Physics 207 Lecture 23. Lecture 23

Transcription

1 Goals: Lecture 3 Chapter 6 Use the ideal-gas law. Use pv diagrams for ideal-gas processes. Chapter 7 Employ energy conservation in terms of 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 HW9, Due Wednesday, Apr. 5 th HW0, Due Wednesday, Apr. nd (9 AM) Physics 07: Lecture 3, Pg Thermodynamics: A macroscopic description of matter Recall 3 Phases of matter: Solid, liquid & gas All 3 phases exist at different p,t conditions Triple point of water: p = 0.06 atm T = 0.0 C Triple point of CO : p = 5 atm T = -56 C Physics 07: Lecture 3, Pg Page

2 Modern Definition of Kelvin Scale Water s triple point on the Kelvin scale is 73.6 K One degrees Kelvin is defined to be /73.6 of the temperature at the triple point of water Accurate water phase diagram Triple point Physics 07: Lecture 3, Pg 3 Transferring energy to a solid (ice). Temperature increase or. State Change If a gas, then V, p and T are interrelated.equation of state Physics 07: Lecture 3, Pg 4 Page

3 Special system: Water 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 ρ(kg/m 3 ) Water has its maximum density at 4 C. Density T ( o C) Reason: Alignment of water molecules Physics 07: Lecture 3, Pg 5 Exercise 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) 3 o C (D) 00 o C (E) o C Physics 07: Lecture 3, Pg 6 Page 3

4 Temperature scales Three main scales Farenheit Celcius Kelvin Water boils T F T C Water freezes Absolute Zero 9 o 5 o = TC + 3 F TC = ( TF 3 F) 5 9 = T 73.5 K T = T C K Physics 07: Lecture 3, Pg 7 Some interesting facts In 74, 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. Later, Fahrenheit moved the freezing point of water to 3 (so that the scale had 80 increments). In 745, Carolus Linnaeus of Upsula, Sweden, described a scale in which the freezing point of water was zero, and the boiling point 00, making it a centigrade (one hundred steps) scale. Anders Celsius (70-744) used the reverse scale in which 00 represented the freezing point and zero the boiling point of water, still, of course, with 00 degrees between the two defining points. T (K) 0 8 Hydrogen bomb Sun s interior Solar corona Sun s surface Copper melts Water freezes Liquid nitrogen Liquid hydrogen Liquid helium Lowest T~ 0-9 K Physics 07: Lecture 3, Pg 8 Page 4

5 Ideal gas: Macroscopic description Consider a gas in a container of volume V, at pressure P, and at temperature T Equation of state Links these quantities Generally very complicated: but not for ideal gas 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 PV = nrt In SI units, R =8.35 J / mol K R is called the universal gas constant n = m/m : number of moles Physics 07: Lecture 3, Pg 9 Boltzmann s constant Number of moles: n = m/m m=mass M=mass of one mole One mole contains N A =6.0 X 0 3 particles : Avogadro s number = number of carbon atoms in g of carbon In terms of the total number of particles N PV = nrt = (N/N A ) RT PV = N k B T k B = R/N A =.38 X 0-3 J/K k B is called the Boltzmann s constant P, V, and T are the thermodynamics variables Physics 07: Lecture 3, Pg 0 Page 5

6 The Ideal Gas Law pv = nrt What is the volume of mol of gas at STP? T = 0 C = 73 K p = atm =.0 x 0 5 Pa V = = nrt P 0.04 ( mol K ) 8.3 J / =.0 0 m 3 = 5 73 K Pa.4 l Physics 07: Lecture 3, Pg Example A spray can containing a propellant gas at twice atmospheric pressure (0 kpa) and having a volume of 5.00 cm 3 is at 7 o C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 37 o C, what is the pressure inside the can? Assume any change in the volume of the can is negligible. Steps. Convert to Kelvin (From 300 K to 600 K). Use P/T = nr/v = constant P /T = P /T 3. Solve for final pressure P = P T /T Physics 07: Lecture 3, Pg Page 6

7 Example problem: Air bubble rising A diver produces an air bubble underwater, where the absolute pressure is p = 3.5 atm. The bubble rises to the surface, where the pressure is p =.0 atm. The water temperatures at the bottom and the surface are, respectively, T = 4 C, T = 3 C What is the ratio of the volume of the bubble as it reaches the surface,v, to its volume at the bottom, V? (Ans.V /V = 3.74) Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture. Physics 07: Lecture 3, Pg 3 Example problem: Air bubble rising A diver produces an air bubble underwater, where the absolute pressure is p = 3.5 atm. The bubble rises to the surface, where the pressure is p = atm. The water temperatures at the bottom and the surface are, respectively, T = 4 C, T = 3 C What is the ratio of the volume of the bubble as it reaches the surface,v, to its volume at the bottom, V? (Ans.V /V = 3.74) 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 is more soluble in blood. As divers ascend, N dissolved in their blood stream becomes gaseous again and forms N 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 (00 feet). The diver in this question only went down 5 meters. How do we know that? Physics 07: Lecture 3, Pg 4 Page 7

8 Example problem: Air bubble rising A diver produces an air bubble underwater, where the absolute pressure is p = 3.5 atm. The bubble rises to the surface, where the pressure is p = atm. The water temperatures at the bottom and the surface are, respectively, T = 4 C, T = 3 C What is the ratio of the volume of the bubble as it reaches the surface,v, to its volume at the bottom, V? (Ans.V /V = 3.74) pv=nrt pv/t = const so p V /T = p V /T V /V = p T / (T p ) = / (77 ) If thermal transfer is efficient. [More than likely the expansion will be adiabatic and, for a diatomic gas, PV γ = const. where γ = 7/5, see Ch. 7 & 8] Physics 07: Lecture 3, Pg 5 Buoyancy and the Ideal Gas Law 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 0.0 m. If the balloon is launched on a day with conditions of.0 atm and 73 K, how hot would you have to heat the air inside (assuming the density of the surrounding air is. kg/m 3 and the air behaves and as an ideal gas) in order to keep the balloon at a constant altitude? 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 ρ T = const. =. x 73 kg K/m 3 Physics 07: Lecture 3, Pg 6 Page 8

9 Buoyancy and the Ideal Gas Law m balloon g = ρ air at 73 K V g ρ air at T V g m balloon = ρ air at 73 K V ρ air at T V m balloon = (. 330 / T) V 700 / 400 =. 330 / T 330 / T = (. - 0.) T = 330 K 57 C Physics 07: Lecture 3, Pg 7 PV diagrams: Important processes Isochoric process: V = const (aka isovolumetric) Isobaric process: p = const pv Isothermal process: T = const = constant T Pressure Isochoric p p = T T Pressure Isothermal p = V pv Pressure Isobaric V V = T T Volume Volume Volume Physics 07: Lecture 3, Pg 8 Page 9

10 Work and Energy Transfer (Ch. 6) K reflects the kinetic energy of the system K =W conservative + W dissipative + W external W conservative = - U (e.g., gravity) W dissipative = - E Thermal W external Typically, work done by contact forces K + U + E Th = W external = E sys Physics 07: Lecture 3, Pg 9 Work and Energy Transfer (Ch. 7) K + U + E Th = W external = E sys But we can transfer energy without doing work Q thermal energy transfer K + U + E Th = W + Q = E sys If K + U = E Mech = 0 E Th = W + Q Physics 07: Lecture 3, Pg 0 Page 0

11 st Law of Thermodynamics E th =W + Q W & Q with respect to the system Thermal energy E th : Microscopic energy of moving molecules and stretching molecular bonds. E th depends on the initial and final states but is independent of the process. Work W : Energy transferred to the system by forces in a mechanical interaction. Heat Q : Energy transferred to the system via atomic-level collisions when there is a temperature difference. Physics 07: Lecture 3, Pg st Law of Thermodynamics E th =W + Q W & Q with respect to the system Work W and heat Q depend on the step process by which the system is changed (path dependent). The change of energy in the system, E th depends only on the total energy exchanged W+Q, not on the process. Physics 07: Lecture 3, Pg Page

12 st Law: Work & Heat Work done on system (an ideal gas) final Won = p dv = (area under curve) initial W on system < 0 Moving left to right [where (V f > V i )] If ideal gas, PV = nrt, and given P i & V i fixes T i W by system > 0 Moving left to right Physics 07: Lecture 3, Pg 3 Work: st Law: Work & Heat 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 Physics 07: Lecture 3, Pg 4 Page

13 st Law: Work ( Area under the curve) Work depends on the path taken in the PV-diagram : 3 3 (a) W a = W to + W to 3 (here either P or V constant) W a (on) = - P i (V f - V i ) + 0 > 0 (b) W b = W to + W 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 Physics 07: Lecture 3, Pg 5 Recap Assignment HW9, Due Wednesday, Apr. 5 th HW0, Due Wednesday, Apr. nd (9 AM) Physics 07: Lecture 3, Pg 6 Page 3