College Physics for AP® Courses

# Chapter 13

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) $ΔTºF = T2ºF−T1ºF = 95T2ºC+32.0º−95T1ºC+ 32.0º = 95T2ºC−T1ºC=95ΔTºC ΔTºF = T2ºF−T1ºF = 95T2ºC+32.0º−95T1ºC+ 32.0º = 95T2ºC−T1ºC=95ΔTºC$

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:

$V≈L031+3αΔT= L 0 3 +3αL03ΔT.V≈L031+3αΔT= L 0 3 +3αL03ΔT. size 12{V approx L rSub { size 8{0} } rSup { size 8{3} } left (1+3αΔT right )=L rSub { size 8{0} rSup { size 8{3} } } +3αL rSub { size 8{0} } rSup { size 8{3} } ΔT} {}$

So writing the length terms in terms of volumes gives $V=V0+ΔV≈V0+3αV0ΔT,V=V0+ΔV≈V0+3αV0ΔT, size 12{V=V rSub { size 8{0} } +ΔV approx V rSub { size 8{0} } +3αV rSub { size 8{0} } ΔT,} {}$ and so

$ΔV=βV0ΔT≈3αV0ΔT,orβ≈3α.ΔV=βV0ΔT≈3αV0ΔT,orβ≈3α. size 12{ΔV=βV rSub { size 8{0} } ΔT approx 3αV rSub { size 8{0} } ΔT,"or"β approx 3α} {}$

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/mol⋅K)(K)= JnRT=(mol)(J/mol⋅K)(K)= J size 12{ ital "nRT" = $$"mol"$$ $$"J/mol" cdot K$$ $$K$$ =" J"} {}$

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

(c) $nRT = (mol)(L⋅atm/mol⋅K)(K) = L⋅atm=(m3)(N/m2) = N⋅m=J nRT = (mol)(L⋅atm/mol⋅K)(K) = L⋅atm=(m3)(N/m2) = N⋅m=J$

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) $3.7×10−17Pa3.7×10−17Pa size 12{3 "." 7 times "10" rSup { size 8{ - "17"} } "Pa"} {}$

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

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

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) $8.80×10−2 g8.80×10−2 g size 12{ size 11{8 "." "80" times "10" rSup { size 8{ - 2} } " g"}} {}$

(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.

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

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.

1.

(a), (c)

3.

(d)

5.

(b)

7.

(a) 7.29 × 10-21 J; (b) 352K or 79ºC