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Thermal PhysicsIB Physics Topic 3: Ideal gases
Ideal GasesUnderstand and Apply the following.Pressure. Gas Laws (by name)PV = nRTKinetic Molecular TheoryExplain PressureWilliamThompson (Lord Kelvin)
PressurePressure is defined as force per unit areaNewtons per square metre or N/m21 Nm-2 = 1 Pa (pascal)The weight of the person is the force applied to the bed and the small area of each nail tip combines to make an overall large area.The result is a small enough pressure which does not puncture the person.Click on Me
Atmospheric PressureBasically weight of atmosphere!Air molecules are colliding with you right now!Pressure = 1.013 x105 N/m2 = 14.7 lbs/in2!Example: Sphere w/ r = 0.1 mDemoA = 4 p r2 = .125 m2F = 12,000 Newtons (over 2,500 lbs)!21
Qualitative Demonstration of Pressurey16Force due to molecules of fluid colliding with container.Force Impulse = DpAverage Pressure = F / A
PressurePressure is defined as force per unit areaNewtons per square metre N/m2The pressure exerted by a gas results from the atoms/ molecules bumping into the container wallsMore atoms gives more bumps per second and higher pressureHigher temperature means faster atoms and gives more bumps per second and higher pressureAt sea level and 20C, normal atmospheric pressure is1atm 1 x 105 N/m2
GasesGases (as we will see) can behave near perfectly.NA= 6.02 x 1023 molecules mol-1Molecules size ~ 10-8 m to 10-10 mExample: How molecules are there in 6 grams of hydrogen gas?We have 3 moles, H2 has 2 grams mol-13 x 6.02 x 1023 = 1.81 x 1024 molecules.
ExampleA glass contains about 0.3 L of water, which has a mass of about 300 g. Since the molar mass of water (H2O) is 18 g mol-1 Hence, 300g/18g mol-1 ~ 17 mol ~ 1025 molecules Make a rough estimate of the number of water molecules in an ordinary glass of water.
The Boyle-Mariotte Law Gases (at constant temperature) decrease in volume with increasing pressure.P =F/AV = r2 h
Figure 17-3Increasing Pressure by Decreasing Volume
The Boyle-Mariotte Law Gases (at constant temperature) decrease in volume with increasing pressure. Isothermal Process PV = constant P1V1 = P2V2
ExampleThe pressure of gas is 2 atm and its volume 0.9 L if the pressure is increased to 6 atm at constant temperature, what is the new volume? Answer:P1V1 = P2V2
2 x 0.9 = 6 x V; hence V = 0.3 L
The volume-temperature law Charles & Gay-Lussac Isobaric Process V/T = constant V1 / T1 = V2 / T2 At absolute zero a gas would take up zero volume. In reality they liquefy when they get really cold!
The pressure-temperature Law Gases (at constant volume) increase in temperature with increasing pressure. Isochoric Process P/T= constant P1/T1 = P2/T20100200-200-100temp. Cpressure
ExampleA bottle of hair spray is filled to a pressure of 1 atm at 20CWhat is the canister pressure if it is placed into boiling water, and allowed to reach thermal equilibrium?P1/ T1 = P2/ T2 orp1 T2 = p2 T11 / 293 = p2 / 373 p2 = 373/293p2 = 1.27 atm
Absolute zeroIdeal gas has zero volumeResistance of metal drops to zero (actually superconductivity cuts in above 0K)Brownian motion ceases! (kinetic energy = 3/2 kT)But lowest temperature attained is 10-9K0-273 Ctemp. Cpressure
Equations of stateState, identifies whether solid liquid or gasKey parameters or state variablesVolume, V (m3)Pressure, p (N/m2)Temperature, T (K)Mass, M (kg) or number of moles, nEquation of state relates V, p , T, m or n
Bulk vs moleculesConsider force between two molecules At absolute zeroNo thermal energyMolecules sit at r0Above absolute zeroSome thermal energyMolecules are at r> r0 (thermal expansion)At high temperatureThermal energy > binding energyMolecules form a gasrforceenergyr0repulsionattractionbindingenergythermal energy
Equation of state for a gasAll gases behave nearly the samepV = nRTR = 8.3 J/(mol K) for all gases (as long as they remain a gas)T is in K!!!!!!
ExampleWhat is the mass of a cubic metre of air?Molecular weigh of air 32gpV = nRT
Atmospheric pressure = 105 N/m2Atmospheric temp. = 300K
For a volume of 1 m3
n = pV/RT = 105 / (8.3 x 300) = 40 moles
M = 40 x 0.032 = 1.3kg
Constant mass of gasFor a fixed amount of gas, its mass or number of moles remains the samepV/T = nR = constantComparing the same gas under different conditionsp1V1/T1 = p2V2/T2 Hence can use pressure of a constant volume of gas to define temperature (works even if gas is impure - since all gases the same)Must use T in K!!!!!!
ExampleA hot air balloon has a volume of 150m3If heated from 20C to 60C how much lighter does it get?Molecular weight of air 32gpV/T = nRn = pV/RT
Balloon has constant volume and constant pressure
ncool =105x150 / (8.3 x293) = 61680
nhot =105x150 / (8.3 x333) = 54271 Dn = 7409 moles
DM = 7409 x 0.032 = 237kg
Molecules have finite sizeCannot reduce volume of gas to zero!When you try, it becomes a liquidSlightly increases the measured volumeAtoms/ molecules always attract each otherSlightly reduces the measured pressureVan de Waals forces
p-V diagrams (for gases)Useful to consider the pressure/volume changes at constant temperatureIsotherms are p-V values for a fixed amount of gas at constant volumep a 1/VvolumePressureIncreasing temperature
Kinetic theory of gasesA gas consists of a large number of molecules.Molecules move randomly with a range of speeds. (Maxwell's Distribution)The Volume of the molecule is negligible compared with the volume of the gas itself.Collisions are elastic (KE conserved)No inter-molecular forces.Collision time with walls are very smal.Molecules obey Newtons Laws of Mechanics
Molecules in a gasGas atoms/molecules move in a straight linevelocity due to thermal energyKE = 1/2 m vx2 3/2 kTKEavg absolute tempRMS (Root mean squared) Pressure results from collisions with the walls of the container. (NOT collisions between moleculesFimpact = p/t = (2m vx)/t
ExampleHow fast does a typical gas molecule (travel at room temperature? Lets take Nitrogen-14! (k = 1.38x10-23J/K) KE = 1/2 mv2 = 3/2 kT v = (3kT/m)1/2 v = [(3)(1.38x10-23 x 293/(4.65x10-26)]1/2
v = 511 m/sec
ExampleIf it takes 2 mins for your kettle to begin boiling how much longer does it take to boil dry?Assume kettle is 3kWStarting temp of water 20C
Work done by kettle = power x time = 2 x 60 x 3000 = 360 000J
= Work to boil water of mass M = DT x M x cwater 360 000J = 80 x M x 4190
Mass of water = 1.07kg
Energy to boil water = M x Lv (water) = 1.07 x 2256 x103 = 2420 000J
Time required = Energy /power = 2420 000/3000 = 808 s 13mins
P=F/A