PHYSICS 231 PH 2 1 Review 3 PHY 231 1 For Mid Term Exam PHY 231 2 Pressure Pressure F A N m2 Pa Pascal Same Force different pressure PHY 231 3 The Young s modulus tensile stress Y tensile strain tensile stress F A N m 2 Pascal Pa tensile strain L L0 F L0 F A P Y L L0 A L L L0 Pressure P F A Beyond the elastic limit an object is permanently deformed it does not return to its original shape if th stress the t is i removed d PHY 231 4 Density M V 3 kg m specific material water 4 PHY 231 o C 5 Pascal s Principle If we exert pressure on any part of an incompressible fluid this pressure will be t transferred f d to t the th restt of f the th fluid fl id with ith no losses l This concept is called Pascal s Principle and can be stated When there is a change in pressure at any point in a confined fluid there is an equal change in pressure at every point in the fluid fluid Pascal s Principle is the basis for many modern hydraulic y devices in use today y PHY 231 6 Pascal s principle In other words then before a change in pressure applied to a fluid that is enclosed in transmitted to the whole fluid and all the walls of the container that hold the fluid fluid P F1 A1 F2 A2 If A2 A A1 then th F2 F1 So if we apply a small force F1 we can exert a very large Force F2 Hydraulic press demo PHY 231 7 Pressure and Depth Pdepth h Pdepth 0 gh P0 h Where Pdepth h the p pressure at depth p h Pdepth 0 the pressure at depth 0 density of the liquid g 9 81 m s2 g 9 81 h depth P Pdepth 0 Patmospheric 1 013x10 1 013x105 Pa 1 atm 760 Torr From Pascal s principle If P0 changes then the pressures att all ll depths d th changes h with ith the th same value l PHY 231 8 Archimedes ss Principle Archimedes THE MAGNITUDE OF THE BUOYANT FORCE IS THE WEIGTH OF THE FLUID DISPLACED BY THE OBJECT It is true whether the object is completely or partially submerged in the fluid PHY 231 9 Pressure at depth h P P0 fluidgh h distance between liquid surface and the point where you measure P P0 h P Buoyantt force B f f for submerged b d object bj t B fluidVobjectg Mfluidg wfluid The buoyant y force equals q the weight g of the amount of water that can be put in the volume taken by the object If object is not moving B wobject object fluid b b fl d Buoyant force for floating object The buoyant force equals the weight of the amount of water that can be put in the part of the volume of the object that is under water objectVobject waterVdisplaced h h objectVobject waterA object fluid PHY 231 B h w 10 Buoyant force For floating object object fluid Vdisplaced Mobject water Mobject objectVobject For submerged object object orr fluid Vdisplaced Vobject Vobject Mobject object PHY 231 11 Bernoulli s equation P1 v12 gy1 P2 v22 gy2 P v2 gy constant The sum of the pressure P P the kinetic energy per unit volume v2 and the potential energy per unit volume gy is constant at all points along a path of flow Note that for an incompressible fluid A1v1 A2v2 This is called the equation of continuity PHY 231 12 Buoyant force For floating object object fluid Vdisplaced Mobject water Mobject objectVobject For submerged object object orr fluid Vdisplaced Vobject Vobject Mobject object PHY 231 13 Bernoulli s equation P1 v12 gy1 P2 v22 gy2 P v2 gy constant The sum of the pressure P P the kinetic energy per unit volume v2 and the potential energy per unit volume gy is constant at all points along a path of flow Note that for an incompressible fluid A1v1 A2v2 This is called the equation of continuity PHY 231 14 Temperature scales Conversions Tcelsius Tkelvin 273 15 Tfahrenheit 9 5 Tcelcius 32 We will use Tkelvin If Tkelvin 0 0 the th atoms molecules t l l have no kinetic energy and every substance is a solid it is called the Absolute zero point Celsius Fahrenheit Kelvin PHY 231 15 Thermal expansion length L L Lo T surface f A A A Ao T T 2 2 volume V V V Vo T 3 3 L0 coefficient of linear expansion different for each material Some examples examples 24E 06 1 K Aluminum 1 2E 04 1 K Alcohol PHY 231 T T0 T T0 T 16 example An architect wants to design a 5m high circular pillar with a radius of 0 5 m that holds a bronze statue that weighs 1 0E 04 1 0E 04 kg He chooses concrete for the material of the pillar Y 1 0E 10 Pa How much does the pillar compress F A Y L L0 2 M statue g R ppillar L L0 R 0 5 m L0 5m Y 1 R 0 5 Y 1 0E 10 0E 10 Pa M 1 M 1 0E 04 0E 04 kg 5m L 6 2E 05 m PHY 231 17 Number of particles mol 1 mol of particles 6 02 x 1023 particles A Avogadro s d number b NA 6 02x10 6 02 1023 particles ti l per moll It doesn t m matter what w kind of fp particles 1 mol is always NA particles PHY 231 18 Boyle Charles Gay Lussac IDEAL GAS LAW PV T nR n number of particles in the gas mol R universal gas constant 8 31 J mol K P pressure Pa P P 1 Liter 10 3 m3 1000 cm3 V volumn m3 T temperature p K If no molecules are extracted from or added to a system PV constant T PHY 231 P1V1 P2V2 T1 T2 19 A small matter of definition Ideal gas law PV T nR PV T N NA R PV T N N n number of mols N number um of fm molecules u NA number of molecules in 1 mol Rewrite ideal gas law PV T NkB where kB R NA 1 38x10 23 J K Boltzmann s constant PHY 231 20 Example An ideal gas occupies a volume of 1 0cm3 at 200C at 1 atm A How many molecules mo cu s are ar in n the th volume o um B If the pressure is reduced to 1 0x10 11 Pa while the temperature drops to 00C how many molecules remained in the volume A PV T nR so n PV TR R 8 31 J mol K T 200C 293K P 1atm 1 013x10 P 1atm 1 013x105 Pa V 1 0cm V 1 0cm3 1x10 66m3 n 4 2x10 5 mol n 4 2x10 5 NA 2 5x1019 molecules 11 Pa B T 0 T 00C 273K C 2 3K …
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