

20 Reading Quiz What is the mass, in u, of a molecule of carbon dioxide, CO2? A. 12 B. 24 C. 32 D. 36 E. 44 m O = 16µ m C = 12µ m CO2 = = 44µ Slide 12-12

Kinetic properties of an ideal gas Consider the gas volume at shown on the right. A molecule with mass m is elastically scattered by the wall which supplies an impulse force F 1.

21 Kinetic properties of an ideal gas Consider the gas volume at shown on the right. A molecule with mass m is elastically scattered by the wall which supplies an impulse force F 1. 2mvx F Δt = mv m v ) = 2mv F = 1 x ( x x 1 Δt After a time Δt=2d/(v x ), the mass bounce off the far wall and return to bounce of this wall again. If we use this value for Δt in the expression above, we can get the average force for much large time intervals. F = = x ( 2d / vx ) mvx mv / d Multiplying by the number of molecules in the gas and dividing by the area A, we get the pressure on the wall. N 2 N 2 P= mvx = mvx Ad V This would be correct if each molecule was moving the x direction with exactly velocity v x. To correct for the differences in velocities, we use the average <v x2 > instead. Also <v 2> = <v x2 >+ <v y2 >+ <v z2 >=3 <v x2 >. Thus N 2 2 N N P = m v / 3 = m v = < KE > V 3 V 2 3 V wall A d wall

22 1 2 mv 2 rms Average kinetic energy of gas molecules If we combine the last expression with the ideal gas law equation of state, we get a useful expression for the means kinetic energy of gas molecules: 2N n R N P= < KE > P = RT N = nn A kb P= kbt 3V V NA V 2N/ N KE kbt 3V < >= / < KE >= m v 2 mv 2 rms = kbt / V/ k B =1.38x10-23 J/K note: increasing temperature increases thermal (KE) energy Example: If the translational rms speed of water molecules (H 2 O) in a volume of air is 648 m/s, what is the translational rms speed of carbon dioxide (CO 2 )? Both gases are at the same temperature. (v rms =[<v 2 >] 1/2 ) 3k T B v rms,h2o = = m H = 3 2O 2 k T = m H 2 O B m m HO CO2 m CO2 m H2 m O CO2 2 v rms,co2 = 3k B T 3k B T 3k B T m H2 O v = rms,co m m HO CO 2 v rms,h O N m = A H2O v ( ) rms,h2o 648m / s N m A CO = = m / s

Physics 231 Topic 12: Temperature, Thermal Expansion, and Ideal Gases Alex Brown Nov

Physics 231 Topic 12: Temperature, Thermal Expansion, and Ideal Gases Alex Brown Nov 18-23 2015 MSU Physics 231 Fall 2015 1 homework 3 rd midterm final Thursday 8-10 pm makeup Friday final 9-11 am MSU