Ch 19.3 #1

A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a temperature of –80.0°C) and in boiling ethyl alcohol (78.0°C). The two pressures are 0.900 atm and 1.635 atm.

(a) What Celsius value of absolute zero does the calibration yield?

What is the pressure at (b) the freezing point of water and (c) the boiling point of water?

PV = nRT (V is constant) T is directly proportional to P y = mx + b; Plot T vs P So find slope, then y-intercept m = (78 - -80) / (1.635 - 0.9) m = 214.966 °C/atm Obtain y-int y = m x + b 78 = 214.966(1.635) + b b = -273.47 °C

(a) y = mx + b y = 214.966(0) + -273.47 0 K = -273.47 °C (b) y = 214.966(x) + -273.47 0 = 214.966(x) + -273.47 x = 1.27 atm

(c) y = m x + b T = 214.966 P + -273.47 y = 214.966(x) + -273.47 100 = 214.966(x) + -273.47 x = 1.74 atm

Ch 19.4 #13

The active element of a certain laser is made of a glass rod 30.0 cm long by

1.50 cm in diameter. If the temperature of the rod increases by 65.0°C, what is the increase in (a) its length, (b) its diameter, and (c) its volume? Assume that the average coefficient of linear expansion of the glass is 9.00 ´ 10^–6(°C)^–1.

(a) ΔL = α L ΔT ΔL = 9´10^–6(300 mm)65 ΔL = 0.176 mm

(b) area is πr^2…but diameter is just 2πr so still linear ΔL = α L ΔT ΔL = 9´10^–6(15 mm)65 ΔL = 0.00878 mm

(c) Volume is Lπr^2 (cubic)…so 3 α ΔV = 3α V ΔT ΔV = 3*9´10^–6(Lπr^2) 65 ΔV = 3*9´10^–6(30π(1.5/2)^2) 65 ΔV = 0.0930 mm^3

Ch 19.4 #15

A brass ring of diameter 10.00 cm at 20.0°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0°C. Assuming the average coefficients of linear expansion are constant, (a) to what temperature must this combination be cooled to separate them? Is this attainable? (b) What If? What if the aluminum rod were 10.02 cm in diameter?

Why this equation?…we know at a certain temp their total length after expansion (negative) will be equal to each other.

LAl + DLAl = LBr + DLBr LAl + LAl aAl DT = LBr + LBr aBr DT Solve for DT DT = (LAl - LBr) / (LBr aBr - LAl aAl) DT = (10.01 - 10)/(10*19x10^-6 – 10.01*24x10^-6) DT = -199 C° 20°C + -199 C° > -273.15 °C; yes…it can be done

b) DT = (LAl - LBr) / (LBr aBr - LAl aAl) DT = (10.02-10)/(10*19x10^-6 –10.02*24x10^-6) DT = -396 C° 20°C + -396 C° < -273.15 °C; no, we can’t go below absolute zero

Alpha varies slightly depending on source…lab manual a_Al = 25x10^{-6 lecture} book a_Al = 24x10^-6

Also…Brass is an alloy…so alpha will vary also…

Ch 19.5 #33 The mass of a hot-air balloon and its cargo (not including the air inside) is 200 kg. The air outside is at 10.0°C and 101 kPa. The volume of the balloon is 400 m^3. To what temperature must the air in the balloon be heated before the balloon will lift off? (Air density at 10.0°C is 1.25 kg/m^3.) Hint: Since n/V is a measure of density we have n/V = (P/R) 1/T Pressure is constant since open hole…so n/V is inversely proportional to Temperature
mballoong – FB = 0 mballoong - (Vρairg -Vρhot airg) = 0 mballoon = Vρair - Vρhot air mballoon/V = ρair - ρhot air ρhot air = 1.25 – (200/400) ρhot air = 0.75 kg/m^3	ρhot air/ρair = T/Thot 0.75/1.25 = 283.15/Thot Thot = 472 K

Ch 19.5 #57

Consider an object with any one of the shapes displayed in Table 10.2. What is the percentage increase in the moment of inertia of the object when it is heated from 0°C to 100°C if it is composed of (a) copper or (b) aluminum? Assume that the average linear expansion coefficients shown in Table 19.1 do not vary between 0°C and 100°C.

I = ∫r^2dm Δr = ri α ΔT rf – ri = ri α ΔT rf = ri + ri α ΔT rf = ri (1 + α ΔT) If / Ii = ∫rf^2dm / ∫ri^2dm If / Ii = ∫ri^2(1 + αΔT)^2dm /∫ri^2dm If / Ii = ∫ri^2dm(1 + αΔT)^2/∫ri^2dm	If / Ii = ∫ri^2dm(1 + αΔT)^2/∫ri^2dm If / Ii = (1 + α ΔT)^2 If / Ii = 1^2 + 2 α ΔT + α^2ΔT^2 Since α^2ΔT^2 << 1, it cancels out If / Ii = 1 + 2 α ΔT If / Ii - 1 = 2 α ΔT (If - Ii) / Ii = 2 α ΔT
(a) αCu = 17.0 x 10^-6 (If - Ii) / Ii = 2(17.0x10^-6)100 (If - Ii) / Ii = 0.340 %	(b) αAl = 24.0 x 10^-6 (If - Ii) / Ii = 2(24.0x10^-6)100 (If - Ii) / Ii = 0.480 %