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1 Dr. Fritz Wilhelm page of C:\physics\3 lecture\ch9 temperature ideal gas law.docx; Last saved: /5/ 6:46: PM; Last printed:/5/ 6:46: PM Problems: See website, Table of Contents: 9. Temperature as a Measure of an average Kinetic Energy; 9.,3 Constant-Volume Gas Thermometer; Thermal Expansion of Solids and Liquids; 5 9.4a nomaly of water with respect to expansion with temperature; Macroscopic Description of an Ideal Gas; 7 9.5a Van der Waals equation; 9 9.5b arometric Formula; 9

2 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page of 9. Temperature as a Measure of an average Kinetic Energy. When objects at different temperatures are brought into contact heat is exchanged between them. They are said to be in thermal contact. Once they are at the same temperature they are said to be in thermal equilibrium. We can only talk about temperature T in a situation with extremely large numbers of molecules involved. Every mole of a substance, i.e. the molecular or atomic mass in grams, contains vogadro s number of molecules or atoms: 3 molecules (9.) = 6. mole Caution: n = number of moles, 3 = number of molecules, = vogadro s number = number of molecules in mole=6., = n These individual molecules have a kinetic energy of translation (9.) v KE = m The molecules have an average translational kinetic energy: mv m v v v (9.3) + + i= i= KE = mv = = ( ) i i i ix iy iz This average kinetic energy is a direct measure of the temperature T. Usually we write this relationship as: 3-3 kt = m v ; k =.38 oltzmann constant (9.4) J Joules [ k ] = = K Kelvin If we multiply (9.4) with vogadro s number we get the relationship for moles rather than for individual molecules. 3 3 kt v = v = m RT= M = k m -3 3 (9.5) k = =R universal gas constant R=8.34 J/mol K =.86 atm L/mol K If we take the squareroot of the mean squared velocity we get the so-called rms (root-meansquared) speed of the molecule.

3 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 3 of v (9.6) Problem: Calculate the rms speed of an oxygen molecule at C. i i vrms = v for molecules From this definition it is also obvious that absolute. temperature cannot be achieved, as that would require absolute kinetic energy and speed. This is not possible as long as Heisenberg s uncertainty relations (quantum physics) are correct, and we believe they are correct under all circumstances in our physical universe. (9.7) x p ll substances maintain a fundamental vibration consistent with a zero point energy. The smallest possible energy of an oscillator is given by (this is a result from quantum-physics, which I mention here to give you a more complete picture of the fundamental laws of physics.) hf (9.8) E = ω Quantum physics states that any energy is quantized, and consists of a finite number of multiple energy quanta, which, in the case of an oscillator, is equal to ω. h 34 Where (9.9) ( hbar) = =.5 Js = Planck's constant/π π (The horizontal bar symbol means division by π.) Example: For a spring with spring constant./m and an amplitude of oscillation of. cm, we get a total energy of.5 k = J With a mass of.kg we get an angular frequency of k = 7.7s The fundamental oscillation has an energy of m ω = 7.46 J = 4.66 ev 9 ev ( electron Volt) =.6 J This is obviously a very small number, but it is not absolute. 9.,3 Constant-Volume Gas Thermometer. Historical Temperature Scales, Kelvin, Fahrenheit, Celsius. The point of the Kelvin scale (or the absolute point) can be obtained through extrapolation of the pressure-temperature relationship of an ideal gas, whose pressure is measured at the freezing point of water (triple point) at degree Celsius and the steampoint of water at degree Celsius. This is done with the constant volume gas thermometer. nr PV = nrt P = T (9.) V T in Kelvin degrees, K = K

4 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 4 of ccording to this Ideal Gas Law, pressure in an ideal gas, maintained at constant volume, is directly proportional to temperature. The slope of the P(T) graph depends on the number n of moles of the gas and its volume. The graph passes through P()= only if you use Kelvin degrees. Otherwise there will be an intercept depending on whether you Celsius or Fahrenheit degrees. If temperatures are measured with gas thermometers containing different gases at different initial pressures, the extrapolation of all straight lines intercepts the same point of pressure at -73ºC. Obviously, the experiment can only be conducted with a very diluted regular gas, and as long as the gas does not turn liquid, or solid. The graph shows several lines for different substances with different slopes, all converging to the absolute K temperature point. Pressure P P=mT+P m=slope -73ºC (,) ºC Temperature T in Celsius Problem: For dry ice (CO ) solidification occurs at -8.ºC and.9 atm, for ethyl alcohol solidification occurs at 78.ºC and.635 atm. These two exactly reproducible points are used as calibration points for the same amount of substance at the same volume, i.e. having the same slope m=nr/v. We get the slope of the straight line by P P atm atm m = = = Having the slope we can easily find the intercept value T T 78 ( 8) C C.635 = m(78 C + ) P P =.7atm 3 atm (9.) P( T ) = 4.65 T +.7atm C This allows us to find the pressure of these gases at any temperature, and vice versa. Specifically, we find that atm pressure is obtained at -73ºC. t a ºC the gases have a pressure of 7.737atm. The most common temperature measurements use Cº which are related to the scientific units of Kelvin through: point Kelvin =-73.6 Celsius. 9 (9.) TF = TC + 3 F; T F = TC; TC = TF 3C = TF C

5 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 5 of The zero point for the Fahrenheit scale is the freezing point for water (triple point, to be more exact), set by historical convention at 3 degree Fahrenheit, or degree Celsius. The boiling point for water is set at F degree, and Celsius degree. From these definitions it is easy to construct the functional relationship between Fahrenheit and Celsius. (Two points define a straight line.) Let T F be the y-axis, and T C the x axis: T F (ºC, ºF)=(x,y) (ºC, 3ºF) (ºC, ºF) (ºC, ºF) T C (9.3) 9 y = mx + b = x + 3 = x Thermal Expansion of Solids and Liquids. Most solids and liquids expand slightly when the temperature is increased. The relative change in one dimension (linear expansion) is (for small temperature changes) directly proportional to the change in temperature. The average coefficient of expansion is denoted by α. L = α T L = αli T Li (9.4) [ α ] = C If you write this as a differential relationship and integrate it you see that the above formula is an approximation: dl = αdt ln L = αt + c; c is the constant of integration L (9.5) integrate ( ) αt+ c L=e ; initial condition: L T= = L αt L= Le If you expand the exponential formula using the McLaurin series: n 3 x x x x (9.6) e = = + x n= n!! 3! we get:

6 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 6 of (9.7) ( α ) L= L + T = L + LαT L = αt L Remember when we talked about stress and defined Young s modulus, or the elastic modulus through a similar relationship: F L (9.8) stress= = Y = Yα T (Y Cu = E /m ) L i α T This means that thermal expansion increases the stress inside of materials. Example: Calculate the thermal stress in a copper bar of length.m, if the temperature is increased from to C. (.87E8 Pa) The copper bar expands by.7mm. ote that if you have material with holes in it, the holes expand with the material, as if there were no holes in it. For example, if you heat a copper ring with inner diameter 5. cm from degree to degree Celsius, the diameter will increase according to 6 7 L = α Li T = 5.cm 8C =.5cm C (9.9) αcu L =.3% L For a volume expansion of isotropic materials we want to find the relative expansion knowing the expansion in the linear directions. We use what we have learnt about calculating relative uncertainties, which is the same mathematics. Here is a little review: The relative change of any function is given by the total derivative of the natural log of that function. See the handout supplement on uncertainty calculations: df (9.) = dln f f (9.) f ( xyz,, ) f f f df = dx + dy + dz grad f dr,, dx, dy, dz x y z x y z In cases where the function is a simple polynomial the change of the function is particulary simple to calculate by using the logarithmic derivative process of (9.)

7 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 7 of (9.) f = π kx y z ln f = ln( π k) + ln x+ ln y+ ln z dx dy dz dln f = + + x y z ow, let us apply this to the volume expansion. ssume that we have homogeneous material which expands in all directions in the same way with α. dv dx dy dz V = xyz; = + + V x y z (9.3) dx dy dz = = = α T for isotropic material x y z dv = 3α T V 9.4a nomaly of water with respect to expansion with temperature: Water expands with temperature above 4º C, but it contracts below 4º Celsius. To put it differently, the density of water increases from to 4.ºC, and it decreases above 4ºC. It has a maximum for the density/temperature curve at 4ºC g ρ( T = 4. C) =. 3 (9.4) cm g ρ( T 4. C) <. 3 cm This means that water at º Celsius has a lesser density than water at 5º Celsius. It will therefore move to the top of a lake. (The buoyant force is larger than its weight.) With further cooling it will turn into ice which has an even smaller density. Therefore ice will form at the top of a lake and not at the bottom, which would be the case if water at ºCelsius had a greater density than water at 4º C. 9.5 Macroscopic Description of an Ideal Gas. The molar mass of each chemical element is the atomic mass, expressed in grams per mole. The same definition is applied to molecules. Thus, the atomic mass of water would be 8, and its molar mass would be 8 grams. Every mole, or 8 grams, of water (vapor, ice, or liquid) contains vogadro s number of molecules. The density of water depends on its state (liquid, vapor, solid), its temperature, and the pressure it is under. The relationship expressing this is called the equation of state, which, for water is very complicated. The same relationship is relatively simple for an idealized gas, in which the gas particles have no chemical or other interactions with their neighbors. The ideal gases like Helium, Radon, rgon come close to this, so do Hydrogen and Oxygen molecules over some range of temperature and at low densities. We use the ideal gas law to illustrate and approximately calculate many of the thermodynamic quantities of gases. The equation of state for an ideal gas is given by:

8 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 8 of (9.5) PV = nrt for n moles; R=universal gas constant 8.34 J R = ; [ ] [ ] [ ].3 mol K L atm = L atm [ ] ( ) E = P V J = Pa m = atm L 3 R=8.34 J= P is pressure in Pascals, V is volume in m 3 or in atmospheres and liters.(atm=.3e5pa) ote that the product of volume and pressure has the dimension of energy, Joules. In order to obtain the gas law for particles, rather than n moles, we have to make use of the fact, that the number of moles is equal to the number of particles, divided by vogadro s number (9.6) 3 = 6. mol lso, the universal gas constant for a single atom is given by oltzmann s constant k. (9.7) n= and R= k (9.8) PV = nrt = kt = kt ote that in this equation, stands for total number of individual particles, not moles. (9.9) k R = J K 3.38 oltzmann's constant Pressure P, volume V, temperature T (and number of molecules) are the typical thermodynamic variables, very much like x, y, z in mechanics. The major thermodynamic functions like heat Q(P,V,T), internal energy U, entropy S can all be expressed in terms of such variables. These functions are all scalar functions. Pressure, volume, and temperature can be functions of space and time coordinates. s we are always dealing with very large numbers of particles, we need to ultimately employ statistical models. Problem: Calculate the average pressure of outer space (universe) where we have typically about four protons per m 3 at a temperature of K. Solution: 3 3 PV nrt kt P 4.38 = = = 5 Pa

9 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page 9 of a Van der Waals equation: The behavior of real gases requires a modification of the ideal gas law. Experiment and theory confirm that we need to add an additional pressure term, and an additional volume term. This equation allows among other things to understand the liquification of gases when the pressure is increased and the temperature decreased. (See chapter.) a P + ( V ) mol b = RT ( V ) mol a is an internal pressure deriving from the attraction of the molecules to each-other Vmol (9.3) b is the volume occupied by a single finite mole of particles 6 a [ a] pa m pa mol = = 6 [ a] = Vmol m mol 3 m [ b] = mol For example: for carbondioxide we have a=.36 pa m 6 /mol, and b=4.3-5 m 3 /mol We can use formula (9.5) to determine the density ρ of a gas at a particular pressure and temperature: PV= kt P= kt = nkt V V (9.3) nv = = number density; V M ρ = nv m= m= ; m= mass of individual particle;m=total mass in the volume V. V V The number density n V is related to the conventional mass density ρ by: (9.3) number of atoms ρ = nv m= mass of one atom volume y using the mass density we get in (9.3): (9.33) ρ ρ P= nkt V = kt = m Mmol RT Solving for the density, we get: (9.34) M mol m ρ = P= P RT k T Problem: Find the density of air at degree Celsius and atmospheric pressure: ir consists to 8% of nitrogen and % of oxygen. This gives us the molar mass of air at C as: (9.35) M =.8 8g+. 3g = 9g air

10 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page of (9.36) ρ air atm 9 g / mole g kg = =.9 =.9 73C.8 atm L / K mole L m ( ) 3 9.5b arometric Formula. Problem: How does the pressure in air vary with height? ssume that air behaves like an ideal gas. We know that the pressure in any liquid changes according to: (9.37) P = P ρgy dp = ρgdy In this formula we choose the pressure point at the bottom of the liquid. Therefore the pressure decreases with increasing values for y. From (9.34) we see that at constant temperature T, pressure P is proportional to the density ρ: Pmmolecule (9.38) ρ = kt With this substitution for ρ in (9.37) we get from dp = ρgdy (9.39) P mmolecule dp mg P mg dp = gdy = dy ln P = y + constant P (9.4) kt P kt kt mmoleculeg y Mmol g y RT kt (9.4) P= Pe = Pe oth pressure and density decrease exponentially with the potential energy like potential energy of one molecule kt e (9.4) m molecule Mmole The ratios = obviously being equal. k R Therefore, we can also write: = e (9.43) ( ) potential energy RT mmoleculeg y Mmol g y kt RT ρ y = ρ e = ρ e We will see later that, in general, the probability to find a molecule in a particular energy E state is proportional to: E kt (9.44) e E is the total of all energies, potential and kinetic of the molecules.

11 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page of Problem: t what height will the pressure of nitrogen gas drop to 8% of the original pressure, ºC. (9.45) m m moleculeg kg y Mmol g y s y kt RT J = = =.8P Pe Pe Pe ln.8 = y y = = 85m E 4 hot air at T hot and pressure P mass of hot air =m = ρ V air at T cold and pressure P; density ρ mass of displaced cold air: m =ρ V Problem: hot air balloon consists of a balloon with volume of 5 m 3. The ballast, i.e. everything of the system with the exception of the air inside the balloon, has a mass of 5kg. The surrounding air is at a temperature of ºC. To which temperature does the air inside of the balloon have to be raised for the balloon to take off? This is a buoyancy problem and an ideal gas problem. Let us call the mass of air in the balloon m and the remaining mass of the baloon and the equipment m. mass of balloon +equipment = m The weight of the displaced cold air with density ρ must be equal to the weight of the hot air plus the additional weight mg.when the air inside of the balloon is heated, its density and weight decrease. m g = ρ Vg (9.46) air will lift the balloon: The buoyancy of the hot air, which is equal to the weight of the displaced cold (9.47) ρ Vg = ρ Vg + m g buoyancy= weight of the hot remaining weight weight of displaced air in the balloon cold air We need to figure out how the density varies with temperature and assume that air behaves like an ideal gas. molecule mol Formula (9.34) ρ = Pm PM kt = RT is applicable. In our problem pressure is constant (the balloon is open to the outside air), so, the density varies inversely with the temperature. The ratio between the densities at hot and cold temperatures is therefore: (the constants cancel)

12 Dr. Fritz Wilhelm: Temperature and the Ideal Gas Law page of (9.48) ρ ρ T = ρ = ρ hot cold cold hot cold cold Thot Thot T Inserting the density of hot air ρ hot in equation (9.47) we get: (9.49) T ρ Vg = ρvg + m g = ρ Vg + m g; ρ =.9 kg / m cold cold Thot T ρ V = ρ V + m ρ VT = ρ T V + m T cold hot cold hot Thot ρ T V = = = 446 = 73 ( ρ ) cold Thot V m ρtcoldv Thot K C ρv m 3

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