Exam 1 Review Sheet - version 1

Exam 1 - Chapter 5 - Gases

What sections did we cover? All sections of Chapter 5 except sections 5.8, 5.9, and 5.12 were covered. Section 12 is an interesting section on atmospheric chemistry but we will not cover it. Also sections 6-7 have lots of technical proofs that will NOT be on the exam - just focus on the KEY parts of those two sections which are the 2 boxed formulas, one for kinetic energy and the other for rms particle velocity. Also, LOOK at the plot of a Maxwell-Boltzmann distribution and how the curve shifts with temperature (Figure 5.16). Know your definitions of effusion and diffusion plus the boxed equations relating velocity (rate) and mass and temperature (all of this is ON this review sheet too).

NO Equations will be given on the exam! So KNOW your equations for the exam. I will always provide constants like R and other specific constants where needed.

Fundamentals (Stoichiometry)

Know how to calculate molar mass. Know your diatomic gases (hydrogen, oxygen, nitrogen, fluorine, and chlorine). Be able to calculate amounts of reactants or products when given amounts of reactant or products – this is called stoichiometry – KNOW IT. See chapters 1-3 Know how to determine the limiting reactant and how much product it leads to. Be able to incorporate percent yield into a problem if necessary. All that is new is moles to \(P\) or \(V\) instead of grams.

Check out the "Learning Modules" on our Companion Site

The companion site has lots of good help on the topics we've covered, including videos. That site is at: http://ch301.cm.utexas.edu

Gas Laws

Know the NAMES and the law associated with each name (scientist).

Boyle’s Law: Pressure is inversely proportional to volume (assuming constant temperature and amount of gas, moles). Any units will work here:

$$P\,V = k \hskip12pt{\rm (constant)}$$

Charles’ Law: Volume is directly proportional to absolute temperature (assuming constant pressure and amount of gas, moles). Any units for volume but remember, T must be Kelvin:

$${V\over T} = k \hskip12pt{\rm (constant)}$$

Avogadro’s Law: Volume is directly proportional to amount of gas in moles. (assuming constant temperature and pressure):

$${V\over n} = k \hskip12pt{\rm (constant)}$$

Combined Gas Law: Most books and people refer to this as Boyle’s Law + Charles’ Law which is:

$${P_1V_1\over T_1} = {P_2V_2\over T_2} = k \hskip12pt{\rm (constant)}$$

However, our book includes Avogadro’s Law also which leads to:

$${P_1V_1\over n_1T_1} = {P_2V_2\over n_2T_2} = k \hskip12pt{\rm (constant)}$$

which is fine, except that this is really just the Ideal Gas Law in disguise. The constant, k that is defined by this version of the combined gas law is actually the Universal Gas constant, R. So by definition:

$${PV \over nT} = R$$

which rearranges to give

$$PV = nRT$$

 which is the Ideal Gas Law we all know and love. When you use the ideal gas law be sure and use the right units on \(P\), \(V\), \(n\), and \(T\). You have no choice on \(n\) and \(T\), they must be in moles and Kelvin. Now you’ve got choices for units on \(P\) and \(V\) and THAT’s why we’ve got different values (OK, mostly different) of \(R\) listed in the book. Here are my favorites (feel free to find one as your favorite):

\[R = 0.08206\quad {\rm L\;\;atm/mol\;K} \]

\[R = 0.08314\quad {\rm L\;\;bar/mol\;K} \]

\[R = 8.314 \quad{\rm Pa\;\;m^3/mol\;K} \]

\[R = 62.36 \quad{\rm L\;\;torr/mol\;K} \]

See, it’s not so bad and it's really not that confusing. But I know what you’re thinking. Do I (state your name) need to memorize all those values of R? Why not? Drop that value of R into your next conversation and watch your popularity skyrocket. Oh well, I will still print those values on the cover sheet to the exam for you to refer to while your brain is slowly melting during the exam. One more thing... the values of R are also given on the back of all the Quest bubblesheets (There's a periodic table there too!).

STP vs SATP

A standard is whatever you chose as standard. Sometimes we chose different standards – and yes, I think that defeats the point in the first place but there you go. Know the difference in STP (standard temperature and pressure) and SATP (standard ambient temperature and pressure):

STP: 0˚C and 1 atm (old school, most of our problems)

SATP: 25˚C and 1 bar (new school, newer problems although some still have atm as the pressure unit)

Realize that you WILL need to convert ˚C into Kelvin when using the ideal gas law.

Let’s MIX it up a bit

All those laws are great and work really well for most gases at reasonable pressures and temperatures. You can even apply those laws to mixtures of gases and just treat the whole mix as one big family of gas - that is you COUNT ALL the moles regardless of the type. When you do this you are calculating TOTAL pressure due to ALL the gases present. After you do that, I’m still going to want to know what the pressure is for each individual gas. The pressure exerted by say helium in a mixture of helium, neon, and xenon is the partial pressure of helium, PHe. Now the big lesson (and name!),

Dalton’s Law of Partial Pressures: The total pressure of a system is equal to the sum of all the partial pressures within the system. Mathematically:

\[P_{\rm total}=P_{\rm A}+P_{\rm B}+P_{\rm C}+\cdots \]

Now also realize that each of those partial pressures will have it’s own version of the ideal gas law. What I mean is that \(P_{\rm A}V = n_{\rm A}RT\). See, I’m only counting moles of A in this equation and I therefore get out \(P_{\rm A}\). One of the other neat results of this law is that the partial pressure’s ratio to the total pressure exactly matches its mole fraction. Uh-oh, what’s a mole fraction? The number of moles of A divided by the total moles of gas in the mixture. Mole fraction of A is given by the symbol \(x_{\rm A}\) (OK, so it’s \(\chi_{\rm A}\) in the book – much too greek for me). So by definition:

\[x_{\rm A} = { {\rm mol\;A}\over {\rm mol\;A + mol\;B + mol\;C + \cdots }} \]

and you can set that equal to the pressure ratios:

\[x_{\rm A} = {P_{\rm A}\over P_{\rm total}}\quad{\rm \;\;\;and\; \; \;therefore\quad} P_{\rm A} = x_{\rm A}P_{\rm total}\]

So you can use pressures to get mole fractions OR use mole fractions to get pressures. Cool.

Kinetic Molecular Theory of Gases

This theory describes the actually physical behavior of gas molecules as they zip around in space. The book points out 4 major points:

It is important to realize that REAL gases have forces between them. Those forces are called intermolecular forces (IMFs) and are the subject of Chapter 16 sections 1 and 2. We WILL get there eventually. Real gases also have real volumes of the particles.

Finally, if we do follow the model of KMT, we can (OK, smart mathmatical proof types of people) derive all the relationships in the ideal gas law. Do take home from this section in the book the equations for kinetic energy. There are 2 equations. The first is simply the phyical view of a moving particle which is:

$$ E_{\rm k} = {1\over 2} mv^2 $$

Where \(E_{\rm k}\) is kinetic energy (in J), \(m\) is mass (in kg), and \(v\) is velocity (in m/s). The second equation is certainly more relevant to us chemists. It shows the direct proportionality of temperature with kinetic energy for a gas:

$$ E_{\rm k} = {3 \over 2} RT $$

\(R\) is 8.314 J/mol K and the temperature must be in Kelvin. KNOW both of these equations.

The confusion on effusion

Effusion of a gas is when a gas slips through a very small hole (or holes) from a high pressure to a low pressure. This IS what is going on when your once large and buoyant helium balloon (fun) is now much smaller and is now riding around on the floor (not as fun). Helium effuses through the pores of the latex balloon. How fast does the helium effuse? Graham knows.

Graham’s Law of Effusion: At constant temperature, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

$${\rm rate\;of\;effusion\;} = \sqrt{1\over M}$$

and so if you compare two gases at the same temperature you’ll get

$${{\rm rate\;of\;effusion\;of\;A}\over {\rm rate\;of\;effusion\;of\;B}} = \sqrt{M_{\rm B}\over M_{\rm A}}$$

Temperature also has a square root dependence, but it is a direct square root dependence to the rate of effusion or speed. For a given molar mass at two differents temperatures:

$${{{\rm velocity\;at\;}T_2}\over {\rm velocity\;at\;}T_1} = \sqrt{T_{\rm 2}\over T_{\rm 1}}$$

All these relationships are derived in the textbook in section 5.6. Lot's of mathmatical work there but the main equation is in a nice call-out box and it is the formula for the root mean square velocity:

$$v_{\rm rms} = \sqrt{3RT\over M}$$

You don’t know how it feels – to be me

Ah yes, Tom Petty sang it so well and he never realized just how much all the real gases love that song. How DOES it FEEL to be REAL? Well for starters, you’ve got a real volume your-OWN-self. Real gas molecules DO have real diameters and take up space – ideal gases do NOT have a real diameter – they are just a point in space. Speaking of point, Tom Petty says in the same song, “let’s get to the point…” See he was talking about ideal gases right? Actually, now that I think about it, and sing the rest of that chorus… he’s talking about a whole other gas. But WAIT! Yes, he IS talking about real gases now that I think about it! Another line, “… think of me what you will, I’ve got a little space to fill”. SEE? That’s what gases do – fill space. I digress.

A great way to show non-ideality is too calculate a gases actual (real) molar volume, Vm, and compare it to the molar volume of an ideal gas, Vm,ideal. When compared as a ratio, you get the compression factor, Z:

$$Z = {V_{\rm m,real}\over V_{\rm m,ideal}}$$

Now that particular version of the compression factor is not shown in Zumdahl's textbook. However, there is something that is exactly the same. The compression factor for Zumdahl is:

$$Z = {PV \over nRT}$$

Zumdahl never bothers showing a "Z" though, you can see the plot of it none-the-less in your book in Figure 5.22. An ideal gas will have Z = 1 under ALL conditions. Real gases have real forces at play. A value of Z < 1 corresponds to attractive forces being dominant. A value of Z > 1 corresponds to repulsive forces being dominant. Note that as zero pressure is approached, all real gases take on ideal behavior (Z = 1). Real gases aren’t that far off the line at 1 atm either, which is why the ideal gas law works so well most of the time.

How to improve on an already IDEAL equation?

It is possible to come up with an equation that attempts to include the effects of attractions and repulsions between the gas particles. It does not predict the exact behavior of a real gas, but it comes closer. The main problem is that you must PICK the real gas you want the "law" to work for. All gases are NOT the same in the real world of gases, especially at high pressures and low temperatures – that is where the ideal gas law is at it’s worst. We can tweak the KMT of ideal gases a bit to better model real behavior. One way is to adjust the volume to be the "empty space" volume. This "correction" simply subtracts the volume of the gas particles (spheres) themselves. This correction is the \(nb\) term in the formula below:

The Hard Sphere Model:

$$ P\;\underbrace{(V-nb)}_{\rm empty\;space} = nRT$$

Now we can go one betterf and adjust for the attractive forces that real gases have. This is handled in the van der Waals equation (section 5.10 in your book).

The van der Waals equation:

$$\left(P+a\left({n\over V}\right)^2\right)(V-nb) = nRT$$

Now we’ve got a fixed up equation that has added two new paramaters. One to deal with attractions, \(a\), and one to deal with the hard sphere model \(b\). Note the hard sphere model really is adjusting for repulsions between molecules. These are the van der Waals parameters and they are NOT temperature dependent – yea! You DO need to look up the values for \(a\) and \(b\) for every gas you want to work with though. The parameter \(a\) is an “adjustment” for attractive forces between molecules. The bigger \(a\) is, the smaller the pressure will be. The parameter \(b\) is the same volume correction term from the hard sphere model – it is roughly the volume of a mole of condensed gas (OK, it's really a bit bigger that that). It is there to account for the repulsions between the gas particles which get very important at high concentrations (pressures).

Know About the Hard Sphere Model for the Exam! The van der Waals equation will NOT be on the exam.

Life’s a Gas – now you know more about it.

Yes, read your book. Try some of those problems in the back of the chapter. You’ll feel so GOOD about yourself.

Errors?

Well I'd like to hope that there are no errors in this Chapter review. But there can always be a few that slip through. Feel free to send me an email to correct any errors you find. Remember, this document is not anything that you can wave and point at after the exam. YOU have to cross check information and formulas here with the textbook and make sure they are right.