 # Internal Energy Physics 202 Professor Lee Carkner Lecture 14.

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Internal Energy Physics 202 Professor Lee Carkner Lecture 14

PAL #13 Kinetic Theory  Which process is isothermal?   Since T is constant, nRT is constant and thus pV is constant  Initial pV = 20, A: pV = 20, B: pV = 21   3 moles at 2 m 3, expand isothermally from 3500 Pa to 2000 Pa  For isothermal process: W = nRTln(V f /V i )   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   Since T is constant,  E = 0, Q = W = 3917 J

Ideal Gas  pV=nRT v rms = (3RT/M) ½ K ave =(3/2) k T

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  Strictly true only for monatomic gasses   Note that this is the total internal energy, not the change in internal energy

Molar Specific Heats  If we add heat to something, it will change temperature, depending on the specific heat   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 for an ideal gas

Specific Heat and Internal Energy  If C V = (3/2)R we can find the internal energy in terms of C V  E int = nC V  T   True for any process (assuming monatomic gas)

Specific Heat at Constant Pressure  We can also find the molar specific heat at constant pressure (C p )   E int = nC V  T W = p  V = nR  T C p = C V + R  C p is greater than C v   At constant pressure, you need more heat since you are also doing work

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:   Each degree of freedom (f) has associated with it energy equal to ½RT per mole  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)  It is always true that C p = C V + R

Oscillation  At high temperatures we also have oscillatory motion   So there are 3 types of microscopic motion a molecule can experience:  translational -- l  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  Remember also that  E int =-W since Q=0

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

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

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

Next Time  Read: 21.1-21.4  Note: Test 3 next Friday, Jan 20

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)v A = (1/√2) v 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