1. Physics 207: Lecture 23, Pg 1 Lecture 23 Goals: Chapter 16 Chapter 16  Use the ideal-gas law.  Use pV diagrams for ideal-gas processes. Chapter 17 Chapter 17  Employ energy conservation in terms of 1 st law of TD  Understand the concept of heat.  Relate heat to temperature change  Apply heat and energy transfer processes in real situations  Recognize adiabatic processes. Assignment Assignment  HW9, Due Wednesday, Apr. 15 th  HW10, Due Wednesday, Apr. 22 nd (9 AM)

2. Physics 207: Lecture 23, Pg 2 Thermodynamics: A macroscopic description of matter l Recall “3” Phases of matter: Solid, liquid & gas l All 3 phases exist at different p,T conditions l Triple point of water: p = 0.06 atm T = 0.01°C l Triple point of CO 2 : p = 5 atm T = -56°C

3. Physics 207: Lecture 23, Pg 3 Modern Definition of Kelvin Scale l Water’s triple point on the Kelvin scale is 273.16 K l One degrees Kelvin is defined to be 1/273.16 of the temperature at the triple point of water Triple point Accurate water phase diagram

4. Physics 207: Lecture 23, Pg 4 Transferring energy to a solid (ice) 1. Temperature increase or 2. State Change If a gas, then V, p and T are interrelated….equation of state

5. Physics 207: Lecture 23, Pg 5 Special system: Water l Most liquids increase in volume with increasing T  Water is special  Density increases from 0 to 4 o C !  Ice is less dense than liquid water at 4 o C: hence it floats  Water at the bottom of a pond is the denser, i.e. at 4 o C Water has its maximum density at 4 C.  (kg/m 3 ) T ( o C) l Reason: Alignment of water molecules

6. Physics 207: Lecture 23, Pg 6 Exercise l Not being a great athlete, and having lots of money to spend, Bill Gates decides to keep the pool in his back yard at the exact temperature which will maximize the buoyant force on him when he swims. Which of the following would be the best choice? (A)0 o C (B) 4 o C (D) 32 o C(D) 100 o C (E) 212 o C

7. Physics 207: Lecture 23, Pg 7 Temperature scales l Three main scales 212 Farenheit 100 Celcius 320273.15 373.15 Kelvin Water boils Water freezes 0-273.15-459.67 Absolute Zero

8. Physics 207: Lecture 23, Pg 8 Some interesting facts l In 1724, Gabriel Fahrenheit made thermometers using mercury. The zero point of his scale is attained by mixing equal parts of water, ice, and salt. A second point was obtained when pure water froze (originally set at 30 o F), and a third (set at 96 ° F) “when placing the thermometer in the mouth of a healthy man”.  On that scale, water boiled at 212.  Later, Fahrenheit moved the freezing point of water to 32 (so that the scale had 180 increments). l In 1745, Carolus Linnaeus of Upsula, Sweden, described a scale in which the freezing point of water was zero, and the boiling point 100, making it a centigrade (one hundred steps) scale. Anders Celsius (1701-1744) used the reverse scale in which 100 represented the freezing point and zero the boiling point of water, still, of course, with 100 degrees between the two defining points. T (K) 10 8 10 7 10 6 10 5 10 4 10 3 100 10 1 0.1 Hydrogen bomb Sun’s interior Solar corona Sun’s surface Copper melts Water freezes Liquid nitrogen Liquid hydrogen Liquid helium Lowest T~ 10 -9 K

9. Physics 207: Lecture 23, Pg 9 Ideal gas: Macroscopic description l Consider a gas in a container of volume V, at pressure P, and at temperature T l Equation of state  Links these quantities  Generally very complicated: but not for ideal gas PV = nRT R is called the universal gas constant In SI units, R =8.315 J / mol·K n = m/M : number of moles l Equation of state for an ideal gas  Collection of atoms/molecules moving randomly  No long-range forces  Their size (volume) is negligible  Density is low  Temperature is well above the condensation point

10. Physics 207: Lecture 23, Pg 10 Boltzmann’s constant l In terms of the total number of particles N l P, V, and T are the thermodynamics variables PV = nRT = (N/N A ) RT k B is called the Boltzmann’s constant k B = R/N A = 1.38 X 10 -23 J/K PV = N k B T l Number of moles: n = m/M  One mole contains N A =6.022 X 10 23 particles : Avogadro’s number = number of carbon atoms in 12 g of carbon m=mass M=mass of one mole

11. Physics 207: Lecture 23, Pg 11 What is the volume of 1 mol of gas at STP ? T = 0 °C = 273 K p = 1 atm = 1.01 x 10 5 Pa The Ideal Gas Law

12. Physics 207: Lecture 23, Pg 12 Example l A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm 3 is at 27 o C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 327 o C, what is the pressure inside the can? Assume any change in the volume of the can is negligible. Steps 1. Convert to Kelvin (From 300 K to 600 K) 2. Use P/T = nR/V = constant  P 1 /T 1 = P 2 /T 2 3. Solve for final pressure  P 2 = P 1 T 2 /T 1 http://www.thehumorarchives.com/joke/WD40_Stupidity

13. Physics 207: Lecture 23, Pg 13 Example problem: Air bubble rising l A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm. The bubble rises to the surface, where the pressure is p 2 = 1.0 atm. The water temperatures at the bottom and the surface are, respectively, T 1 = 4°C, T 2 = 23°C l What is the ratio of the volume of the bubble as it reaches the surface,V 2, to its volume at the bottom, V 1 ? (Ans.V 2 /V 1 = 3.74) l Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture.

14. Physics 207: Lecture 23, Pg 14 Example problem: Air bubble rising l A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm. The bubble rises to the surface, where the pressure is p 2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T 1 = 4°C, T 2 = 23°C l What is the ratio of the volume of the bubble as it reaches the surface,V 2, to its volume at the bottom, V 1 ? (Ans.V 2 /V 1 = 3.74) l Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture. This is addition to “the bends”, or decompression sickness, which is due to the pressure dependent solubility of gas. At depth and at higher pressure N 2 is more soluble in blood. As divers ascend, N 2 dissolved in their blood stream becomes gaseous again and forms N2 bubbles in blood vessels, which in turn can obstruct blood flow, and therefore provoke pain and in some cases even strokes or deaths. Fortunately, this only happens when diving deeper than 30 m (100 feet). The diver in this question only went down 25 meters. How do we know that?

15. Physics 207: Lecture 23, Pg 15 Example problem: Air bubble rising l A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm. The bubble rises to the surface, where the pressure is p 2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T 1 = 4°C, T 2 = 23°C l What is the ratio of the volume of the bubble as it reaches the surface,V 2, to its volume at the bottom, V 1 ? (Ans.V 2 /V 1 = 3.74) l pV=nRT  pV/T = const so p 1 V 1 /T 1 = p 2 V 2 /T 2 V 2 /V 1 = p 1 T 2 / (T 1 p 2 ) = 3.5 296 / (277 1) If thermal transfer is efficient. [More than likely the expansion will be “adiabatic” and, for a diatomic gas, PV  = const. where  = 7/5, see Ch. 17 & 18]

16. Physics 207: Lecture 23, Pg 16 Buoyancy and the Ideal Gas Law l A typical 5 passenger hot air balloon has approximately 700 kg of total mass and the balloon itself can be thought as spherical with a radius of 10.0 m. If the balloon is launched on a day with conditions of 1.0 atm and 273 K, how hot would you have to heat the air inside (assuming the density of the surrounding air is 1.2 kg/m 3 and the air behaves and as an ideal gas) in order to keep the balloon at a constant altitude? l Hint: Remember the weight of the air inside the balloon. Balloon weight = Buoyant force – Weight of hot air Ideal gas law: pV = nRT  nT= pV/R = const. or  T = const. = 1.2 x 273 kg K/m 3

17. Physics 207: Lecture 23, Pg 17 Buoyancy and the Ideal Gas Law m balloon g =  air at 273 K V g –  air at T V g m balloon =  air at 273 K V –  air at T V l m balloon = (1.2 – 330 / T) V l 700 / 4200 = 1.2 – 330 / T l 330 / T = (1.2 - 0.2) l T = 330 K  57 C

18. Physics 207: Lecture 23, Pg 18 PV diagrams: Important processes l Isochoric process: V = const (aka isovolumetric) l Isobaric process: p = const l Isothermal process: T = const Volume Pressure 1 2 Isochoric Volume Pressure 1 2 Isobaric Volume Pressure 1 2 Isothermal

19. Physics 207: Lecture 23, Pg 19 Work and Energy Transfer (Ch. 16) l K reflects the kinetic energy of the system l ΔK =W conservative + W dissipative + W external  W conservative = - ΔU (e.g., gravity)  W dissipative = - ΔE Thermal  W external  Typically, work done by contact forces ΔK + ΔU + ΔE Th = W external = ΔE sys

20. Physics 207: Lecture 23, Pg 20 Work and Energy Transfer (Ch. 17) ΔK + ΔU + ΔE Th = W external = ΔE sys But we can transfer energy without doing work Q ≡ thermal energy transfer ΔK + ΔU + ΔE Th = W + Q = ΔE sys If ΔK + ΔU = ΔE Mech = 0  ΔE Th = W + Q

21. Physics 207: Lecture 23, Pg 21 1 st Law of Thermodynamics l Thermal energy E th : Microscopic energy of moving molecules and stretching molecular bonds. ΔE th depends on the initial and final states but is independent of the process. l Work W : Energy transferred to the system by forces in a mechanical interaction. l Heat Q : Energy transferred to the system via atomic-level collisions when there is a temperature difference. ΔE th =W + Q W & Q with respect to the system

22. Physics 207: Lecture 23, Pg 22 1 st Law of Thermodynamics l Work W and heat Q depend on the step process by which the system is changed (path dependent). l The change of energy in the system, ΔE th depends only on the total energy exchanged W+Q, not on the process. ΔE th =W + Q W & Q with respect to the system

23. Physics 207: Lecture 23, Pg 23 1 st Law: Work & Heat l Work done on system (an ideal gas) l W on system V i )] l If ideal gas, PV = nRT, and given P i & V i fixes T i l W by system > 0 Moving left to right

24. Physics 207: Lecture 23, Pg 24 1 st Law: Work & Heat l Work:  Depends on the path taken in the PV-diagram (It is not just the destination but the path…)  W on system > 0 Moving right to left

25. Physics 207: Lecture 23, Pg 25 1 st Law: Work (“Area” under the curve) l Work depends on the path taken in the PV-diagram : (a) W a = W 1 to 2 + W 2 to 3 (here either P or V constant)  W a (on) = - P i (V f - V i ) + 0 > 0 (b) W b = W 1 to 2 + W 2 to 3 (here either P or V constant)  W b (on) = 0 - P f (V f - V i ) > W a > 0 (c) Need explicit form of P versus V but W c (on) > 0 2 1 3 1 2 3

26. Physics 207: Lecture 23, Pg 26 Recap Assignment Assignment  HW9, Due Wednesday, Apr. 15 th  HW10, Due Wednesday, Apr. 22 nd (9 AM)