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Chem 221 Notes | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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1 Gases 2 Microscopic Energies 3 First Law 4 2nd & 3rd Law 5 Phase transitions 6 Mixtures 7 Phase Diagrams 8 Equilibrium 9 Molecular Interactions |
Properties of Gases.The Perfect GasIn first year, we learned about several gas laws that applied to a kind of gas called an "ideal gas", These gas laws were described as phenomenological equations, whose origins come from pure experimental results with little or no theoretical background. Our text book (Atkins and de Paula) uses a different term (Perfect Gas) to refer to what you formerly knew as an ideal gas. Personally, I don't like the term perfect gas as it implies that all other gases are somehow flawed. But for the sake of consistency with the text, I will use perfect gas in this course. A generalized equation that describes the state of a gas is p = f(T,V,n), 1.1 where pressure is defined once we know T, V, and n. This kind of function is not generally known but for special cases it can be determined exactly. One such case is the perfect gas, for which we know it to be p = nRT/V 1.2 PressurePressure is defined as force/area. In SI units, it has units of Pa or N·m-2 or kg·m-1·s-2. Standard pressure is 1 bar or 100 kPa. Note that formerly, standard pressure was defined as 1 atm. This old standard is close to the current one (1 atm = 1.01325 bar = 101.325 kPa) and is still used to day in many text books and references. Generally, we measure pressure of a gas by measuring the force it exerts on a known area. This can be accomplished with mechanical or electronic devices but in the past, it was done by measuring the height of a column of liquids supported by the gas. If we have a column of liquid of area A and height h with density r. then the force it exerts is ρAhg and we can calculate pressure via: TemperatureTemperature is a bit tougher to get a handle on. It is really related directly to the molecular dynamics of the material in question but we can measure it and get ideas of relative temperatures without knowing the concepts behind it. For example, we know that a material that is hot has a higher temperature than something that is cold. However, the terms hot and cold themselves are meaningless on their own. An object that feels hot to your body might be so cold to, say, a chemical process occurring at 600ºC that if the two contact each other, the chemical process would be quenched (terminated) It is useful to define the kind of boundary that can occur if two bodies touch each other.
Thus if two bodies contact each other via a diathermic boundary then; if their state changes, it must mean that the temperatures were not the same, if their states remain unchanged then they are at thermal equilibrium, i.e., they are at the same temperature. If we have three bodies, A, B, C, where A and B are in thermal equilibrium and B and C are also in thermal equilibrium, then if follows that A and C are in thermal equilibrium. This is the Zeroth Law of Thermodynamics. The zeroth law is the basic principle that allows us to use a thermometer to tell the temperature of objects. If one of the objects, A, is a thin column (capillary) of a liquid that expands with temperature then by measuring the length of the column when it is in contact with B and again with C, we can say that C and B are in thermal equilibrium if the length of the capillary is the same in both cases and that the state of B and C will not change if they come in contact with each other via a diathermic boundary. Similarly, we can mark off certain lengths of the capillary to denote temperatures.
For example, a mark at when the thermometer is in contact with ice and another when it is in contact with boiling water were used to make the Celsius scale. The symbol for temperature in Celsius that we will use here is q and has units of ºC. Generally, a liquid thermometer is not always the most accurate tool, since the coefficient of expansion of the liquids generally do not remain constant with temperature (they expand different amounts depending on the temperature) and different liquids expand by different amounts and so are not universal. If we use a perfect gas, the expansion is always directly proportional to the temperature, in fact, the perfect gas temperature scale is identical to the thermodynamic temperature scale and is measured in units of kelvin (K). To convert from Celsius to Kelvin we use the following relationship:
Gas LawsWe will assume these three gas laws are familiar to you.
These are called limiting laws in that they are only strictly valid in certain limiting situations. In this case, in the limit, as p → 0. we can combine them into a single law called the Perfect Gas law (a.k.a. the Ideal Gas Law). This law covers the relationship between temperature, pressure, volume and number of moles of an Ideal gas. After some consideration and algebra, we arrive at V = koverall nT/P 1.7 where koverall turns out to be the universal gas constant, R = 8.314 510 J mol-1K-1, in SI units. We're more familiar with the equation written as:
This is the equation of state for any gas that behaves like a perfect
gas
Molecular LevelOK, so we've seen the relationships that describe the state of a gas that behaves like a perfect gas. But what is a perfect gas? Why does a perfect gas behave the way it does? To get at those answers, we need to look at the molecular level for an explanation. Let's assume that a perfect gas is made up of molecules that have no intermolecular interactions whatsoever. that makes the math (and the concepts) quite easy. For example, a given sample of gas in a container will have a pressure that is created because the molecules are colliding with the walls of the container at a fixed rate and therefore exerting force on the walls. if we reduce the volume to half, it is reasonable to expect that there would be twice as many collisions happening per second, hence, twice the pressure, exactly as Boyle's Law states. If we heat the sample then the average kinetic energy of the molecules increases (they move faster) and hence both the collision frequency and the force of the collisions is increased, hence, the pressure will increase (at fixed volume), as Charles' Law predicts. A more mathematically rigorous approach involves a model called the Kinetic Molecular Theory of gases. Kinetic Molecular TheoryTo begin developing the model KMT, we need to start with a few assumptions (postulates). For a more thorough discussion of these along with an interesting history, check out the wiki page http://en.wikipedia.org/wiki/Kinetic_molecular_theory. The three postulates that we will find most useful for our derivation are:
Consider a cubic container of length l with a gas in it. Pressure is defined as a force exerted over an area. In a collision, we assume the molecule retains it's total kinetic energy (elastic collision) and that the only change is the direction of travel (like a pool ball bouncing off a cushion) and the only part of the direction that changes is the component directly perpendicular to the wall. consider the wall whose perpendicular is the x axis; then the x component of motion vx reverses when a molecule collides with it. The molecules momentum (along the x axis) changes from mvx to -mvx. So the change in momentum is Δpi = 2mvxi, where i is an index to the particular molecule. If the length of the side along the x direction is l then the total time for a particle to travel from a collision with one wall across the box and back again to re-collide is 2l/vx and hence, the frequency of collisions with the wall for this molecule is the reciprocal, vx/2l. thus, the force exerted by a single molecule (i) on the wall is (rate of change of momentum) The total force exerted on the wall by all molecules would thus be: Thus pressure (force divided by area) is The overall pressure will thus be the average of all the component pressures This particular formalism is not too useful since we would need to know the speed of every individual molecule. Let's replace the sum of the individual speeds with the square of the root mean squared speed <v2> multiplied by the number of molecules N. or Note that since the kinetic energy is defined as 1/2 mv2 we can calculate the average total kinetic energy as Thus, we can see that the pressure-volume product is always equal to two thirds of the total kinetic energy of the gas. Now, from statistical mechanics (later) we can derive the formula So we can equate the two pV equations (1.14 and 1.16) and get Now realize that we can get Ek on the left hand side if we multiply by 3/2 so
(note (1) that Nkb = R if and only if N is Avagadro's number) By combining equations 1.14 and 1.15 , we find that
We have derived a very important relationship in equation 1.18. The kinetic energy is directly proportional to the absolute temperature, which in turn (1.19) is proportional to the pV state of the gas. Further, by combining equations 1.18 and 1.15, we can calculate the average speed knowing nothing more than the temperature and the mass of the molecules.
or 1.20
We can now see that the RMS speed (molecular level) is easily calculated from simple macroscopic measurables. Also note that we have derived several of the ideal gas equations here. Equation 1.19 clearly shows that pV is a constant for a given temperature (Boyle's Law). In fact, combining 1.18 and 1.19 clearly gives us the perfect gas relationship (for one mole) itself. It is simple to understand that the number of moles factors in simply and thus, Avagadro's Law is explained. and our Perfect gas law is now complete as pV = nRT 1.21 Real GasesSo we've explored the simplest gas model, the "perfect gas" and we were able to use molecular level reasoning to justify the equations that were initially derived experimentally. The problem is that this simple model only works in the limit of zero pressure. In real situations with real gases, the model will work with an accuracy that is dependent on the conditions involved. Let's look again at the postulates of the perfect gas. That assumption actually involves two things: molecular complexity and type of intermolecular forces. If the molecule has many atoms, there is a greater possibility that some of the kinetic energy in the collisions will be converted into vibrational and rotational energy (or vice versa), making the collisions less than perfectly elastic. Additionally, if the molecular kinetic energy is not significantly higher than the energy of the intermolecular interactions then it is likely that the trajectories of the molecules will curve, rather than remain straight. Both of these effects will be more noticeable at low temperatures when the molecular kinetic energies are low.
An easy measure of how much deviation occurs for a gas can be obtained from a simple measure of the actual volume of a gas divided by the volume a perfect gas would have under the same conditions.
Where the subscript indicates we're using molar volumes. Since the molar volume of the perfect gas is simply Vºm = RT/p , we can substitute and rearrange to get pVm = RTZ 1.22a or pV = nRTZ 1.22b Clearly, if Z is 1 then the gas we are measuring behaves exactly like a perfect gas. if Z deviates from 1 then the behaviors of the real gas deviate from the perfect gas model. A challenge at this point is to try to get an
equation that predicts how a given real gas will behave under
changes in p, V and/or T. There are several approaches to
this question. One, the Virial equation, is to simply fit
a polynomial expansion of one of the variables to the
experimentally determined compressibility data. pVm = RT(1 + B'p + C'p2 + ... ) 1.24 pVm = RT(1 + B/Vm + C/Vm2 + ... ) 1.25
or
a perfect gas has a slope of zero in these limits but a real gas has a slope of B (or B'). But it's not that simple yet. Since the virial coefficients depend on temperature, there may be a temperature when the slope is zero (when B is zero). That temperature is called the Boyle temperature (TB) at this temperature, the real gas and the perfect gas do have identical properties as p → 0. Clearly, a real gas will behave like the perfect gas model over a wider range of pressures when at (or near) the Boyle temperature since the term B/Vm is zero and the other terms C/Vm2 ... are very small and only become significant as p gets larger (Vm gets smaller.)
Let's assume for a minute another model of what happens when two molecules collide or approach each other.
We can also view the deviation from perfect gas model by plotting pV isotherms at different temperatures, similar to what we did above for a perfect gas.
We can easily see that for a given gas, the deviation from perfect behavior depends on temperature. At high temperature, the real gas behaves much like a perfect gas but as the temperature lowers, the pV isotherm deviates more and more from perfect behavior. Note that at a temperature q = 31.04ºC, the real isotherm for carbon dioxide levels off to a slope of zero and then resumes it's decent as volume increases. This one temperature is the critical temperature, Tc . At any temperature above Tc the slope of the curve is always negative. At Tc, the slope is zero at exactly one point (T = Tc, p = pc and V = Vc) and below zero, there is a discontinuity in the slope where it suddenly breaks from negative to zero (over a range of volumes) back to negative again. The horizontal portion of an isotherm (between points E and C) are the pressure/volume regions where there are two phases, liquid and solid. The horizontal portion of the line occurs at the vapor pressure of the liquid at that temperature. Explore the applet: Can you identify the PV regions of the more familiar phase diagram with the regions in this isotherm diagram. Clearly, the behavior of the gas as we compress it is quite different above Tc than below it. At T > Tc, we have only one phase (supercritical fluid), at T < Tc we encounter a region along the isotherm where there are two phases in equilibrium with each other.
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1.28a |
or, by using molar volumes:
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1.28b |
Where parameters a and b are called the van der Waals parameters. This two-parameter equation cannot possibility fit the experimental data better than the virial equation if it uses more than two parameters. However, we have a meaning attached to the two parameters and hence, by fitting this equation to experimental isotherms, we get some understanding of the molecular properties of the gas that gives rise to the non-perfect gas behavior. In this case, b is approximately the molar volume of the molecules themselves and parameter a is a measure of the intermolecular forces that would act on molecules passing each other (even if not colliding) that would make their velocities not constant as the KMT assumed.
The model does a pretty good job of fitting to real gas isotherms when T > Tc and it is reasonable for the gas phase and the liquid phase separately for temperatures below the critical temperature. However, it falls apart completely in the phase change region, where it does not predict that the pressure remains constant as volume is reduced. In fact it predicts a multi valued function (which is impossible) in the phase change region.
![]()
Fig. 1.8 (from
Atkins): A region of the
p,V,T surface
of a fixed amount of perfect gas. The points
forming the surface represent the only
states of the gas that can exist.
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![]()
Fig. 1.17 (from
Atkins) The surface of possible states
allowed by the van der Waals equation.
Compare this surface with that shown in
Fig. 1.8.
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Look at the isotherms drawn by the following applet to explore this.
Resources/VderWSpreadsheet.xlsx
![]() Fig. 1.18 Van der Waals isotherms at several values of T/Tc. Compare these curves with those in Fig. 1.15. The van der Waals loops are normally replaced by horizontal straight lines. The critical isotherm is the isotherm for T/Tc = 1.
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We see that the isotherms below the critical temperature (reduced temperature = 1) in figure 1.18. seem to predict that as we increase pressure the volume will actually increase (in some regions). We accommodate for this obvious error in the model via a procedure called Maxwell's construction, whereby we draw a horizontal line through that region such that the area in the loop above the line equals the area in the loop below the line.
Look again at the van der Waals equation.
Note that if Vm is large enough, the parameters a and b are negligible and we get back the perfect gas equation.
In the region where the loops occur the first term (kinetic energy and repulsive forces) and the second term (cohesive forces) are comparable. This explains why we can have two phases in these regions.
We can use the critical isotherm to find the
critical constants. The inflection point, where
(T = Tc,
p = pc and
V = Vc) have both the first and second
derivative equal to zero.

From the solution to these two equations in conjunction with
van der Waals equation itself, we get
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1.29 |
We can use equation 1.22 to determine the compressibility at the critical point
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1.30 |
This implies that all gases will have the same compressibility at their respective critical points and leads us to another interesting concept called the principle of corresponding states. This implies that if we recast the pressure, volume and temperatures as relative to their critical values that the equation of state would coincide for all substances.
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1.31 |
![]() Fig. 1.19 The compression factors of four of the gases shown in Fig. 1.14 plotted using reduced variables. The curves are labeled with the reduced temperature Tr = T/Tc. The use of reduced variables organizes the data on to single curves. |
We can rewrite the van der Waals equation using these reduced variables

Now, substitute in equations 1.29 to get rid of the critical variables

and after some algebra... :)
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1.32 |
This is the van der Waals equation using reduced variables and, within the limits of the principles of corresponding states, can be used for any van der Waals gas. Gases with strong polarity or very asymmetric shapes deviate furthest from this equation.
Last updated:
07-Apr-2010