Chapter 12

# Chapter 12 - CHAPTER 12 Quick Quizzes 1 Situation(a(b(c...

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C HAPTER 12 Quick Quizzes 1. Situation System Q W U (a) Rapidly pumping up a bicycle tire Air in the pump near 0 + + (b) Pan of room-temperature water sitting on a hot stove Water in the pan + near 0 + (c) Air quickly leaking out of a balloon Air originally in balloon near 0 (a) Because the pumping is rapid, no energy enters or leaves the system by heat. Because when work is done on the system, it is positive here. Thus, must be positive. The air in the pump is warmer. 0 W > UQW ∆=+ (b) No work is done either by or on the system, but energy flows into the water by heat from the hot burner, making both Q and U positive. (c) Again no energy flows into or out of the system by heat, but the air molecules escaping from the balloon do work on the surrounding air molecules as they push them out of the way. Thus W is negative and U is negative. The decrease in internal energy is evidenced by the fact that the escaping air becomes cooler. 2. A is isovolumetric, B is adiabatic, C is isothermal, D is isobaric. 3. C, B, A. A 700 K 1 0.300 1000 K e =− = , B 500 K 10 375 800 K e = . , C 300 K 1 0.500 600 K e = , so . CB eee >> A 4. (b). because in any adiabatic process. 0 S ∆= 0 Q = 5. The number 7 is the most probable outcome because there are six ways this could occur: 1- 6, 2-5, 3-4, 4-3, 5-2, and 6-1. The numbers 2 and 12 are the least probable because they could only occur one way each: either 1-1, or 6-6. Thus, you are six times more likely to throw a 7 than a 2 or 12. 373

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CHAPTER 12 Problem Solutions 12.1 From kinetic theory, the average kinetic energy per molecule is B A 33 22 molecule R KE k T T N ⎛⎞ == ⎜⎟ ⎝⎠ For a monatomic ideal gas containing N molecules, the total energy associated with random molecular motions is A N U N KE RT nRT N =⋅ = = Since for an ideal gas, the internal energy of a monatomic ideal gas is found to be given by PV nRT = 3 2 UP V = . 12.2 The work done by the gas is ( ) by gas WP V = +∆ , so ( ) fi f i V P V n R T n R T n R =−= − =∆ T , Thus, () by gas 20.0 J 0.0241 mol 8.31 J mol K 100 K W n = ∆⋅ , and ( ) ( ) 0.0241 mol 4.00 g mol 0.0963 g mn M = = 96.3 mg 12.3 The number of molecules in the gas is A N nN = and the total internal energy is ( ) AB U N KE nRT = 3J 3.0 mol 8.31 303 K 2m o l K 4 1.1 10 J × Alternatively, use the result of Problem 12.1, Vn R T , just as found above. 374
CHAPTER 12 12.4 (a) The work done by the gas on the projectile is given by the area under the curve in the PV diagram. This is ( ) ( ) ( )( ) ( ) ( )( ) () ( ) by gas 000 0 3 53 63 triangular area rectangular area 11 22 11 1.0 10 Pa 40.0 8.0 cm 19 J 21 ff f f W PPV V PV V PPV V =+ =− + − ⎛⎞ ⎡⎤ × = ⎣⎦ ⎜⎟ ⎝⎠ 0 m 0 c m From the work-kinetic energy theorem, 2 1 0 2 WK Em v = ∆= where is the work done on the projectile by the gas. Thus, the speed of the emerging projectile is W () 3 1 9 J 40.0 10 kg net W v m == = × 31 m s (b) The air in front of the projectile would exert a retarding force of ( ) ( ) ( ) 52 2 4 2 1.0 1.0 cm 1 m 10 cm 10 N ra i r FPA ==× = on the projectile as it moves down the launch tube. The energy spent overcoming this retarding force would be ( ) 10 N0 .

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## This note was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 12 - CHAPTER 12 Quick Quizzes 1 Situation(a(b(c...

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