Gay+Lussac's+Law

Gay Lussac's law states that the pressure of a fixed mass of gas a constant volume varies directly with the kelvin temperature. This is true because temperature is a measure of how much kinetic a substance has, and as the kinetic energy of a gas increases, the particles will crash against the walls of the container causing more pressure to be exerted. The equation to represent this law looks like:    P1/T1=P2/T2. P1 and T1 represent the initial or given conditions of pressure and temperature, and P2 and T2 represent a different set of conditions. Given three of the four values, one can calculate the remaining value for a system at constant volume.

= = The gas in a container is at a pressure of 3.00 atm and at 25ºC. The label says not to store in temperatures exceeding 52º C. What would the pressure be if it was stored above 52º C?
 * Example Problem:**

__Step One:__ Given: P1 of gas = 3.00atm; T1 of gas = 25ºC + 273 = 298K; T2 of gas =52ºC + 273 = 325K Looking For: P2 of gas in atm

__Step 2__: Because tje gasepis cpmtemts re,aom at tje cpmstamt volume of the container, an increase in the temperature will cause an increase in pressure. Rearange Gay-Lussac's law to solve for P2: P2 = (P1)(T2) / T1

__Step 3:__ Substitute the values for the equation P2 = (3.00atm)(325K) / 298K = 3.27atm

A temperature increase at constant volume causes the pressure of the contents in the container to increase. Home Page
 * Conclusion:**