General Physics (PHY 2130)

Transcription

1 General Physics (PHY 2130) Lecture 30 Thermal physics Thermal expansion Gases. Absolute temperature Ideal Gas law Exam 3 review

2 Lightning Review Last lecture: 1. Sound Pipes. Doppler s effect 2. Thermal physics Temperature. Temperature scales. 0 th law of thermodynamics. Review Problem: When a hole is made in the side of a container holding water, water flows out and follows a parabolic trajectory. I the container is dropped in free fall, the water flow 1. diminishes 2. stops altogether 3. goes out in a straight line 4. curves upward Recall: pressure ~ ρgh in a container frame of reference

3 Comparing Temperature Scales T T C F = T K = T + 32 C 5 9 Δ T = ΔT F 5 C

4 4 Example: On a warm summer day, the air temperature is 84 F. Express this temperature in (a) C and (b) kelvins. Fahrenheit/Celsius T T F C = = ( 1.8 F/ C) T F F/ 0 0 F C T C + 32 F Absolute/Celsius T = T C

5 5 Example: (a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? Idea: use relations among temperature scales: T F = 1.8T T C C + 32 = = 40 C T C (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees? T T F F = 1.8T = 1.8 = 1.8 C = 574 F + 32 ( T 273) ( T 273) F

6 Thermal Expansion The thermal expansion of an object is a consequence of the change in the average separation between its constituent atoms or molecules At ordinary temperatures, molecules vibrate with a small amplitude As temperature increases, the amplitude increases This causes the overall object as a whole to expand

7 Linear (area, volume) Expansion For small changes in temperature ΔL = α L o ΔT The coefficient of linear expansion,, depends on the material Similar in two dimensions (area expansion) and in three dimensions (volume expansion) α ΔA = γ A o ΔT, γ = 2α ΔV = β V o ΔT for solids, β = 3α

8 8 Area Expansion, once again How does the area of an object change when its temperature changes? The blue square has an area of L 02. With a temperature change ΔT each side of the square will have a length change of ΔL = αδtl 0. L 0 L 0 +ΔL In other words, Thus, A = ( L 0 +αδtl )( 0 L 0 +αδtl ) 0 = L αΔTL 0 2 +α 2 ΔT 2 L 0 2 L αΔTL 0 2 = A αΔT ( ) ΔA = A A 0 = γ A o ΔT, with γ = 2α

9 Expansion joints permit the roadbed of a bridge to expand and contract as the temperature changes 9

10 Example A copper telephone wire has essentially no sag between poles 35.0 m apart on a winter day when the temperature is 20.0 C. How much longer is the wire on a summer day when T C = 35.0 C? Assume that the thermal coefficient of copper is constant throughout this range at its room temperature value.

11 Applications of Thermal Expansion 1. Thermostats Use a bimetallic strip Two metals expand differently 2. Pyrex Glass Thermal stresses are smaller than for ordinary glass 3. Sea levels Warming the oceans will increase the volume of the oceans

12 12 Molecular Picture of a Gas The number density of particles = the number of particles in a unit volume = N/V where N is the total number of particles contained in a volume V. If a sample contains a single element, the number of particles in the sample is N = M/m. N is the total mass of the sample (M) divided by the mass per particle (m).

13 Ideal Gas Properties of gases A gas does not have a fixed volume or pressure In a container, the gas expands to fill the container Ideal gas: Collection of atoms or molecules that move randomly Molecules exert no long-range force on one another Molecules occupy a negligible fraction of the volume of their container Most gases at room temperature and pressure behave approximately as an ideal gas

14 Moles It s convenient to express the amount of gas in a given volume in terms of the number of moles, n n = mass molar mass One mole is the amount of the substance that contains as many particles as there are atoms in 12 g of carbon-12

15 Avogadro s Hypothesis Equal volumes of gas at the same temperature and pressure contain the same numbers of molecules Corollary: At standard temperature and pressure, one mole quantities of all gases contain the same number of molecules This number is called N A Can also look at the total number of particles: N = n N A

16 Avogadro s Number The number of particles in a mole is called Avogadro s Number N A =6.02 x particles / mole The mass of an individual atom can be calculated: m = atom molar mass N A

17 Equation of State for an Ideal Gas Boyle s Law At a constant temperature, pressure is inversely proportional to the volume Charles Law At a constant pressure, the temperature is directly proportional to the volume Gay-Lussac s Law At a constant volume, the pressure is directly proportional to the temperature

18 Ideal Gas Law Summarizes Boyle s Law, Charles Law, and Guy-Lussac s Law PV = n R T R is the Universal Gas Constant R = 8.31 J / mole K R = L atm / mole K P V = N k B T k B is Boltzmann s Constant k B = R / N A = 1.38 x J/ K

19 Question An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples? a. 1/9 b. 1/3 c. 3.0 d. 9.0

20 Question An ideal gas is confined to a container with adjustable volume. The number of moles and temperature are constant. By what factor will the volume change if pressure triples? a. 1/9 b. 1/3 c. 3.0 d. 9.0

21 Example: Incandescent lightbulbs are filled with an inert gas to lengthen the filament life. With the current off (at T = 20.0 C), the gas inside a lightbulb has a pressure of 115 kpa. When the bulb is burning, the temperature rises to 70.0 C. What is the pressure at the higher temperature? 21

22 Exam 3 Review

23 Exam 3 Review Chapter 9: Solids and fluids density and pressure buoyant force Archimedes principle Fluids in motion Note: no problems on viscosity

24 Example: A flat-bottomed barge loaded with coal has a mass of kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline? 24 FBD for the barge y F B w x Apply Newton s 2 nd Law to the barge: ρ m w F w ρ V F = B g w = = = F w B w( Ad ) = mb ( ρ V ) m b w w = 0 w g = m b g d = mb ρ A w = kg ( kg/m )( 20.0 m*10.0 m) 5 = 1.5 m