PV = n R T = N k T. Measured from Vacuum = 0 Gauge Pressure = Vacuum - Atmospheric Atmospheric = 14.7 lbs/sq in = 10 5 N/m

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1 PV = n R T = N k T P is the Absolute pressure Measured from Vacuum = 0 Gauge Pressure = Vacuum - Atmospheric Atmospheric = 14.7 lbs/sq in = 10 5 N/m V is the volume of the system in m 3 often the system is in cylinder with piston Force on the piston does work on world T,P,V,n Does work on world P=0 outside

2 PV = n R T = N k T chemists vs physicists Mole View (more Chemical) = nrt R is gas constant J/mole/K Molecular View (physicists) = NkT N is number of molecules in system K is Boltzmann s Constant 1.38x10-23 J/K

3 Using PV=nRT Recognize: it relates state variables of a gas Typical Problems Lift of hot air balloon Pressure change in heated can of tomato soup Often part of work integral

4 Heat and Work are Processes Processes accompany/cause state changes Work along particular path to state B from A Heat added along path to B from A Processes are not state variables Processes change the state! But Eq. Of State generally obeyed

5 Ideal Gas Law Derivation: Assumptions Gas molecules are hard spheres without internal structure Molecules move randomly All collisions are elastic Collisions with the wall are elastic and instantaneous

6 Gas Properties N number of atoms in volume n m moles in volume m is atomic mass ( 12 C = 12) nm n N A m m V V V mass density ρ = m T = = Avogadro s Number N A = molecules mole 1

7 Motion of Molecules Assume all molecules have the same velocity (we will drop this latter) The velocity distribution is isotropic

8 Collision with Wall Change of î : p = mv x f mv momentum ĵ: =mv x, x,0 p y y, f mv y,0 Elastic collision v y, f = v y,0 v x v x, f = v x,0 Conclusion î : p x = 2mv x

9 Momentum Flow Tube Consider a tube of cross sectional area A and length v t x In time t half the molecules in tube hit wall ρ ρ Mass enclosed that m = Volume = Av t x hit wall 2 2

10 Pressure on the wall Newton s Second Law G G 2 p 2 mv x 2ρ Av x t F î = 2î wall, gas = = t t 2 t î = ρ Av x Third Law G G F 2î gas, wall = F wall, gas = ρ Av x Pressure P pressure = G F gas, wall 2 A = ρv x

11 Average velocity Replace the square of the velocity with the average of the square of the velocity (v 2 ) x ave random motions imply () = v v ( ) +(v ) +(v ) = 3(v ) ave x ave y ave z ave x ave Pressure n m N A 1 P = ρ () v = mv 3 3 V 2 () 2 pressure ave ave

12 Degrees of Freedom in Motion Three types of degrees of freedom for molecule 1. Translational 2. Rotational 3. Vibrational Ideal gas Assumption: only 3 translational degrees of freedom are present for molecule with no internal structure

13 Equipartition theorem: Kinetic energy and temperature Equipartition of Energy Theorem 1 (#degrees of freedom) 3 mv ( 2 ) ave = kt = kt Boltzmann Constant k = J K Average kinetic of gas molecule defines kinetic temperature

14 Ideal Gas Law Pressure 2 n m N A n n m N = mv m N A 3 p () kt = A pressure = kt 3 V 2 ave 3 V 2 V Avogadro s Number N A = molecules mole 1 Gas Constant R = N A k = 8.31 J mole 1 K 1 Ideal Gas Law pv = n RT m

15 Ideal Gas Atmosphere Equation of State PV = n m RT Let M molar be the molar mass. Mass Density ρ = n M m molar = M molar P V RT Newton s Second Law AP ( ( z ) Pz (+ z)) ρga z = 0

16 Isothermal Atmosphere Pressure Equation Differential equation Integration Solution Exponentiate P( z + z ) P( z) = ρg z dp M = ρg = molar g P dz RT Pz ( ) z dp = M g molar dz P 0 p P z ln P 0 z =0 RT () = M molar g RT ()= P 0 exp M g molar P z z RT z