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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 = 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

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 +L03 Δ T . V L 0 3 1 + 3αΔT = L 0 3 +L03 Δ T .

So writing the length terms in terms of volumes gives V=V0+ΔVV0+V0ΔT,V=V0+ΔVV0+V0ΔT, 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 .
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

(b) nRT=(mol)(cal/molK)(K)= calnRT=(mol)(cal/molK)(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

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) 3.7×1017Pa3.7×1017Pa

(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) 1.20×1019J1.20×1019J

(b) 1.24×1017J1.24×1017J

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) 8.80×102 g8.80×102 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.

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