Chapter 18 1 From Eq 18 6 we see that the limiting value of the pressure ratio is the same as the absolute temperature ratio 373 15 K 273 16 K 1 366 2 We take p3 to be 80 kPa for both thermometers According to Fig 18 6 the nitrogen thermometer gives 373 35 K for the boiling point of water Use Eq 18 5 to compute the pressure The hydrogen thermometer gives 373 16 K for the boiling point of water and a The difference is pN pH 0 056 kPa b The pressure in the nitrogen thermometer is higher than the pressure in the hydrogen thermometer 3 Let TL be the temperature and pL be the pressure in the left hand thermometer Similarly let TR be the temperature and pR be the pressure in the right hand thermometer According to the problem statement the pressure is the same in the two thermometers when they are both at the triple point of water We take this pressure to be p3 Writing Eq 18 5 for each thermometer we subtract the second equation from the first to obtain First we take TL 373 125 K the boiling point of water and TR 273 16 K the triple point of water Then pL pR 120 torr We solve 841 N3373 35K 80kPa 109 343kPa 273 16K273 16KTpp H373 16K 80kPa 109 287kPa 273 16Kp 0 06 kPa 33 273 16K and 273 16K LRLRppTTpp 3 273 16K LRLRppTTp 3120torr373 125K273 16K 273 16K p 842 CHAPTER 18 for p3 The result is p3 328 torr Now we let TL 273 16 K the triple point of water and TR be the unknown temperature The pressure difference is pL pR 90 0 torr Solving the equation for the unknown temperature we obtain TR 348 K 4 a Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y Then For x 71 C this gives y 96 F b The relationship between y and x may be inverted to yield Thus for y 134 we find x 56 7 on the Celsius scale 5 a Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y Then If we require y 2x then we have which yields y 2x 320 F b In this case we require and find which yields y x 2 12 3 F 6 We assume scales X and Y are linearly related in the sense that reading x is related to reading y by a linear relationship y mx b We determine the constants m and b by solving the simultaneous equations which yield the solutions m 40 00 500 0 8 000 10 2 and b 60 00 With these values we find x for y 50 00 90 0torr273 16K 273 16K 328torrRT 9532yx 59 32 xy 9532yx 9232 5 32 160C5xxx 12yx 19 10 32 32 24 6C2513xxx 70 00125 030 00375 0mbmb 50 0060 001375X 0 08000ybxm 843 7 We assume scale X is a linear scale in the sense that if its reading is x then it is related to a reading y on the Kelvin scale by a linear relationship y mx b We determine the constants m and b by solving the simultaneous equations which yield the solutions m 100 170 53 5 0 858 and b 419 With these values we find x for y 340 8 The increase in the surface area of the brass cube which has six faces which had side length L at 20 is 9 The new diameter is 10 The change in length for the aluminum pole is 11 The volume at 30 C is given by where we have used 3 12 a The coefficient of linear expansion for the alloy is Thus from 100 C to 0 C we have The length at 0 C is therefore L L L 10 015 cm 0 0188 cm 9 996 cm 373 15 53 5 273 15 170 mbmb 34041992 1X 0 858ybxm 22262b26 6121212 1910 C 30cm 75C20C 11cm ALLLLLLT 601 1 2 725cm 1 2310 C 100 0C0 000C 2 731cm ADDT 601 33m 2310 C 15 C 0 011m AT 363 1 13 50 00cm 13 29 0010 C 30 00C60 00C 49 87cmVVTVT 510 015cm10 000cm1 8810 C 10 01cm 100C20 000C LLT 52 10 015cm 1 8810 C 0C100C 1 88 10cm LLT 844 CHAPTER 18 b Let the temperature be Tx Then from 20 C to Tx we have giving T 48 C Thus Tx 20 C 48 C 68 C 13 THINK The aluminum sphere expands thermally when being heated so its volume increases EXPRESS Since a volume is the product of three lengths the change in volume due to a temperature change T is given by V 3 V T where V is the original volume and is the coefficient of linear expansion see Eq 18 11 ANALYZE With the volume of the sphere given by V 4 3 R3 where R 10 cm is the original radius of the sphere and then The value for the coefficient of linear expansion is found in Table 18 2 LEARN The change in volume can be expressed as where is the coefficient of volume expansion For aluminum we have 14 a Since A D2 4 we have the differential dA 2 D 4 dD Dividing the latter relation by the former we obtain dA A 2 dD D In terms of s this reads We can think of the factor of 2 as being due to the fact that area is a two dimensional quantity Therefore the area increases by 2 0 18 0 36 b Assuming that all dimensions are allowed to freely expand then the thickness increases by 0 18 c The volume a three dimensional quantity increases by 3 0 18 0 54 d The mass does not change e The coefficient of linear expansion is 510 009cm 10 000cm 1 8810 C 10 000cm LLTT 62310 C 336343 2310 C410cm100C29cm 3VRT VVT 636910 C 2 for 1 ADDADD 250 18101 810C 100CD DT 845 15 After the change in temperature the diameter of the steel rod is Ds Ds0 sDs0 T and the diameter of the brass ring is Db Db0 bDb0 T where Ds0 and Db0 are the original diameters s and b are the coefficients of linear expansion and T is the change in temperature The rod just fits through the ring if Ds Db This means Ds0 sDs0 T Db0 bDb0 T Therefore The temperature is T 25 00 C 335 0 C 360 0 C 16 a We use m V and The percent change in density is b Since L L T 0 23 10 2 100 C 0 0 C 23 10 6 C the metal is aluminum using Table 18 2 17 THINK Since the aluminum cup and the glycerin have different coefficients of thermal expansion their volumes would change by a different amount under the same T EXPRESS If Vc is the original volume of the cup a is the …
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