Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 14: 18 February


Gas Laws, cont.

LN Demos

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


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, & 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; and 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.

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. 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 was the pressure in the container when it was sealed in the lab at 25°C?

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)

= 5.133 atm = 5.13 atm.


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© R A Paselk

Last modified 18 February 2011