### Gay-Lussac's LawDiscussion and Ten Examples

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Discovered by Joseph Louis Gay-Lussac in the early 1800's. That is pretty much all the ChemTeam knows. Maybe I'll learn more of the details someday.

Gay-Lussac's Law gives the relationship between pressure and temperature when volume and amount are held constant. In words:

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.

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:

1) Change 25.0 °C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation to get:

 97.0 kPa 101.325 kPa ––––––– = ––––––––– 298.0 K x

x = 311.3 K

2) 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:

1) Convert to Kelvin and insert:

P1 / T1 = P2 / T2

745.0 mmHg / 295.0 K = x / 273 K

2) Cross multiply and divide for the new pressure. Be careful to cross-multiply correctly when you see a set up like the one just above. It's easy to get the numbers mixed up.

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.

Example #3: What is the new pressure (in atm) when a constant volume of gas is heated from 25.1 °C to 37.5 °C? The starting pressure is 755.0 mmHg.

Solution:

1) Set it up and solve:

 755.0 mmHg x –––––––––– = ––––––– 298.1 K 310.5 K

x = 786.4 mmHg

2) Convert to atm:

(786.4 mmHg) (1 atm / 760.0 mmHg) = 1.035 atm

You could have converted the 755.0 mmHg to atm first, if so desired.

By the way, the real value for standard temperature is 273.15 K. However, when used in gas law problems, 273 K tends to be used. In addition, 273 tends to be the value used when converting Celsius to Kelvin. The technically correct value to use is 273.15, but it is used far less often than 273.

Example #4: A constant volume of gas has a pressure of 0.350 atm at 50.0 °C. What is the pressure when the temperature is doubled?

Comment: the problem statement gives Celsius, so it seems obvious that Celsius is intended to double. However, what if Kelvin was what the question writer had in mind?

Solution for doubling Celsius:

1) Convert temperatures:

50.0 °C ---> 323 K
100.0 °C ---> 373 K

2) Solve:

(0.350 atm) (373 K) = (x) (323 K)

x = 0.404 atm

Solution for doubling Kelvin:

P1T2 = P2T1

(0.350 atm) (646 K) = (x) (323 K)

x = 0.700 atm

The P1T2 = P2T1 equation is not often used. Be careful, it may just show up without any explanation.

Example #5: If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be?

Solution:

15.0 atm / 298 K = 16.0 atm / x

x = 317.9 K

Changing it to Celsius and three sig figs gives 45.0 °C for the final answer.

Example #6: If the absolute temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?

Solution:

1) Let's start here:

 P1 P2 ––– = ––– T1 T2

2) Let's allow the absolute temperature to double:

 P1 P2 ––– = ––– 1 2

3) Solve it:

P2 = 2P1

P2 will be twice the value of P1. In other words, the pressure will double when the absolute temperature doubles (given constant volume).

4) You can also assign values to the pressure, as follows:

 1 x ––– = ––– 1 2

Note how the absolute temperature doubles from 1 to 2. It can be any two values, just as long as the second one is double the first one.

This next example uses Boyle's Law as well as Gay-Lussac's Law. Twice the fun!

Example #7: Two containers of identical volume are connected by a tube with a stopper to prevent gases inside the tanks from mixing with each other. Container A has a pressure of 2.00 atm and a volume of 10.0 L while container B has a pressure of 4.00 atm and a volume of 10.0 L. Both containers have an initial temperature of 27.0 °C and were then heated to 127 °C. Upon reaching 127 °C, the stopper was released. What is the new pressure in both containers after the stopper was released?

Solution:

1) Determine the pressure in each tank at 127 °C by using Gay-Lussac's Law:

P1 / T1 = P2 / T2

(A)

2.00 atm / 300. K = x / 400. K

x = 2.67 atm

(B)

4.00 atm / 300. K = y / 400. K

y = 5.33 atm

2) When the stopper is removed, each tank acts like its volume has gone from 10.0 L to 20.0 L. Use Boyle's Law to calculate the new pressure in each tank.

P1V1 = P2V2

(A)

(2.67 atm) (10.0 L) = (x) (20.0 L)

x = 1.33 atm

(B)

(5.33 atm) (10.0 L) = (y) (20.0 L)

y = 2.67 atm

3) Add the two pressures together.

1.33 atm + 2.67 atm = 4.00 atm

4) A second way to solve this problem would be to do the reverse of the above sequence. Use Boyle's Law to calculate the new pressures in A and B as they each change from 10.0 L to 20.0 L. Then, combine the two pressures. Finally, use Gay-Lussac's law on the one volume (which would be 20.0 L) to determine the new pressure after heating to 127.0 °C. You may amuse yourself doing the second way, if you so desire.

Example #8: If a car tire is filled to a pressure of 32.2 lbs/in2 measured at 79.0 °F, what will be the tire pressure if the tires heat up to 120. °F during driving?

Solution:

1) Supposing 32.2 lbs/in2 is a gauge pressure:

(32.2 psi / 14.6959 psi/atm) − (1 atm) = 1.19109 atm

"psi" stands for pounds per square inch.

2) Temperature conversions:

79.0 °F = 299.261 K
120. °F = 322.039 K

The Fahrenheit to Kelvin conversion is not one the ChemTeam taught in his classroom. I won't delve into the conversion here.

3) Use Gay-Lussac's Law to calculate P2:

P1 / T1 = P2 / T2

(P1) (T2 / T1) = P2

(1.19109 atm) (322.039 K / 299.261 K) = P2

P2 = 1.28175 atm <--- this is the absolute pressure in the tire

4) Convert back to gauge pressure, then psi:

1.28175 atm + 1 atm = 2.28175 atm

(2.28175 atm) (14.6959 psi/atm) = 33.5324 psi

To three sig figs, 33.5 psi

Example #9: If the temperature of a gas in a container is doubled on the Kelvin scale, what will happen to the pressure of the gas?

Go back to Example #6. When I say "absolute temperature," this means Kelvins. This question is exactly like Example #6.

The pressure doubles.

Example #10: The temperature of a sample of gas, already at 80 kPa, was quadrupled at constant volume. Calculate the final pressure.

Solution:

1) For no particular reason, I want to use the cross-multiplied form:

P1T2 = P2T1

2) We know that the temperature quadruples from T1 to T2. We'll use "fake" values:

(P1) (4 K) = (P2) (1 K)

3) Now, enter the pressures and solve:

(80 kPa) (4 K) = (x) (1 K)

x = 320 kPa

Bonus Example: A rigid vessel containing a 3:1 mole ratio of carbon dioxide and water vapor is held at 240 °C where it has a total pressure of 2.10 atm. If the vessel is cooled to 10. °C so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

Solution:

1) We will also neglect the fact that some small amount of CO2 dissolves into the water, forming a solution of about pH 5.5.

2) 3:1 molar ratio means that the mole fraction of the CO2 is 0.75.

3) The pressure of the CO2 at the 240 °C condition is:

2.10 atm times 0.75 = 1.575 atm

4) The container cools from 240 °C to 10 °C, but does not change in volume. This is a pressure-temperature relatonship, governed by Gay-Lussac's Law:

P1 / T1 = P2 / T2

1.575 atm / 513 K = x / 283 K

x = 0.87 atm

 Boyle's Law Combined Gas Law Gay-Lussac's Probs 1-10 Charles' Law Ideal Gas Law Return to KMT & Gas Laws Menu Avogadro's Law Dalton's Law Diver's Law Graham's Law No Name Law