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Discovered by Joseph Louis Gay-Lussac in the early 1800's. However, the ChemTeam's knowledge of this is much less sure than concerning Boyle's or Charles' Law.
Gives the relationship between pressure and temperature when volume and amount are held constant.
If the temperature of a container is increased, the pressure increases.
If the temperature of a container is decreased, the pressure decreases.
Suppose the temperature is increased. This means gas molecules will move faster and they will impact the container walls more often. This means the gas pressure inside the container will increase, since the container has rigid walls (volume stays constant).
Gay-Lussac's Law is a direct mathematical relationship. This means there are two connected values and when one goes up, the other also increases.
The mathematical form of Gay-Lussac's Law is: P ÷ T = k
This means that the pressure-temperature fraction will always be the same value if the volume and amount remain constant.
Let P1 and T1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T2, then the pressure will change to P2. Keep in mind that when volume is not discussed (as in this law), it is constant. That means a container with rigid walls.
As with the other laws, the exact value of k is unimportant in our context. It is important to know the PT data pairs obey a constant relationship, but it is not important for us what the exact value of the constant is. Besides which, the value of K would shift based on what pressure units (atm, mmHg, or kPa) you were using.
We know this: P1 ÷ T1 = k
And we know this: P2 ÷ T2 = k
Since k = k, we can conclude that P1 ÷ T1 = P2 ÷ T2.
This equation of P1 ÷ T1 = P2 ÷ T2 will be very helpful in solving Gay-Lussac's Law problems.
This graphic simply restates the above in a way HTML cannot do.
Notice the similarities to the Charles' Law graphic. This is because both laws are direct relationships.
Make sure to convert any Celsius temperature to Kelvin before using it in your calculation.
Example #1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?
Solution: change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation (the ChemTeam will use the left-hand one in the graphic above) and get:
The answer is 311.3 K, but the question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C. Notice that the volume never enters the problem. This is because the problem is asking about the relationship between pressure and temperature; the volume (as well as the moles) remains constant.
Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?
Solution: convert to Kelvin and insert:
Cross multiply and divide for the new pressure.
Sometimes a problem will give you one pressure in one unit and ask for the new pressure in a different unit. In that case, simply do the problem and then convert the pressure to the different unit.
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