Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 15: 25 February

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Gas Laws. cont.

Gas Laws describe the relationships between the four properties characterizing any gas:

Boyle's Law

Boyle's Law describes the relationship between pressure and volume when the temperature and amount of substance are held constant.

PV = c @ constant T & n

Plotting pressure volume data (keeping n and T constant) gives a graph for a hyperbola (xy = c), as seen below:

Boyles Law Plot

boyle's law animation

public domain image courtesy of Wikipedia commons

Notice that we can rearrange this equation to give a straight-line relationship:

Divide both sides by V: (PV)/V = c/V

P = c (1/V)

This is now in the form of a straight line: y = ax + b, where b = 0

Thus, "At constant temperature the volume of any quantity of gas is inversely proportional to its pressure." V = k (1/P) & multiplying both sides by P get P1V1 = k = P2V2, so P1V1 = P2V2.

(Aside on straight-line plots: Very popular in science. Traditionally, we will do almost anything to get a straight line. Why? Because straight lines easy to recognize and evaluate. Also easy to evaluate statistically.)

Charles' Law

The relationship between volume and temperature was determined much later because accurate thermometers had to be developed first. But once thermometers were available a number of workers determined that volume is directly proportional to temperature. Plotting data for the relation of volume of a gas to temperature between 0° C and 100 ° C gives a plot similar to that below:

Charles Law Plot

charles law animation

public domain image courtesy of Wikipedia commons

Extrapolating this data to V = 0 we can find an absolute minimum value of temperature on the assumption that negative volumes can't exist:

Charlesw Law Plot extrapolated to zero volume

The intercept on the volume axis is then taken as absolute zero = -273.15 °C = 0 K for an ideal or "perfect" gas with particles of zero volume and no interactions other than collisions.

Algebraically we then find that V = k'T, & dividing both sides by T, V1/T1 = k' = V2/T2, or V1/T1 = V2/T2.

Combined Gas Law

We can combine Boyle's and Charles' relationships (T was part of the constant for Boyle's Law and P is part of the constant for Charles' Law) to give:

(PV)/T = constant.

Avogadro's Law

V = an, where n = moles of stuff. So we have a linear relation between volume and moles. Dividing both sides by n get V1/n1 = a = V2/n2, or V1/n1 = V2/n2.

Ideal Gas Law

Ideal Gas Law ("Perfect Gas Law"): The constant for the combined law includes amount of stuff, and breaking that out we then get

(PV)/T = nR, or PV = nRT

where R = the gas constant with units appropriate to the various measurements. We will use atm, L, K, and moles, so that

R = 0.0821 L*atm/mole*K

I will base all of my examples on this equation because that requires a minimum of memorization. However you may find it easier to memorize a series of equations such as the "combined gas law equation" etc.

As an example, let's find the density of sulfur hexafluoride at a temperature of 25 °C and a pressure of 767 mmHg. The first thing we should do is determine what we want to know, mainly the mass of gas in a one Liter sample, since density = D for gases is in g/L.

So we want mass and volume from our gas law equation. To do this we need to find 1) moles/L and 2) grams/mole. So let's write out the Ideal Gas Law:

PV = nRT

Rearranging to find moles/L,

n/V = P/RT

Next let's list what we know in this equation:

Looking at these values we see that we need to convert P to atm and T to K in order to match the units in R!

Plugging these values into the gas law equation

n/V = P/RT = (1.009atm)/(0.0821L*atm/mole*K)(298.15K)

n/L = 4.122 x10-2 moles/one liter

The MW of SF6 = [32.07 + 6(19.00)] g/mol = 146.07 g/mol. By unit analysis we see can get rid of moles by multiplying mol/L by g/mol. Multiplying mol/L by g/mol then gives the density, D:

D = (4.122 x10-2 mol/L)(146.07 g/mol) = 6.0210 g/L = 6.02 g/L

 

Example: In an experiment the pressure of nitrogen inside a sealed, rigid container is found to be 15.6 atm at 557 °C. What will the pressure in the container be when it is cooled back to a lab temperature of 25°C? (Is the final P higher or lower??)

First look at the problem to see what changes and what doesn't.

Next write out the gas law equation,

PV = nRT

Rearranging to put the variables on one side:

P/T = nR/V, and since nR/V is constant, Po/To = Pf/Tf

Rearranging: Pf = (Po)(Tf/To)

Next, convert temperatures into Kelvins,

Pf = (15.6 atm)(298.15 K)/(830.15 K)

= 5.6028 atm = 5.60 atm.

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© R A Paselk

Last modified 25 February 2015