V = constant/P
Then in 1787 Charles discovered that at constant P, the volume of a gas was directly proportional to temperature:
V = constant*T
Combining these two equations we see
V = constant * (T/P)
A standard temperature and pressure is called STP and is T = 00C = 273.15 K and P = 1 atm.
Gay-Lussac discovered that when gases react with one another they do so by volumes that are in the ratios of small whole numbers.
In 1811 Avogadro developed the hypothesis:
Avogadro's hypothesis was proven 50 years later by Cannizzaro and then people accepted it and considerable progress in understanding the molecular world began. Avogadro's hypothesis led people to determine that at STP 22.4 liters of any gas will weigh the gram molecular weight of the gas: 22.4 liters of H2 will weigh 2 grams, 22.4 liters of O2 will weigh 32.00 grams, etc. You see that Avogadro's hypothesis led the scientists to understand that the volume of a gas is proportional to the number of moles of the gas.
So now we have:
If we combine all of these we will have
V = constant*(nT/P) and if we let the constant be R then we have
V = nRT/P or PV = nRT ....... the ideal gas equation of state.
The value of R depends upon the units:
R = PV/(nT)
R = 0.082056 latm/molK = 82.056 cm3atm/molK
R= 1.987 cal/molK
R = 8.3144 J/molK
So what value we use for R will depend upon the problem that we are working.
So we have an equation of state (EOS): PV=nRT. You note that I called this the ideal gas equation of state. Why did I call it the ideal gas equation of state?
Good question. Solve for V and we get
V = nRT/P
Now what happens to V as P goes to infinity or as T goes to zero? The volume goes to zero right? So what happened to our gas? It disappeared as the pressure increased towards infinity!! This means that the gas does not have any size!! Clearly this is not for a real gas. Thus we call our equation of state an ideal gas equation of state(IG EOS). We will explore this further later.
Go here for practice using the IG EOS.
What if we have a mixture of gases rather than just one gas? Dalton discovered that the pressure is equal to the sum of the partial pressures of the gases in the mixture. The partial pressure is the pressure that the gas would exert if it were alone in the same container in which the mixture is present. The partial pressure of the ith gas is then
Pi = niRT/V and the total pressure by Dalton's law is
P = SUM Pi = P1 + P2 + P3 + ...
P = SUM niRT/V = RT/V SUM ni = (RT/V)ntotal
We can get a relationship between the total pressure and the partial pressures quite easily. Look at the ratio of the partial pressure to the total pressure:
Pi/Ptotal = (niRT/V)/(ntotalRT/V) = ni/ntotal = Xi .. the mole fraction of the ith gas. Therefore we have
Pi = XiPtotal
Let's work an application of what we have learned so far with Dalton's law and the relationship between partial pressures and total pressure. What is the total pressure of a mixture of 1.0 g of hydrogen gas and 5.0 g of helium gas confined to a vessel of volume 5.0 liters at 20 0? Also what are the partial pressures of each gas?
We know that P = ntotalRT/V so we need the total number of moles of the gases in the mixture.
ntotal = 1.0 g H2*(1 molH2/2.0 g H2) + 5.0 g He*(1 mol He/4.0 g He) = 0.5 moles H2 + 1.25 moles He = 1.75 moles
So the pressure is
P = ntotalRT/V = 1.75 moles*(0.082 latm/molK)293K/5.0 l
P = 8.4 atm
Now how about the partial pressures? From the moles of the individual gases, we can calculate the mole fraction of the gases:
Xhydrogen = 0.5/1.75 = 0.29
Xhelium = 0.71
Phydrogen = 0.29 * 8.4 atm = 2.4 atm
Phelium = 0.71 * 8.4 atm = 6.0 atm
Check to see that the sum of the partial pressures is the total pressure and we see that 2.4 + 6.0 = 8.4.
We can also develop now a useful equation that will help in dealing with some situations. From the ideal gas EOS we can solve for R and get
R = PV/nT
Now since this is a constant it has the same value for one set of state variables as it does for another set of state variables. That means that
R = P1V1/n1T1
This equation would be useful for solving problems where there are two sets of conditions which are somewhat different and on variable is unknown. For example, what is the volume of a baloon at temperature 500 K and pressure .750 atm when you know that the volume of the baloon at temperature 300 K and pressure 1.00 atm is 2 liters (n will be the same for we don't change the contents of the baloon. Give that problem a try. The answer is 4.44 liters.
One further extension of the IG EOS is helpful which just involves a little rearrangement and the definition of density and moles. So
PV = nRT = (wt/MW)RT where we used the definition of mole. Now divide both sides by V:
P = (wt/V)RT/MW and recognize that weight over volume is density, D:
P = DRT/MW so now you can use density data in the IG EOS.
Here is a site to help you practice. You enter numbers and it calculates the answer for you -- this is intended to help you check your answers.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
|You must use your real name when it asks for a name.|
|The test will only submit when you have answers all of the questions correctly.|
|If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.|
You can either take a break or go to the next section,
Web Author: Dr. Leon L. Combs
Copyright ©1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED