1. Gay-Lussac’s Law

2. Gay-Lussac’s Law
The relationship among pressure and temperature, at constant volume, can be mathematically represented by an equation known as Gay-Lussac’ law.
P1 = P2
T T2
where:
P1 is the initial pressure and P2 is the new pressure.
T1 is the initial temperature and T2 is the new temperature.
V1 and V2 are the same (constant volume.)

3. Gay-Lussac’s Law 22.4 L At constant volume, when temperature
is increased,
the pressure will increase.
At constant
volume, when temperature
is decreased,
the pressure will decrease.
22.4 L

4. Gay-Lussac’s Law P = k T P T (K)
As the Kelvin temperature of the gas increases, the pressure of the gas increases and vise versa.
P = k T P P1 •
What
temperature
is this? P2 •
T (K) T2 T1
-273oC

5. Gay-Lussac’s Law
Ex. (1) If a rigid container of He at STP were cooled to 200. K, then what would be the new pressure in atmospheres?
P1V1 P1 =
P2V2 P2 T1 T2
(1.00 atm) =
(X)
(273 K)
(200. K) X =
0.733 atm

6. Gay-Lussac’s Law
Ex. (2) If a rigid container of oxygen gas (O2) at atm and 32.8oC were heated to 50.0oC, what would be new pressure?
P1V1 P1 =
P2V2 P2 T1 T2
(1.04 atm) =
(X)
(305.8 K)
(323.0 K) X =
1.10 atm

7. Gay-Lussac’s Law
Ex. (3) What would be the new temperature of a rigid container of a gas at kPa and K if the gas pressure increased to kPa?
P1V1 P1 =
P2V2 P2 T1 T2
(101.3 kPa) =
(115 kPa)
(293 K)
(X) X =
333 K

8. Gay-Lussac’s Law
Ex. (4) What would have been the initial temperature of a rigid container of neon (Ne) at 1.00 atm if the final temperature was K and the final pressure was 1.10 atm?
P1V1 P1 =
P2V2 P2 T1 T2
(1.00 atm) =
(1.10 atm)
(X)
(293 K) X =
266 K