Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 14: 18 February

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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