CHAPTER 17: Temperature, Thermal Expansion, and the Ideal Gas Law

Responses to Questions

2. Properties of materials that could be exploited in making a thermometer include:

a. thermal expansion, both linear and volume

b. the proportionality between temperature and pressure (for an ideal gas when volume is

held constant)

c. temperature dependence of resistivity

d. frequency of emitted radiation from a heated object (blackbody radiation curve).

(Note: resistivity and blackbody radiation are defined in Volume II.)

5. No. We can only infer that the temperature of C is different from that of A and B. We cannot infer anything about the relationship of the temperatures of A and B.

9. Lower. Mercury has a larger coefficient of volume expansion than lead. When the temperature rises, mercury will expand more than lead. The density of mercury will decrease more than the density of lead will decrease, and so the lead will need to displace more mercury in order to balance its weight.

11. If water is added quickly to an overheated engine, it comes into contact with the very hot metal parts of the engine. Some areas of the metal parts will cool off very rapidly; others will not. Some of the water will quickly turn to steam and will expand rapidly. The net result can be a cracked engine block or radiator, due to the thermal stress, and/or the emission of high temperature steam from the radiator. Water should always be added slowly, with the engine running. The water will mix with the hotter water already in the system, and will circulate through the engine, gradually cooling all parts at about the same rate.

15. Slow. On a hot day, the brass rod holding the pendulum will expand and lengthen, increasing the length and therefore the period of the pendulum, causing the clock to run slow.

16. Soda is essentially water, and water (unlike most other substances) expands when it freezes. If the can is full while the soda is liquid, then as the soda freezes and expands, it will push on the inside surfaces of the can and the ends will bulge out.

17. The coefficient of volume expansion is much greater for alcohol than for mercury. A given temperature change will therefore result in a greater change in volume for alcohol than for mercury. This means that smaller temperature changes can be measured with an alcohol thermometer.

Solutions to Problems 6. Assume that the temperature and the length are linearly related. The change in temperature per unit length change is as follows.

T 100.0  C  0.0  C  9.970C  cm l 21.85 cm 11.82 cm

Then the temperature corresponding to length L is T ll 0.0 C    11.82 cm  9.970C cm .

(a) T 18.70 cm  0.0 C   18.70 cm  11.82 cm 9.970C cm   68.6  C

(b) T 14.60 cm  0.0 C   14.60 cm  11.82 cm 9.970C cm   27.7  C

7. Take the 300 m height to be the height in January. Then the increase in the height of the tower is given by Eq. 17-1a.

6 ll  0 T 12  10 C   300 m  25 C  2  C   0.08m

10. The increase in length of the rod is given by Eq. 17-1a.

ll0.010 ll 0fTT  TTi 25 C+ 6 551.3  C  550  C ll0019 10 C

M 1.00 103 kg 11. The density at 4oC is   . When the water is warmed, the mass will stay the V 1.00m3 same, but the volume will increase according to Eq. 17-2. 6 3 2 3 VVT  0  210  10 C   1.00m  94 C  4  C   1.89  10 m

M 1.00 103 kg The density at the higher temperature is     981kg m3 V 1.00m3  1.89 102 m 3

14. Assume that each dimension of the plate changes according to Eq. 17-1a.

AAA ll  w  w  l w  l w  l w w l l w l w 0 l w  w  l   l  w

Neglect the very small quantity l w .

Al w w l l  w  T  w  ll  T  2  w  T

15. The change in volume of the aluminum is given by the volume expansion formula, Eq. 17-2.

3 634 8.75 cm VVT 0  75  10 C  3 180 C  30  C  3.9cm 2

20. (a) When a substance changes temperature, its volume will change by an amount given by

Eq. 17-2. This causes the density to change.

MMMMMMM 1   f          1 VVVVVVVTVVT0 0   0 0 0  0 0 1 

 11 TT            1 TT 1    1  T  If we assume that T = 1, then the denominator is approximately 1, so     T .

(b) The fractional change in density is

   T   T 87 1063  C  55 C 25 C 6.96  10 

This is a 0.70% increase .

22. (a) The original surface area of the sphere is given by Ar 4 2 . The radius will expand with

2 temperature according to Eq. 17-1b, rnew  r1   T  . The final surface area is Arnew 4 new

, and so the change in area is AAA new  .

A A A4 r2 1   T22 4  r2  4  r2  1   T 1 new    

 4r2  1  2   T   2  T2 1  4 r2 2   T 1 1   T    2 

1 2 If the temperature change is not large, 2 T = 1, and so A 8 r  T

(b) Evaluate the above expression for the solid iron sphere.

2 A8 r2   T 8   60.0  102 m   12  106 C  275  C 15 C 2.8 102 m 2

25. The thermal stress must compensate for the thermal expansion. E is Young’s modulus for the

aluminum.

Stress FAET   25  106 C   70  109 N m 2  35 C  18  C   3.0  107 N m 2

PV 31. Assume the gas is ideal. Since the amount of gas is constant, the value of is constant. T

PV1 1 PV 2 2 PT1 2 331.00 atm 273 38.0  K   VV21   3.80m   1.35m TT1 2 PT2 1 3.20 atm 273 K

33. Assume the nitrogen is an ideal gas. From Example 17-10, the volume of one mole of nitrogen gas at STP is 22.4 1033 m . The mass of one mole of nitrogen, with a molecular mass of 28.0 u, is 28.0 grams. Use these values to calculate the density of the oxygen gas.

M 28.0 103 kg     1.25kg m3 V 22.4 1033 m

37. (a) Assume the nitrogen is an ideal gas. The number of moles of nitrogen is found from the atomic

weight, and then the ideal gas law is used to calculate the volume of the gas.

1 mole N n 28.5 kg  2 1017 mol 28.01 103 kg

nRT 1017 mol  8.314J molg K 273 K  PV nRT  V   22.79 m33  22.8m P 1.013 105 Pa

(b) Hold the volume and temperature constant, and again use the ideal gas law.

1 mole N n 28.5 kg  25.0 kg 2  1910 mol 28.01 103 kg PV nRT nRT 1910 mol  8.314 J molg K 273 K  P   1.90  105 Pa  1.88atm V 22.79 m3

41. We assume that the gas is ideal, that the amount of gas is constant, and that the volume of the gas is constant.

PP P 2.00atm 1 2  TT 2 273.15  20.0 K 586.3K  313.15  C  313  C 21    TT1 2 P1 1.00atm

42. Assume that the air is an ideal gas. The pressure and volume are held constant. From the ideal gas PV law, the value of  nT is held constant. R

nT21273 38  K 311 n1 T 1 n 2 T 2      1.080 nT12273 15  K 288

Thus 8.0% must be removed.

48. From the ideal gas equation, we have PV1 1 nRT 1 and PV2 2 nRT 2 , since the amount of gas is constant. Use these relationships along with the given conditions to find the original pressure and temperature. PV1 1 nRT 1 ; PV 2 2  nRT2  PV1 1  PV 2 2  nRT1  nRT 2  nR T1  T 2  

PV1 1 P 1 450 Pa  V2  nR  T1  T 2   P1  V 1  V 2  V 2 450 Pa  nR  9.0 K   nR9.0 K  V  450 Pa  4.0 mol  8.314 J molg K 9.0 K    0.018m3  450 Pa  2 P1  3 VV12 0.002 m  1.537  1055 Pa  1.5  10 Pa

53 PV 1.537 10 Pa 0.020 m  TT11   92.4 K   181  C 11nR 4.0 mol  8.314 J molg K

52. We assume that the water is at 4oC so that its density is 1000kg m3 .

1033 m 1000 kg 1 mol 1.000 L  55.51 mol 33 1 L 1 m 15.9994 2  1.00794  10 kg

23 6.022 10 molecules 25 55.51 mol3.343 10 molecules 1 mol

53. We use Eq. 17-4.

33 N 1molecule  100 cm 23 J 17 PV NkT  P  kT  1.38 10 3K 4  10 Pa 33    V 1cm  m K

55. Assume the gas is ideal at those low pressures, and use the ideal gas law.

12 2 6 3 NP 1 10 N m 8 molecules 10 m PV NkT    3  10 23 3 3 V kT 1.38 10 J K 273 K  m 1 cm  300 molecules cm3

56. We assume an ideal gas at STP. Example 17-10 shows that the molar volume of this gas is 22.4 L. 3 We calculate the actual volume of one mole of gas particles, assuming a volume of l 0 , and then find the ratio of the actual volume of the particles to the volume of the gas.

23 10 3 V 6.02 10 molecules  3.0 10 m  molecule  4 molecules  33 7.3  10 Vgas 22.4 L  1 10 m 1L The molecules take up less than 0.1% of the volume of the gas.