GAS LAWS: Boyle’s Law, Charle’s Law, and Gay-Lussac’s Law.

GAS LAWS: Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law

The most easily observable macroscopic traits of a gas are its pressure ( P ), volume ( V ), mass ( m ), and temperature ( T ). Pressure is measured in Pascals ( or psi ) and has units of force ( N ) per unit area ( A ) in ( N / m^2 ). Volume has units of cubic meters ( m^3 ). The mass in both kilograms ( kg ) and moles ( mol ) are common to experiments involving gases, and the temperature measurement of utility is the Kelvin ( Tk = Tc + 273.15 ). Among other things, using the Kelvin scale ensures never having to divide anything by 0 C degrees.

The simplest experiments involve changing a single variable to monitor how another variable is affected. As a consequence, Boyle’s Law, Charle’s Law, and Gay-Lussac’s Law were developed from conducting experiments where mass was an invariant quantity. The experiments differ, however, in choices about what other variables ( P, V, or T ) were kept constant. When temperature and mass are kept constant, the following Boyle’s Law observation is made:

( P )( V ) = constant.

Unit analysis reveals that ( P )( V ) = [ ( N / m^2 )( m^3 ) ] = ( N * m ) = J ( Joules ). ENERGY CANNOT BE CREATED OR DESTROYED, so an increase in volume MUST be accompanied by a decrease in pressure ( and vice-versa ) when temperature and mass are invariant. If either ( P ) or ( V ) is kept constant along with mass, some other change in the system will evolve. An examination of Charle’s Law reveals that a gas system with constant pressure and change in volume has a directly proportional change in temperature:

( V / T ) = constant.

Likewise, when the volume of a gas is kept constant along with its mass, a change in pressure is accompanied by a directly proportional change in temperature, and according to Gay-Lussac’s Law:

( P / T ) = constant.

There is a special relationship between energy and temperature. Since the product of ( P )( V ) = energy in Joules ( J ), the ratio of [ ( P )( V ) ] / ( T ) = ( J / K ) = constant. Furthermore, for a system where the mass of a gas is invariant, it follows that [ ( P1V1 ) / ( T1 ) ] = [ ( P2V2 ) / ( T2 ) ]. 

In order to account for the differing masses of gas atoms and molecules, experiments involving variant gas capabilities must be carried out under conditions where large distances separate the gas atoms or molecules involved. As a consequence, a standard temperature and pressure ( STP ) must be maintained, with a temperature of 273.15 K or 0C degrees, and pressure is 0.101 MPa or 1 atm.

A typical gas at STP contains 2.7 * 10^19 molecules per cubic meter ( m^3 ). On average, the space occupied by virtually ANY gas atom or molecule is offset by empty space that is 3,375 times greater in expanse! Thus, each gas species can be assumed to behave as if all gases have uniform mass characteristics. A detailed explanation of the behavior of gases under standard conditions ( STP ) will be given in a separate summary of The Ideal Gas Law.

Q: A container has a volume of 1.00 m^3, and it is filled with air at 0C degrees and to a pressure that is 20.0 times atmospheric pressure. How much volume would this gas occupy at room temperature under 1 atm of pressure?

A: [ ( P1V1 ) / ( T1 ) ] = [ ( P2V2 ) / ( T2 ) ]. Solving for the final volume, we have Vf = [ ( 20 )( 1.00 m^3 )( 293.15 K ) ] / ( 273.00 K ) = 21.5 m^3.

The increase in Vf is mostly due to the decrease in initial pressure ( Look again at P1 / P 2 ). Temperature was a factor, but ( P )( V ) = constant energy in Joules ( J ) when T is fixed. Some other entity must be factored into the study of gases, and that entity is molar mass.