Mr. ShieldsRegents Chemistry U05 L07

Louis Joseph Gay-Lussac (1778 – 1850) - Gay-Lussac’s Law (abt 1807) - Pressure vs. Temp relationship P/T = k (Constant V and n)

Gay-Lussac & Biot 1804 Gay-Lussac Loved Flying In hot air balloons and made Ascents to 7000 meters

We know from the KMT that The avg. KE of a gas is directly proportional to Temp. (in Kelvin!) And since KE = ½ mv 2 as Temp increases Molecular velocity increases If we hold the volume constant then the frequency of Molecular collisions with the wall must be increasing With increasing T since the velocity is increasing. Developing GL’s Law

If the number of collisions with the wall per unit time increases then the pressure increases So as T, vel. and so does P Remember T must always be in Kelvin! Gay-Lussac described this gas behavior as P/T = k Since P inc with inc T (and vice versa) this represents a Direct Relationship between P and T

You probably noticed that the relationship between P and T has the same relationship as the one between V and T P/T = k V/T = k So you might expect a plot of Guy-Lussac’s law to look The same as Charle’s Law … And you’ld be right!

Notice that Gay-Lussac and Charles law both predict that P and V go to zero as T goes to -273 deg C. This temperature is the zero pt. on the Kelvin scale. A plot of Gay-Lussac’s law Direct relationship

PROBLEM 1: A sample of gas in a container at 30 deg C has a pressure of 1 atm. If the temperature is increased to 200 deg. Celsius, at constant V and n, what pressure will the gas have? Let’s do some problems: P 1 /T 1 = P 2 /T 2 1 atm / 303K = P 2 / 473 deg K P 2 = (1 x 473)/303P = 1.56 atm

Problem 2 A fire broke out in a nitrogen storage facility. The Electronic gages on the nitrogen tanks, which at 25 deg C had a normal pressure of 68 atm, recorded the Highest pressure the tanks were subjected to was 285 atm. What was the temperature of the fire in the Nitrogen storage room in Kelvin and deg. C? P 1 = P 2 68 atm = 285 T 1 T 2 298K T 2 T 2 = 1249 K T 2 = 976 C (or 1,788 deg. F)