1. Chapter 19 The Kinetic Theory of Gases To study p, V, E int, T, …etc. from a “molecular” approach 19.1 A new way to look at gases: Warm up: How many moles are in 50 g of helium? How many atoms of hydrogen are there in 2 kg of water?

2. One mole is defined to be the number of atoms in a 12-g sample of carbon-12. We call this number “Avogadro’s number”, N A. N A = 6.02 10 23 mol -1 19.2 Avagadro’s number (N A ): The number of molecules (or atoms) N is related to N A through: N = n N A n = M sam /M Does this equation make sense? where n: number of moles, and M: molar mass of the material. m is the mass of one atom (or molecule) n = M sam /(m N A )

3. 19.3 Ideal Gases: Interestingly, different (real) gases behave similarly (from the physical point of view: P, V, T) when at the (ideal) low density limit with “no” interaction amongst the atoms. In this case, gases tend to obey the relation (called the ideal gas law): p V = n R T p: absolute (not gauge!) pressure T: absolute (not Celsius) temperature R: “Universal” gas constant = 8.314 J/(mol K) n: is number of moles and V is volume

4. Boltzmann’s constant (k) is defined: k = R/N A = 1.38 10 -23 J/K Therefore, n R = n N A k = N k Hence, p V = N k T where N is the number of molecules Example: Oxygen at 0 o C is under one atmospheric pressure. How many liters would on mole occupy? How many molecules are in 11.2 liters, under the same conditions?

5. Work done by an ideal gas at constant temperature: W =  dW =  p dV =  n R T/V dV W = n R T ln (V f /V i ) Example: Sample problem 20-2 What if the expansion is reversed; i.e., what happens if it were a compression instead of an expansion?

6. Work done by an ideal gas at constant pressure: W =  dW =  p dV = p  V Since p V = n R T,  (pV) =  (n R T) Also,  (n R T) = n R  T[if n is constant] But,  (p V) = p  V = W [at constant p] Therefore, W = n R  T [for constant n&p!!]

7. Example: Helium gas is heated at constant pressure from 32 degrees Fahrenheit to 212 degrees Fahrenheit. If the gas does 20.0 Joules of work during the process, what is the number of moles. Ans. 0.024 moles

8. What is the work done by an ideal gas at constant volume? W = ??

9. 19.4 Pressure, Temperature and RMS speed: If a few moles of some gas are confined to a closed box, the gas will have some pressure and temperature. What causes pressure? What is the temperature? One can prove (see page 459-460) that: p = (n M v rms 2 )/(3V) v rms = (3 R T/M) 1/2 Do these equations make sense? Do they go with the spirit kinetic theory?

10. What is v rms ? It is a kind of average speed; v rms is typically a few hundred m/s!! Individual molecules may be faster or slower. Interaction: Find an approximate rms speed of air in the room. Why don’t molecules from perfume reach such high speeds?

11. 19.5 Translational kinetic Energy: At a given temperature T, all ideal gas molecules have the same average translational kinetic energy, namely, 3/2 k T. One can show that the average translational kinetic energy per molecule (K avg ) is: K avg = (3/2) k T Check point #2: A gas mixture consists of molecules of types 1, 2, and 3 with molecular mass m1 > m2 > m3. Rank the three types according to (a) the average kinetic energy, and (b) rms speed, greatest first. Beware, there may be rotational and vibrational energies! [in multi-atomic molecules] Omit 20.6 and 20.7 !! (regardless of their mass!!)

12. 19.8 The Molar Specific Heats of an Ideal Gas If we consider, however, monatomic molecules (He, Xe, Ne, …etc.), we can show that: E int = (3/2) n R T That is, the internal energy (for an ideal gas) is a function of the temperature, but does not depend on pressure or volume!! In complex molecules, there can be rotational and vibrational motions coupled to translation motion through collisions.

13. Molar specific heat at constant volume (c v or C v ): For processes performed under constant volume: Q = n c v  T W = 0  E int = Q (3/2) n R  T = n c v  T (monatomic gas) Therefore,  E int = n c v  T (for all ideal gases) For monatomic gas: c v = (3/2) R = 12.5 J/(mol K)

14. Molar specific heat at constant pressure (c p or C p ): For processes performed under constant pressure: Q = n c p  T W = p  V = n R  T  E int = Q - W n c v  T = n c p  T - n R  T Therefore, c p - c v = R(for all ideal gases) For monatomic molecules: c p = 20.8 J/(mol K)  c p / c c =5/3 = 1.67 [for monatomic molecules]

15. What if there are more degrees of freedom, as in polyatomic molecules?? Let us consider a diatomic molecule: how many degrees of freedom? So what is the internal energy? From the equipartition theorem, energy is shared equally (on the average) by each independent degree of freedom. There is 1/2 k B T per degree of freedom! Five! (not 8, why?!) E int = 5 N (1/2 k B T) c v = 5/2 R

16. c p - c v = R c p = 7/2 R   c p / c c =7/5 = 1.4 Experimentally, some degrees of freedom are frozen at low(er) temperatures. Notice that  is lower for diatomic gases than for monatomic gases. Things become more complicated for more complicated molecules! I told you: Physics makes sense!!

17. Example: Two moles of helium (monatomic) gas are heated from 100 degree Celsius to 250 degree Celsius. How much heat is transferred to the gas if the process is isobaric?

18. 19.11 The Adiabatic Expansion of an Ideal Gas When an (ideal) gas expands (reversibly) adiabatically (Q = 0), one can show that: pV  = some constant T V  = some *other* constant p 1 V 1  = p 2 V 2  T 1 V 1  = T 2 V 2  What happens to the temperature under adiabatic expansion? What happens to the temperature under adiabatic compression?

19. What about free expansions? Non-reversible!!!!!!! Q = 0, W = 0  E int = 0  T= 0 But, p V = n R T p 1 V 1  = p 2 V 2 Example: A diatomic gas occupies a volume of 4.3 liters at a pressure of 1.2 atm and a temperature of 37 o C. It is compressed adiabatically to a volume of 0.76 liters. How many moles are there in the sample? Find the final pressure and temperature.