AP PHYSICS AP ME to TE problem

“The difference between a successful person and others is Wednesday, 1/14/09 AP PHYSICS “The difference between a successful person and others is NOT a lack of strength, NOT a lack of knowledge, but rather a lack of WILL.” Vince Lombardi

Recall: For an ideal MONOATOMIC gas, Wednesday, 1/14/09 AP PHYSICS Recall: For an ideal MONOATOMIC gas, Since this is an ideal gas, the molecules have no PE, so

ΔU = Q + W First Law of Thermodynamics Wednesday, 1/14/09 AP PHYSICS First Law of Thermodynamics Whenever heat is added to a system, it transforms to an equal amount of some other form of energy (*Conservation of Energy specifically for thermal energy*) ΔU = Q + W ΔU – change in internal (thermal) energy Q – heat added (+) or heat removed (-) W – work done ON the system (+) or work done BY the system (-)

(* Heat will only flow one way without external work being done.*) p. 420 #17 A gas is enclosed in a container fitted with a piston of cross-sectional area 0.150m2. The pressure of the gas is maintained at 6000Pa as the piston moves inward 20.0cm. (a) Calculate the work done by the gas. (b) If the internal energy of the gas decreases by 8.00J, find the amount of heat removed from the system during the compression. 2nd Law of Thermodynamics Heat will never of itself flow from a cold to a hot object. -180J -188J a) -180J b) -188J (* Heat will only flow one way without external work being done.*)

Recall from yesterday: Work done to get from an initial state to a final state is path dependent. Today: The heat added or removed to get from an initial state to a final state is also path dependent. Both expansions have the same initial P and V and the same final P and V, but one added heat and the other did not. Heat reservoir – a body whose heat capacity is so large that its temperature does NOT change when heat is added or removed.

ΔU = Q + W Q is process dependent. W is process dependent. ΔU IS INDEPENDENT OF THE PROCESS!!!! It is completely determined by the initial and final states! remember yesterday’s graphs.

4 types of processes you need to know 1) isobaric – 2) isovolumetric – 3) isothermal – 4) ADIABATIC - pressure remains constant (yesterday) Since W = PΔV, W ≠ 0 and Q ≠ 0. (V1/T1 = V2/T2) volume remains constant Since W = 0, ΔU = Q. temperature remains constant Since ΔU = 0, Q = -W. (P1V1 = P2V2) no heat enters or leaves Since Q = 0, ΔU = W. draw graphs!!, (isotherms!!!) Justify Q for isobaric!!!

p. 421 #26 One mole of gas is initially at a pressure of 2.00atm, a volume of 0.300L, and an internal energy equal to 91.0J. In its final state the gas is at a pressure of 1.50atm and a volume of 0.800L, and its internal energy equals 180J. For the paths IAF, IBF, and IF, calculate (a) the work done ON (?) the gas and (b) the net heat transferred to the gas in the process. 21 a) W iaf = 76J, ibf = 100J, if = 88.6J b) Q iaf = 165J, Q ibf = 190J, Q if = 178J HW 45: ws. Ch. 12 P/ 11 – 14, 18*

12) W > 0, U = 0, Q < 0 13) a) 152J b) -248J 11) 6P0V0 HW 45 inc. for today weds. cyclic process and heat engines thurs. carnot enigines fri. practice ap questions and more for hw!! HW 45