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College Physics for AP® Courses

Chapter 13

College Physics for AP® CoursesChapter 13

Problems & Exercises

1.

102 º F 102 º F size 12{"102"°F} {}

3.

20.0ºC20.0ºC size 12{"20" "." 0°C} {} and 25.6ºC25.6ºC size 12{"25" "." 6°C} {}

5.

9890 º F 9890 º F size 12{"9890"°F} {}

7.

(a) 22.2ºC22.2ºC size 12{"22" "." 2°C} {}

(b) ΔTºF = T2ºFT1ºF = 95T2ºC+32.0º95T1ºC+ 32.0º = 95T2ºCT1ºC=95ΔTºC ΔTºF = T2ºFT1ºF = 95T2ºC+32.0º95T1ºC+ 32.0º = 95T2ºCT1ºC=95ΔTºC

9.

169.98 m

11.

5 . 4 × 10 6 m 5 . 4 × 10 6 m size 12{5 "." 4 times "10" rSup { size 8{ - 6} } " m"} {}

13.

Because the area gets smaller, the price of the land DECREASES by ~$17,000.~$17,000. size 12{ "~" $"17","000" "." } {}

15.

V = V 0 + ΔV = V 0 ( 1 + βΔT ) = ( 60 . 00 L ) 1 + 950 × 10 6 / º C 35 . 0 º C 15 . 0 º C = 61 . 1 L V = V 0 + ΔV = V 0 ( 1 + βΔT ) = ( 60 . 00 L ) 1 + 950 × 10 6 / º C 35 . 0 º C 15 . 0 º C = 61 . 1 L

17.

(a) 9.35 mL

(b) 7.56 mL

19.

0.832 mm

21.

We know how the length changes with temperature: ΔL=αL0ΔTΔL=αL0ΔT size 12{ΔL=αL rSub { size 8{0} } ΔT} {}. Also we know that the volume of a cube is related to its length by V=L3V=L3 size 12{V=L rSup { size 8{3} } } {}, so the final volume is then V=V0+ΔV=L0+ΔL3V=V0+ΔV=L0+ΔL3 size 12{V=V rSub { size 8{0} } +ΔV= left (L rSub { size 8{0} } +ΔL right ) rSup { size 8{3} } } {}. Substituting for ΔLΔL size 12{DL} {} gives

V = L 0 + αL 0 ΔT 3 = L 0 3 1 + αΔT 3 . V = L 0 + αL 0 ΔT 3 = L 0 3 1 + αΔT 3 . size 12{ V= left (L rSub { size 8{0} } +αL rSub { size 8{0} } ΔT right ) rSup { size 8{3} } =L rSub { size 8{0} rSup { size 8{3} } } left (1+αΔT right ) rSup { size 8{3} } "." } {}

Now, because αΔTαΔT size 12{αΔT} {} is small, we can use the binomial expansion:

V L 0 3 1 + 3αΔT = L 0 3 +L03 Δ T . V L 0 3 1 + 3αΔT = L 0 3 +L03 Δ T . size 12{V approx L rSub { size 8{0} } rSup { size 8{3} } left (1+3αΔT right )=L rSub { size 8{0} rSup { size 8{3} } } +3αL rSub { size 8{0} } rSup { size 8{3} } ΔT} {}

So writing the length terms in terms of volumes gives V=V0+ΔVV0+V0ΔT,V=V0+ΔVV0+V0ΔT, size 12{V=V rSub { size 8{0} } +ΔV approx V rSub { size 8{0} } +3αV rSub { size 8{0} } ΔT,} {} and so

Δ V = βV 0 Δ T V 0 Δ T , or β . Δ V = βV 0 Δ T V 0 Δ T , or β . size 12{ΔV=βV rSub { size 8{0} } ΔT approx 3αV rSub { size 8{0} } ΔT,`"or"`β approx 3α} {}

22.

1.62 atm

24.

(a) 0.136 atm

(b) 0.135 atm. The difference between this value and the value from part (a) is negligible.

26.

(a) nRT=(mol)(J/molK)(K)= JnRT=(mol)(J/molK)(K)= J size 12{ ital "nRT" = \( "mol" \) \( "J/mol" cdot K \) \( K \) =" J"} {}

(b) nRT=(mol)(cal/molK)(K)= calnRT=(mol)(cal/molK)(K)= cal size 12{ ital "nRT" = \( "mol" \) \( "cal/mol" cdot K \) \( K \) =" cal"} {}

(c) nRT = (mol)(Latm/molK)(K) = Latm=(m3)(N/m2) = Nm=J nRT = (mol)(Latm/molK)(K) = Latm=(m3)(N/m2) = Nm=J

28.

7 . 86 × 10 2 mol 7 . 86 × 10 2 mol size 12{7 "." "86" times "10" rSup { size 8{ - 2} } `"mol"} {}

30.

(a) 6.02×105km36.02×105km3 size 12{6 "." "02" times "10" rSup { size 8{5} } `"km" rSup { size 8{3} } } {}

(b) 6.02×108km6.02×108km size 12{6 "." "02" times "10" rSup { size 8{8} } `"km"} {}

32.

73 . 9 º C 73 . 9 º C size 12{ - "73" "." 9°C} {}

34.

(a) 9.14×106N/m29.14×106N/m2 size 12{9 "." "14" times "10" rSup { size 8{7} } `"N/m" rSup { size 8{2} } } {}

(b) 8.23×106N/m28.23×106N/m2 size 12{8 "." "23" times "10" rSup { size 8{7} } `"N/m" rSup { size 8{2} } } {}

(c) 2.16 K

(d) No. The final temperature needed is much too low to be easily achieved for a large object.

36.

41 km

38.

(a) 3.7×1017Pa3.7×1017Pa size 12{3 "." 7 times "10" rSup { size 8{ - "17"} } `"Pa"} {}

(b) 6.0×1017m36.0×1017m3 size 12{6 "." 0 times "10" rSup { size 8{"17"} } `m rSup { size 8{3} } } {}

(c) 8.4×102km8.4×102km size 12{8 "." 4 times "10" rSup { size 8{2} } `"km"} {}

39.

1 . 25 × 10 3 m/s 1 . 25 × 10 3 m/s size 12{ size 11{1 "." "25" times "10" rSup { size 8{3} } `"m/s"}} {}

41.

(a) 1.20×1019J1.20×1019J size 12{ size 11{1 "." "20" times "10" rSup { size 8{ - "19"} } `J}} {}

(b) 1.24×1017J1.24×1017J size 12{ size 11{1 "." "24" times "10" rSup { size 8{ - "17"} } `J}} {}

43.

458 K 458 K size 12{ size 11{"458"`K}} {}

45.

1 . 95 × 10 7 K 1 . 95 × 10 7 K size 12{ size 11{1 "." "95" times "10" rSup { size 8{7} } `K}} {}

47.

6 . 09 × 10 5 m/s 6 . 09 × 10 5 m/s size 12{ size 11{6 "." "09" times "10" rSup { size 8{5} } `"m/s"}} {}

49.

7 . 89 × 10 4 Pa 7 . 89 × 10 4 Pa size 12{ size 11{7 "." "89" times "10" rSup { size 8{4} } " Pa"}} {}

51.

(a) 1.99×105 Pa1.99×105 Pa size 12{ size 11{1 "." "99" times "10" rSup { size 8{5} } " Pa"}} {}

(b) 0.97 atm

53.

3 . 12 × 10 4 Pa 3 . 12 × 10 4 Pa size 12{ size 11{3 "." "12" times "10" rSup { size 8{4} } " Pa"}} {}

55.

78.3%

57.

(a) 2.12×104 Pa2.12×104 Pa size 12{ size 11{2 "." "12" times "10" rSup { size 8{4} } " Pa"}} {}

(b) 1.06%1.06% size 12{ size 11{1 "." "06"%}} {}

59.

(a) 8.80×102 g8.80×102 g size 12{ size 11{8 "." "80" times "10" rSup { size 8{ - 2} } " g"}} {}

(b) 6.30×103 Pa6.30×103 Pa size 12{ size 11{6 "." "30" times "10" rSup { size 8{3} } " Pa"}} {}; the two values are nearly identical.

61.

82.3%

63.

4 . 77 º C 4 . 77 º C size 12{ size 11{4 "." "77"°C}} {}

65.

38 . 3 m 38 . 3 m size 12{ size 11{"38" "." 3``m}} {}

67.

FB/wCuFB/wCu=1.02FB/wCuFB/wCu=1.02 size 12{ { { left ( {F rSub { size 8{B} } } slash {w rSub { size 8{"Cu"} } } right )} over { left ( {F rSub { size 8{B} } } slash {w rSub { size 8{"Cu"} } } right ) rSup { size 8{′} } } } =1 "." "02"} {}. The buoyant force supports nearly the exact same amount of force on the copper block in both circumstances.

69.

(a) 4.41×1010mol/m34.41×1010mol/m3 size 12{ size 11{4 "." "41" times "10" rSup { size 8{"10"} } `"mol/m" rSup { size 8{3} } }} {}

(b) It’s unreasonably large.

(c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used.

71.

(a) 7.03×108m/s7.03×108m/s size 12{ size 11{7 "." "03" times "10" rSup { size 8{8} } `"m/s"}} {}

(b) The velocity is too high—it’s greater than the speed of light.

(c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered.

Test Prep for AP® Courses

1.

(a), (c)

3.

(d)

5.

(b)

7.

(a) 7.29 × 10-21 J; (b) 352K or 79ºC

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