# Gas Laws. cont.

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

• Amount of substance, (in moles)
• Volume, V (in Liters)
• Pressure, P (in atm, though often measured in mmHg; remember 1 atm = 760 mmHg, defined)
• Temperature (in K) Note that Kelvins represent an "absolute" temperature scale with 0 at the lowest possible temperature and degrees the same size as °C. So 0 K = –273.15°C, and K = °C + 273.15.

### 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:

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:

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:

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.

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:

• P = 767 mmHg
• V = 1.000 L (assume one liter since want density in mol/L)
• n = ?
• R = 0.0821 L*atm/mole*K
• T = 25°C

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!

• P = (767 mmHg)(1 atm/760 mmHg) = 1.009 atm
• T = 25 + 273.15 K = 298.15 K

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.

• R is a constant so doesn't change.
• If the can is sealed, then n is constant.
• If it doesn't expand (its rigid), then V is constant.
• T is changed (about how much in absolute terms?).
• P must change in response, since everything else is fixed.

(Is the pressure now about 1,2,3,4 or 5 times higher than it started?)

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,

• T1 = 25 + 273.15 = 298.15 K
• T2 = 557 + 273.15 = 830.15 K

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

= 5.6028 atm = 5.60 atm.

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