Lecture 7: Kinetic Theory of Gases, Part 2 Last lecture, we began to explore the behavior of an ideal gas in terms of the molecules in it We found that the pressure of the gas was: P = N 2 mv x,i! = mn v x i=1 V V Since there s nothing special about the x direction, we should write this in terms of the total velocity of the molecules: v 2 = N! 2 v i i=1 N = = v x 2 + v y 2 + v z 2 N!( v 2 x,i + v 2 2 y,i + v ) z,i i=1 N 2
Since x is not a special direction, v 2 x, v 2 2 y, and v z must all have the same value therefore, we can write: v 2 2 = 3v x which means: P = mn v2 3V We can rewrite this as: P = 2 3 " # % & We do this because we recognize the quantity inside the parentheses: it s the average kinetic energy of a gas molecule N V! 1 2 mv2 $
Temperature So according to Newtonian physics, PV = 2 3 N! " # 1 $ 2 mv2 % & But the ideal gas law (which is experimentally verified!) tells us that: PV = Nk B T The only way both these can be true is if: 2 3 N! 1 $ " # 2 mv2 % & = Nk T B T = 2! 1 $ 3k B " # 2 mv2 % &
So we see that the temperature depends only on the average kinetic energy of the molecules! We can use this fact to find a typical speed of the molecules: v rms = v 2 = 3k T B m = 3RT M Molecular mass Be careful with the interpretation of this! it s not the average velocity, which is 0 (unless the gas as a whole is moving in some direction) For this distribution, the average is 0, but the RMS is as shown by the arrow
Assume we have a monatomic (only one atom per molecule) ideal gas Then the only internal energy associated with the gas is the kinetic energy of the atoms which we ve already shown is related to the temperature: T = 2! 1 $ 3k B " # 2 mv2 % & 1 2 mv2 = 3 2 k T B We could also write this as: 1 2 m v 2 + v 2 2 x y + v z ( ) = 1 2 m ( v 2 + v 2 ) 2 x x + v = 3 x 2 k T B 3 2 mv 2 = 3 x 2 k T B 1 2 mv 2 = 1 x 2 mv 2 = 1 y 2 mv 2 = 1 z 2 k T B
What this says is that for each possible direction of motion, the gas has internal energy equal to 1 per molecule 2 k T B For multi-atomic molecules, there are other ways that energy can be stored in the gas the molecules can rotate or vibrate Any way of storing energy is called a degree of freedom for the molecule The theorem of equipartition of energy says that For every degree of freedom in a system, the system stores 1/2k B T of energy Note also that the internal energy of an ideal gas depends only on the temperature!
Specific Heat What can kinetic theory tell us about the specific heat of a gas? to be precise, we ll talk about the heat capacity per mole, or molar specific heat, defined as Q / n!t Let s first assume that the gas is held at constant volume The 1st Law of Thermodynamics tells us that:!e int = Q +W Since the volume is constant, there s no work being done on the gas, so!e int = Q But we ve already shown that E int is proportional to the temperature of the gas
Assuming the gas is monatomic, we have: E int = 3 2 nrt "!E int =! 3 # $ 2 nrt % & ' = 3 2 nr!t = Q Since we also know that Q = nc V!T, we find that: nc V!T = 3 2 nr!t C V = 3 2 R = 12.5 J mol" K Table 21.2 in the text shows that this values agrees well with what s observed for real monatomic gases
Molar Specific Heat at Constant Pressure Now let s assume that the pressure on the gas is constant as the temperature changes Then the work being done on it is: " W = P V i!v f $ # P ( ) = P nrt i = nr( T i! T ) f =!nr(t! nrt f P % ' & Using the 1st Law, we have: But we already know that!e int = Q +W = Q " nr!t!e int = nc V!T
So we can write: The quantity in red is what we defined as molar specific heat to remind us that this equation applies when pressure is held constant, we write this one as C p This means that for a monatomic gas: And for any gas: nc V!T = Q " nr!t Q = n!t ( C V + R) Q n!t = C + R V C P = C V + R = 5 2 R = 20.8 C P! C V = R J mol! K
C V for More Complex Gases If a gas is not monatomic, then it has additional ways to store internal energy or additional degrees of freedom Consider a diatomic molecule: In addition to moving in any of the three dimensions of space, this molecule can store energy by: Vibrating: Rotating: or: Has both kinetic and potential energy -- 2 degrees of freedom 2 more degrees of freedom
Recall that the theorem of equipartition of energy says that for every degree of freedom, the gas has internal energy of: E int = 1 2 nrt So that our diatomic gas should have E int = 7 2 nrt Repeating our derivation of C V for this gas, we find: C V = 7 2 R = 29.1 J mol! K But this turns out to be a poor approximation -- none of the diatomic gases listed in Table 21.1 have C V this large
What Went Wrong? All our derivation relied on was 1. counting degrees of freedom 2. assuming that each degree of freedom could store any amount of energy Step (1) is pretty safe And so is step 2, as long as the molecules behave the way Newton says they should What we observe, then, is a failure of Newtonian physics The rotational and vibrational degrees of freedom can t take on any energy they please -- there s a minimum amount of energy needed to change them In other words, the energy is quantized This was one of several hints that led to the development of quantum mechanics
Adiabatic Processes We already have derived the relationship between pressure and volume of a gas for isothermal, isobaric, and isovolumetric processes. But what about for adiabatic ones? i.e., processes where no heat flows into or out of the gas Since no heat is involved, the change in internal energy of the gas is equal to the work done on it: de int = nc V dt = dw =!PdV Of course, the gas follows the ideal gas law, so: PV = nrt ( ) = d ( nrt ) d PV PdV +VdP = nrdt
This means we have two expressions for dt: We ve shown before that C V + R = C p dt =!PdV nc V!RPdV = C V! C V + R V P Note that the left side depends only on pressure, while the right side depends only on volume = PdV +VdP nr ( PdV +VdP) ( ) PdV = C V VdP!( C V + R)dV = C V dp
This means we can integrate both sides:!c P dv dp " = " C V V P!C P lnv = C V ln P + c C P lnv + ln P = c# C V lnv ln PV C P C V + ln P = c# C P C V = c# PV C P C V = const To save writing, we define! = C P C V
Distribution of Molecular Energies We now know that the temperature of a gas is proportional to the average energy of each molecule But we also know that all the molecules don t have the same energy some are moving faster and some slower and any moment For any given molecule, we can find the probability for that molecule to have a certain energy E: p( E) = e! E/k B T That means if there are n o molecules in a sample of gas, the number that will have energy E is: n E ( ) = n o e! E/k B T This is is the Boltzmann distribution law