College Physics 2e

# Chapter 13

### Problems & Exercises

1.

$102 º F 102 º F$

3.

$20.0ºC20.0ºC$ and $25.6ºC25.6ºC$

5.

$9890 º F 9890 º F$

7.

(a) $22.2ºC22.2ºC$

(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.

$5 . 4 × 10 − 6 m 5 . 4 × 10 − 6 m$

13.

Because the area gets smaller, the price of the land DECREASES by $~17,000.~17,000.$

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: $ΔL=αL0ΔTΔL=αL0ΔT$. Also we know that the volume of a cube is related to its length by $V=L3V=L3$, so the final volume is then $V=V0+ΔV=L0+ΔL3V=V0+ΔV=L0+ΔL3$. Substituting for $ΔLΔL$ gives

$V = L 0 + αL 0 ΔT 3 = L 0 3 1 + αΔT 3 . V = L 0 + αL 0 ΔT 3 = L 0 3 1 + αΔT 3 .$

Now, because $αΔTαΔT$ is small, we can use the binomial expansion:

$V ≈ L 0 3 1 + 3αΔT = L 0 3 +3αL03 Δ T . V ≈ L 0 3 1 + 3αΔT = L 0 3 +3αL03 Δ T .$

So writing the length terms in terms of volumes gives $V=V0+ΔV≈V0+3αV0ΔT,V=V0+ΔV≈V0+3αV0ΔT,$ and so

$Δ V = βV 0 Δ T ≈ 3α V 0 Δ T , or β ≈ 3α . Δ V = βV 0 Δ T ≈ 3α V 0 Δ T , or β ≈ 3α .$
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) $nRT=(mol)(J/mol⋅K)(K)= JnRT=(mol)(J/mol⋅K)(K)= J$

(b) $nRT=(mol)(cal/mol⋅K)(K)= calnRT=(mol)(cal/mol⋅K)(K)= cal$

(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.

$7 . 86 × 10 − 2 mol 7 . 86 × 10 − 2 mol$

30.

(a) $6.02×105km36.02×105km3$

(b) $6.02×108km6.02×108km$

32.

$− 73 . 9 º C − 73 . 9 º C$

34.

(a) $9.14×106N/m29.14×106N/m2$

(b) $8.23×106N/m28.23×106N/m2$

(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) $3.7×10−17Pa3.7×10−17Pa$

(b) $6.0×1017m36.0×1017m3$

(c) $8.4×102km8.4×102km$

39.

$1 . 25 × 10 3 m/s 1 . 25 × 10 3 m/s$

41.

(a) $1.20×10−19J1.20×10−19J$

(b) $1.24×10−17J1.24×10−17J$

43.

$458 K 458 K$

45.

$1 . 95 × 10 7 K 1 . 95 × 10 7 K$

47.

$6 . 09 × 10 5 m/s 6 . 09 × 10 5 m/s$

49.

$7 . 89 × 10 4 Pa 7 . 89 × 10 4 Pa$

51.

(a) $1.99×105 Pa1.99×105 Pa$

(b) 0.97 atm

53.

$3 . 12 × 10 4 Pa 3 . 12 × 10 4 Pa$

55.

78.3%

57.

(a) $2.12×104 Pa2.12×104 Pa$

(b) $1.06%1.06%$

59.

(a) $8.80×10−2 g8.80×10−2 g$

(b) $6.30×103 Pa6.30×103 Pa$; the two values are nearly identical.

61.

82.3%

63.

$4 . 77 º C 4 . 77 º C$

65.

$38 . 3 m 38 . 3 m$

67.

$FB/wCuFB/wCu′=1.02FB/wCuFB/wCu′=1.02$. The buoyant force supports nearly the exact same amount of force on the copper block in both circumstances.

69.

(a) $4.41×1010mol/m34.41×1010mol/m3$

(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) $7.03×108m/s7.03×108m/s$

(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.

Do you know how you learn best?