Lecture Notes from CHM 1341
13 June 1996


Using PV=nRT

Recall our discussion in lecture about empirical formulae (which have the right ratio of atoms but possibly not the right number of them: HO vs HOOH) being available from elemental analyses (such a burning a hydrocarbon and measuring the amount of carbon dioxide and water produced)? At that time, we lamented that we could turn empirical into molecular formulae if only we knew the molar mass.

Now we do. At least if the molecules are easily vaporized and the resulting gas is ideal...

For now the "n" of PV=nRT permits us to know how many molecules any weight of gas represents. In fact, if we know the density of the gas, that's d=m/V where m is the mass of it that occupies volume V. Since m=nM where M is the molar mass, d=nM/V or n=dV/M making the Ideal Gas equation read PV=dVRT/M which easily rearranges to M=dRT/P. So density, pressure, and temperature determine the molar mass of a gas.

So if elemental analysis determines that ethane's empirical formula is CH3 and it's density is measured at 20 C to be 1.25 g/L, then we can use this alternate form of the ideal gas equation to determine that
    1.25 g/L (0.0821 L atm /[mol K]) (20 + 273)K
M = -------------------------------------------- = 30 g/mol
                    1 atm
which isn't consistent with the empirical formula, M(emp)=[12+3(1)]=15, but rather twice that. So the molecular formula of ethane must be C2H6

So the Ideal Gas equation is a elementary but powerful tool for learning about molecular masses of gases. In other courses, you'll learn about other measurable properties which depend only upon the number of moles in a sample and not its individual composition. So, for example, there are Ideal Solution laws as well as Ideal Gas ones, and such properties as freezing points, boiling points, and vapor pressures above the solutions can be shown to vary in simple ways with the number of moles of constituents. Enormous biological molecules, for example, have their molar mass determined by "osmotic" pressures which can be measured very accurately.

The elementary expressions of the Ideal Gas equation, Boyle's, Charles's, and Avogadro's Laws, assumed that nT, P/n, and V/T, respectively, are held constant. Then the simple relations of the remaining 2 variables make predictions about their changes with respect to one another a snap. But even in situations where only 1 variable is fixed (usually, but not always, the number of moles), the constancy of R makes comparing initial conditions "1" and final conditions "2" simple
   (P1V1)/(n1T1) = (P2V2)/(n2T2)      (eqn 1)

   P1V1 /T1  = P2V2 /T2   if n1 = n2  , say
Then from the 3 initial variables and 2 of the final ones, the 3rd final variable is a snap to find.

Suppose that you fill a tire with air to the manufacturer's recommended 30 psi when the temperature is 52°F (284 K), and during the course of driving, the tire heats up to 152°F (340 K). But unbeknownst to you, your tire has a small leak which has spilled out 10% of the original air. What's the final pressure? Notice that we don't have to know either n or V since the final n/V will be 0.9 times its initial value; V, after all, hasn't changed but n has dropped by 10%. So plugging what we know into eqn. 1 above,
   (P1/T1)(V1/n1) = (P2/T2)(V2/n2)

   (P1/T1)(n2/V2) = (P2/T2)(n1/V1)

   (P1/T1)(0.9)(n1/V1) = (P2/T2)(n1/V1)

   P2 = (0.9) P1(T2/T1)

      = (0.9) 30 psi (340 K / 280 K)

      = 33 psi
Since we're dealing in ratios, we didn't even have to convert pressure into atm. Without that leak, the running pressure would've been 36 psi, 20% over the manufacturer's recommended value. What would've happened if you'd pressurized them to 30 psi when they were hot...and then let them cool off? Without the leak, a 17% loss of pressure. (So manufacturers must build tires with tolerances of this magnitude.)

We can even handle MIXTURES of gases because the gas law doesn't care WHAT the gas is, only how many moles of gas (of all kinds) are within it.

Kinetics of Diffusion

The Ideal Gas law is predicted upon the Kinetic Theory of Gases. But aspects of the Kinetic Theory of Gases are dependent on the nature of the molecules. While the claim is that the average kinetic energy of any gas varies directly with the temperature, we know that molecules of molecular mass m moving with (root mean squared average) velocity v have an average energy of
       ½ m v²
Since two different gases still have average energies proportional to the (absolute) temperature with the same proportionality constant, we know that those average energies are equal to one another.
       ½ m1v1²  =  ½ m2v2²

or

       (v1/v2)² = m2/m1

or

        v1/v2 = (m2/m1)1/2
We can use that middle form to find ratios of molecular weights from (root mean squared) average velocities, but where do we get those? From the measure of time it takes for the two gases to leak out of a vessel (at the same temperature for both gases, naturally). If gas A takes twice as long to leak out (measured by P changes, of course) as does gas B, then A's (rms) average velocity must be half of B's. That equation then tells us that it's molecular weight is four times B's.

Note that the text speaks of rates of change as directly proportional to molecular (rms) averaged speeds, but some questions may be posed in terms of relative TIME to lose the same amount of gas. Velocity and time are inversely related since a rate loss is the number of moles PER UNIT TIME; longer times means slower (smaller) rates.

As an example, the speed of sound in air at sea level and ordinary temperatures is about 760 mph. What would it be if the atmosphere were molecular hydrogen at the same T and P? Since the average M for air is 28.5 g/mol while that for (diatomic) hydrogen is only 2.0 g/mol, we can use that last equation to determine that the speed of sound would likely be
   v(H2)  =  760 mph (28.5/2.0)1/2

          =  2,870 mph !

Return to the CHM 1341 Lecture Notes or Go To Next or Previous Lectures.
Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

   Voice: (214) 883-2485
     Fax: (214) 883-2925
     BBS: (214) 883-2168 (HST) or -2932 (V.32bis)
Internet: parr@utdallas.edu  (sends Chris e-mail.)
Last modified 18 June 1996.