Internal Energy Physics 202 Professor Lee Carkner Lecture 16

Consider two rooms of a house, room A and room B. If the (otherwise identical) molecules in room B have twice as much average kinetic energy than the ones in A, how does the temperature of room A compare to the temperature of room B? A)T A = T B B)T A = 2 T B C)T A = ½ T B D)T A = √2 T B E)T A = (3/2) T B

How does the rms velocity of the molecules in room A compare to the rms velocity of the molecules in room B? A)v A = v B B)v A = 2 v B C)v A = ½ v B D)v A = √2 v B E)v A = (1/√2) v B

PAL #15 Kinetic Theory  Which process is isothermal?  Start with p =4 and V = 5  Since T is constant, nRT is constant and thus pV is constant  Initial pV = 20, A: pV = 20, B: pV = 21  A is isothermal  3 moles at 2 m 3, expand isothermally from 3500 Pa to 2000 Pa  For isothermal process: W = nRTln(V f /V i )  Need T, V f  pV = nRT, T = pV/nR = (3500)(2)/(3)(8.31) = 281 K  V f = nRT/p f = (3)(8.31)(281)/(2000) = 3.5 m 3  W = (3)(8.31)(281)ln (3.5/2) = 3917 J  Since T is constant,  E = 0, Q = W = 3917 J

Ideal Gas  We will approximate most gases as ideal gases which can be represented by: v rms = (3RT/M) ½

Internal Energy  We have looked at the work of an ideal gas, what about the internal energy?  E int = (nN A ) K ave = nN A (3/2)kT E int = (3/2) nRT   Internal energy depends only on temperature   Since monatomic gasses can only have energy of motion

Molar Specific Heats  How does heat affect an ideal gas?   The equation for specific heat is:  From the first law of thermodynamics:   Consider a gas with constant V (W=0),  But  E int /  T = (3/2)nR, so: C V = 3/2 R = 12.5 J/mol K  Molar specific heat at constant volume

Specific Heat and Internal Energy  E int = (3/2)nRT E int = nC V T  E int = nC V  T   True for any process (assuming monatomic gas)

Specific Heat at Constant Pressure  We can also find the specific heat at constant pressure:   E int = nC V  T W = p  V = nR  T  Solving for C p we find: C p = C V + R   For a constant pressure or constant volume situation (assuming a monatomic ideal gas) we can find how much heat is required to produce any temperature change

Degrees of Freedom  Our relation C V = (3/2)R = 12.5 agrees with experiment only for monatomic gases   We assumed that energy is stored only in translational motion   For polyatomic gasses energy can also be stored in modes of rotational motion  Each possible way the molecule can store energy is called a degree of freedom

Rotational Motions Monatomic No Rotation Polyatomic 2 Rotational Degrees of Freedom

Equipartition of Energy  Equipartition of Energy:    We can now write C V as C V = (f/2) R = 4.16f J/mol K  Where f = 3 for monatomic gasses (x,y and z translational motion and f=5 for diatomic gases (3 trans. + 2 rotation)

Oscillation   The atoms oscillate back and forth as if the bonds were springs  So there are 3 types of microscopic motion a molecule can experience:  translational --  rotational --  oscillatory --  If the gas gets too hot the molecules will disassociate

Internal Energy of H 2

Adiabatic Expansion   It can be shown that the pressure and temperature are related by: pV  = constant   You can also write: TV  -1 = constant 

Ideal Gas Processes I  Isothermal Constant temperature Q=W W = nRTln(V f /V i )   Isobaric Constant pressure Q=nC p  T W=p  V 

Ideal Gas Processes II  Adiabatic No heat (pV  = constant, TV  -1 = constant) Q = 0 W=-  E int   Isochoric Constant volume Q= nC V  T W = 0 

Idea Gas Processes III  For each type of process you should know:   Path on p-V diagram  What is constant  Specific expressions for W, Q and  E 

Next Time  Read:  Homework: Ch 19, P: 44, 46, 53, Ch 20, P: 2, 4  Note: Test 2 next Friday, Feb 1