<p>This gas chapter has me seriously confused, and this question in particular really frustrates me.</p>
<p>Given that 1.00 mol nitrogen and 1.00 mol fluorine gas are in separate containers at the same temperature and pressure. Calculate the following ratios.
a. volume F2/ Volume N2
b. Density F2/ Density N2
c. average translational energy F2/ average translational energy N2
d. number of atoms F/ number of atoms N</p>
<p>I have the feeling this is fairly easy, but I am sooooo confused.</p>
<p>a)for the volume, convert from mol to liters remember 1 mol= 22.4 L
b) for density, convert from mol to grams then i would guess divide by the liter you found in part A
c)i don't know
d)convert from mol to molecules (1 mol = 6.02 x 10^23 molecules) then convert from molecules to atoms (like for N2 there would be 2 N atoms)</p>
<p>basically you are using basic dimensional analysis on this. hope it helps. </p>
<p>other ppl: feel free to correct me if i'm wrong</p>
<p>These questions emphasize important conceptual elements from Gas Laws, i.e., which properties depend only on conditions and are independent of the specific gas.</p>
<p>a) it doesn't specify STP, so you can't assume 1 mol=22.4L. Instead, for each gas V=nRT/P. Since nRT/P is the same for both, V1/V2 = 1. (Volume depends only on the number of moles, not the specific gas.)
b) PV=nRT implies PV=(g/MM)RT. Rearranging for density
D=g/L= (PxMM)/RT.
Since all variables are the same except for MM, D1/D2 = MM1/MM2
(Density does depend on the specific gas.)
c) Translational Kinetic Energy (per mole)= (3/2)RT (It's one of the equations on the equation sheet supplied on the AP test.) In other words, KE depends only on temperature, not on the specific gas. Therefore KE1/KE2 = 1.
d) Wonky is right. conceptually, both containers have the same number of moles, therefore the same number of molecules. Since both are diatomic, they also contain the same number of atoms. #atoms1/#atoms2 = 1.</p>
<p>Thanks!</p>
<p>Maybe you could help a bit more... My chem teacher always makes up a couple questions of his own each unit, and usually they're pretty easy, but these don't even make any sense.</p>
<p>In an airport in South America, the elevation is so high that on hot. low barometric pressure days they recommend grounding the private planes. My father in law flew there and was grounded. He knew considering the length of the runway, that a density of .8g/L would be sufficient for take off. As night came, the barometric pressure was 20.0in Hg, To what temperature (fhrenheit) must the surface temperature drop to make the take off safe?</p>
<p>Does this even make sense?</p>
<p>In my distilation apparatus, the water vapor travels as a hot wind through a 1.0 cm diameter orifice. I collect about 7.2 gallons of water in a 24 hour period. Calculate the speed of the steam in miles per hour. The vapor is 97 degrees C and 640 torr.</p>
<p>Thanks to all you chem geniuses!</p>
<p>^anyone?</p>
<p>10 chars</p>
<p>For the distillation problem, convert gallons to mL and divide by 24 to get mL per hour. Since 1mL=1cm^3, you can take the value for mL/hr and divide by 3.14<em>0.5^2 (pi</em>r^2) to get cm/hr. Then convert cm to miles. The temp and pressure numbers are red herrings.</p>
<p>Unless I'm having a complete space-out moment, the airport question is either missing some information or he expects you to get some typical values for atmospheric composition from the textbook. As mentioned above D = .8g/L = (P<em>MM)/RT. If you know the average composition of the atmoshpere, you could find the average MM of the atmoshpere, convert 20inHG to atm, and solve for Temp. (Of course, T comes out it in Kelvin.) There are other ways to approach the problem depending on what other values may be provided, but *some</em> additional info has to be available.</p>