Question

can anyone explain me when to use gay-lussac's law and when not

also explain why the question number e (i) will not be solved with gay lussac's law

Answer 1

1) If the temperature of a container is increased, the pressure increases. 
2) If the temperature of a container is decreased, the pressure decreases.

What makes them true? We can make brief reference to the ideas of kinetic-molecular theory (KMT), which Gay-Lussac did not have access to in the early 1800's. KMT was developed in its modern form about 50 years later.

1) 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).

2) Suppose the temperature is decreased. This means gas molecules will move slower and they will impact the container walls less often. This means the gas pressure inside the container will decrease, 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 (either P or T) goes up, the other (either P or T) also increases. The constant K remains the same value.

The mathematical form of Gay-Lussac's Law is:

P

 

–––

= k

T

 

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

 

–––

= k

T1

 

And we know this:

P2

 

–––

= k

T2

 

Since k = k, we have this:

P1

 

P2

–––

=

–––

T1

 

T2

Be aware that you can also see this equation written as:

P1 / T1 = P2 / T2

or:

P1T2 = P2T1

The second one, of course, resulting from cross-multiplication of the equation in fractional form.

Make sure to convert any Celsius temperature to Kelvin before using it in your calculation.