*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
Chem 107	Fundamentals of Chemistry	Fall 2009
Lecture Notes: 27 October	© R. Paselk 2005

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Gases

Gases: Briefly discussed overall properties of gases (fill container, have mass & volume, compressible, lo density, lo viscosity, etc.). Also noted:

Gas particles are elastic (otherwise would lose energy with collisions and collapse to a liquid).
Gas particles exert pressure.

What is Pressure? Pressure is the force/unit area. Due to collisions of particle with walls of container etc.

Units of Pressure:

mmHg - based on manometers. Two types:

open tube - measures pressure relative to current atmospheric pressure.

closed tube - measures pressure relative to contents of the enclosed volume at the closed end. (A barometer is an example where the enclosed space is "empty", that is it contains a vacuum since the vapor pressure of mercury is very low.)

atm

others include psi (pounds/square inch), pascals, Torr, etc.

Gas Laws

Gases are characterized by four properties. Gas Laws describe the relationships between these four properties:

Amount of substance, n (in moles)
Volume, V (in Liters)
Pressure, P (in atm, though often measured in mmHg)
Temperature (in K)

Boyle's Law

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

PV = c @ constant T & n

Or, "At constant temperature the volume of any quantity of gas is inversely proportional to its pressure." V = k (1/P), or PV = k, & P₁V₁ = P₂V₂.

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

Boyle's law plot of V vs. P showing curved relation

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

Boyle's Law V vs 1/P plot

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

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:

If we extrapolate the plot to zero volume, V = 0, we can find an absolute minimum value of temperature on the assumption that negative volumes can't exist. Looking at the plot we can find a value for the lowest temperature possible, or absolute zero = 0 K:

Charles law plot extrapolated to zero volume showing intercept at absolute zero

The intercept on the volume axis is 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 these 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.

But of course the constant now includes amount of stuff. I f we keep P and T constant we can find the relationship between V and moles:

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" on pg 356 of your text etc.

As an example, let's find the molar volume of a gas under standard conditions of temperature and pressure (STP). STP are defined as: P = 1 atm and T = 0° C. Thus we need to solve the gas equation for 1 mole of gas at 273.15 K (= 0°C+ 273.15) and 1 atm:

PV = nRT

V/n = RT/P

V/(1 mole) = (0.0821 (L*atm)/(mole*K))(273.15 K)/1 atm

V = 22.426 L/mole = 22.4 L/mole = molar volume of an ideal gas.

Let's look at some sample problems.

Example: We can find the MW of a gas given its density. Thus a 1.000 L sample of gas weighed 1.25 g at a temperature of 0.0 °C and a pressure of 1.000 atm. What is the MW of this gas?