PHYS 204 Other, Fall 2010

9 10.1 (a) T, ~gte +32 =—(-273.15)+ 32 =| —460°F a 5 6) T.=5(T,—32)=5 (986-32) = [370°C 5 9 9 Te +32 =—(T — 273.15)+32= 3 (-173.15) +32= — 280°F 10.2 When the volume of a low density gas is held constant, pressure and temperature are related by a linear equation P= AT + 8, where A and B are constants to be determined. For the given constant-volume gas thermometer, P= 0.700 atm when T=100°C = 0.700 atm= A(100°C)+B fl] P=0.512 atm when T=O0°C = 0.512 atm=A(0)+B [2] From Equation [2], B= 0.512 atm. Substituting this result into Equation [1] yields _ 0.700 atm — 0.512 atm 1o0°c A =1.88x10% atm/*C so, the linear equation for this thermometer is: P = (1.88 x10 atm/°C)T +0.512 atm. (a) If P= 0.0400 atm, then _ P-B_ 0.0400 atm—0.512 atm _ T Sa A 1.88 x10° atm/"C — 251°C (b) If T = 450°C, then P =(1.88 x10 atm/°C)(450°C) + 0.512 atm = [1.36 atm 10.3 (a) T,= 21. +32= 2(-196)+ 32 =[-321°F T,, = T, + 273.15 = (-196 + 273.15) K = (b) T, = 27 +32=2(37.0)+ 32= 8.6°F TIK T,, = T+ 273.15 = (37.0 + 273.15) K = 310K 10.4 (a) T= 2 (I, ~32)=3(134-32)= 56.7°C 5 T= hintaan 32)=|-—62.1°C (b) Ty =T, +273.15=(56.7+ 273.15) K =B30 K and T= Te + 273.15 = (-62.1°C + 273.15) K= 211K 10.5 Start with T, =—40°F and convert to Celsius. ‘= (7, —32)= =(-40- 32)=[-40°C Since Celsius and Fahrenheit degrees of temperature change are different sizes, this is the only temperature with the same numeric value on both scales. 10.6 Since we have a linear graph, we know that the pressure is related to the temperature as P=A+BT,, where A and B are constants. To find A and B, we use the given data: 0.900 atm = A + B(-80.0°C) and 1.635 atm = A + B(78.0°C) Solving Equations [1] and [2] simultaneously, we find: A=1.27 atm and B=4.65x10™ atm/°C Therefore, P=1.27 atm+(4.65x 107 atm/°C)T, (a) At absolute zero the gas exerts zero pressure (P = 0), so _ —1.27 atm ¥. 4.65107 atm/°C T, c — 273°C (b) At the freezing point of water, T, = 0 and P=1.27 atm+0=|1.27 atm At the boiling point of water, T,, = 100°C, so P=1.27 atm+(4.65x10™ atm/°C)(100°C) = [1.74 atm (1) [2] 10.7 Apply T, =27T,, +32 to two different Celsius temperatures, (T.), and (T,. e; to obtain (T;),= (Te ), +32 [1] and 9 (Tr), = sll); +32 [2] Subtracting Equation [1] from [2] yields (T,), -(T;), — =[(r), ay (T.), | or AT, = (9/5) AT, 10.8 (a) Using the result of Problem 10.7 above, AT, = 3(AT, )= 3(57.0)°C =BL7°C]. (b) AT, = (Tega $273.15) — (Teas, 273-15) = (Tear — Teun )= AT: = LT K cul 10.9 Ty =2T, +32 =2(43)+32 =(77+32)°F =[109°F]. Yes|. The normal body temperature is 98.6°F, so this patient has a high fever and needs immediate attention. 10.10 a) _ (b) Since temperature differences on the Rankine and Fahrenheit scales are identical, the temperature readings on the two thermometers must differ by no more than an additive constant (Le., 7, = 7, + constant). To evaluate this constant, consider the temperature readings on the two scales at absolute zero. We have 7, = O0°R at absolute zero, and 9 9 T, =—T,, + 32=—(-273.15)+ 32=-459.67°F 5 5 Substituting these temperatures in our Fahrenheit to Rankine conversion gives O° =—459.67° +constant or constant = 459.67° giving |T, =7, + 459.67]. We start with the Kelvin temperature and convert to the Rankine temperature in several stages, using the Fahrenheit to Rankine conversion from part (a) above. Ty =T,.+273.15= =r, —32)+273. 5=[(7,- 459.67) — 32 ]+273.15 = (1, — 491.67)+273.15= oT, =(491 67)+273.15= 21, 273.15+273.15= aT,