Discussion and Ten Examples

Discovered by Joseph Louis Gay-Lussac (the uppermost picture to the right) in 1802. He made reference in his paper to unpublished work done by Jacques Charles (the lower GIF picture to the right) about 1787. Charles had found that oxygen, nitrogen, hydrogen, carbon dioxide, and air expand to the same extent over the same 80 degree interval.

The ChemTeam does admit that Charles looks like he's ready to fail a sobriety test, but you would do well to cut the guy some slack. After all, he did invent the hydrogen-filled balloon and on December 1, 1783, he ascended into the air and became possibly the first man in history to witness a double sunset.

Gay-Lussac was no slouch either in the area of ballooning. On September 16, 1804, he ascended to an altitude of 7016 meters (just over 23,000 feet, about 4.3 miles). This remained the world altitude record for almost 50 years and then was broken by only a few meters. His image is from a stamp France issued in memory of the 100th anniversary of his death in 1950.

Because of Gay-Lussac's reference to Charles' work, many people have come to call the law by the name of Charles' Law. There are some books which call the temperature-volume relationship by the name of Gay-Lussac's Law and there are some which call it the Law of Charles and Gay-Lussac. Needless to say, there are some confused people out there. Most textbooks call it Charles' Law, so that's what the ChemTeam will use.

The same year a 23-year-old Gay-Lussac discovered this law, he had occasion to walk into a linen draper's shop in Paris and there he made a wonderous discovery. He found the 17-year-old shopgirl reading a chemistry textbook while waiting for customers. Needless to say, he was intrigued by this and made more visits to the shop. In 1808, he and Josephine were married and over the years, five little Gay-Lussac ankle-biters were added to the scene.

So ladies, while the ChemTeam cannot guarantee that the study of chemistry will land you the man of your dreams, you just never know what might happen through diligent study of your chemistry.

Charles' Law gives the relationship between volume and temperature if pressure and amount are held constant. In other words, Gay-Lussac found these to be true:

1) If the volume of a container is increased, the temperature increases.

2) If the volume of a container is decreased, the temperature 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 after Gay-Lussac's discovery.

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 (but only for an instant. Think of a short span of time. The span of time the ChemTeam is referring to here is much, much shorter than that. So there.). The greater pressure on the inside of the container walls will push them outward, thus increasing the volume. When this happens, the gas molecules will now have farther to go, thereby lowering the number of impacts and dropping the pressure back to its constant value.It is important to note that this momentary increase in pressure lasts for only a very, very small fraction of a second. You would need a very fast, accurate pressure sensing device to measure this momentary change.

2) Consider another case. Suppose the volume is suddenly increased. This will reduce the pressure, since molecules now have farther to go to impact the walls. However, this does not follow the law; the pressure must remain constant. Therefore, the temperature must go up, in order to get the moecules to the wals faster, thereby overcoming the longer distance and keeping the pressure constant.

Charles' Law is a direct mathematical relationship. This means there are two connected values (the V and the T) and when one goes up, the other also increases. if one goes down, the other will go down as well. The constant k will remain the same value.

The mathematical form of Charles' Law is:

V ––– = k T

This means that the volume-temperature fraction will always be the same value if the pressure and amount remain constant.

Let V_{1} and T_{1} be a volume-temperature pair of data at the start of an experiment. If the volume is changed to a new value called V_{2}, then the temperature must change to T_{2}.

The new volume-temperature data pair will preserve the value of k. The ChemTeam does not care what the actual value of k is, only that two different volume-temperature data pairs equal the same value and that value is called k.

So we know this:

V _{1}––– = k T _{1}

And we know this:

V _{2}––– = k T _{2}

Since k = k, we can conclude this:

V _{1}V _{2}––– = ––– T _{1}T _{2}

Make sure that you realize this:

V_{1}/ T_{1}= V_{2}/ T_{2}

is saying the same thing. As well as this:

V_{1}T_{2}= V_{2}T_{1}

The equations just above will be very helpful in solving Charles' Law problems. (Comment: I did find a problem that was worded in such a way as to use V/T = k for the solution of the problem. I have included it as problem #25.)

Before going to some examples, let's be very clear:

EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVIN, NOT DEGREES CELSIUS.

The ChemTeam hopes you understand this very well. Repeating it does not hurt:

DON'T YOU DARE USE CELSIUS IN A NUMERICAL CALCULATION. USE KELVIN EVERY TIME.

Now, one final nit about symbols:

°C ---> the symbol for temperature in degrees Celsius

K ---> the symbol for temperature in KelvinThere is no such symbol as °K or name such as 'degrees Kelvin.'

There is actually a reason for this; it's related to Kelvin being an absolute scale of temperature and Celsius not being an absolute scale. Don't worry about it too much, just don't use °K or 'degrees Kelvin.' You'll be wrong and you'll run the risk of being told so.

**Example #1:** A gas is collected and found to fill 2.85 L at 25.0 °C. What will be its volume at standard temperature?

**Solution:**

1) Convert 25.0 °C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this:

2.85 L x ––––– = –––– 298 K 273 K Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.

2) Cross-multiply and divide:

x = 2.61 L

**Example #2:** 4.40 L of a gas is collected at 50.0 °C. What will be its volume upon cooling to 25.0 °C?

Comment: 2.20 L is the wrong answer. Sometimes a student will look at the temperature being cut in half and reason that the volume must also be cut in half. That would be true if the temperature was in Kelvin. However, in this problem the Celsius is cut in half, not the Kelvin.

**Solution:**

1) Convert 50.0 °C to 323 K and 25.0 °C to 298 K. Then plug into the equation and solve for x, like this:

4.40 L x ––––– = ––––– 323 K 298 K

2) Cross-multiply and divide:

x = 4.06 L

**Example #3:** 5.00 L of a gas is collected at 100. K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure (as required by Charles' Law)?

**Solution:**

5.00 L 20.0 L ––––– = ––––– 100. K x x = 400. K

Be aware that a problem might ask you for the new temperature to be given in Celsius. Make sure to do the problem in Kelvin, get the new temperature in Kelvin and then convert to Celsius. Like this:

400. minus 273 = 127 °C

**Example #4:** A 2.5 liter sample of gas is at STP. When the temperature is raised to 273 °C and the pressure remains constant, what is the new volume?

**Solution:**

We know the gas starts at standard temperature, zero degrees Celsius. In Kelvins, this is 273 K. Now, note the ending temperature, 273 °C. In Kelvins, that is 546 K.The absolute temperature has doubled! Since Charles' Law is a direct relationship, the volume also doubles, to 5.0 L and that is the answer.

Setting it up mathematically gives this:

5.00 L / 273 K = x / 546 KOn the Internet, you often see problems set up as the above one is. Be very careful reading it and deciding which values to multiply & divide when you cross-multiply and divide.

This is the way you often see the above equation in a textbook:

5.00 L x ––––– = ––––– 273 K 546 Remember also, the pressure remains constant (as does the amount of gas). Consequently, they both simply drop from all consideration in the solving of this problem. After all, both the pressure and the amount remained constant during the entire problem.

**Example #5:** The temperature of a 4.00 L sample of gas is changed from 10.0 °C to 20.0 °C. What will the volume of this gas be at the new temperature if the pressure is held constant?

**Solution:**

(4.00 L) / (283.0 K) = (x) / (293.0 K)This is the classic type of problem a student gets wrong. They see that the Celsius temperature doubled, therefore the volume should also double to 8.00 L. The problem, of course, is that the Kelvin temperature must double for a doubling of the volume. Not the Celsius temperature.

x = 4.14 L

**Example #6:** A 5.0 L container of gas has a temperature change such that the final temperature is 4 times more than the initial. What is the size of the container after the temperature change?

**Solution:**

1) This is a Charles' Law problem because we can't solve it if we assume the pressure or the amount of gas changes:

V_{1}/ T_{1}= V_{2}/ T_{2}V

_{2}= V_{1}T_{2}/ T_{1}V

_{2}= (5.0 L x 4T_{1}) / T_{1}<--- note that T_{2}equals 4T_{1}V

_{2}= 20. L

2) The solution above assumes that the final Kelvin temperature is four times more than the initial temperature. What if it is the Celsius temperature that is four times greater?

I will use a change from 1 °C to 4 °C.

V_{1}/ T_{1}= V_{2}/ T_{2}V

_{2}= V_{1}T_{2}/ T_{1}V

_{2}= (5.0 L x 277 K) / 274 KV

_{2}= 5.0547 LAs you can see, the change is small. So much so, that the properly-rounded off answer is 5.0 L. A 4x increase in the Kelvin value is the usual way the question is understood, if there is any ambiguity.

**Example #7:** A 2.50-L volume of hydrogen measured at -100 degrees Celsius is warmed to 100 degrees Celsius. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.

**Solution:**

1) Charles' Law:

V _{1}V _{2}––– = ––– T _{1}T _{2}

2) Cross multiply:

V_{1}T_{2}= V_{2}T_{1}

3) Isolate V_{2}:

V_{2}= (V_{1}T_{2}) / T_{1}

4) Insert values and solve:

V_{2}= [(2.50 L) (373 K)] / 173 KV

_{2}= 5.39 LNote the conversion from Celsius to Kelvin. Make sure the the proper temperature (the 173 K, for example) is associated with the proper volume (the 2.50 L goes with the 173 K).

**Example #8:** If the temperature of a fixed amount of a gas is doubled at constant pressure, what happens to the volume?

**Solution:**

1) Let's start here:

V_{1}T_{2}= V_{2}T_{1}

2) Let the temperature double:

V_{1}(2 K) = V_{2}(1 K)

3) Notice that V_{2} needs to be twice the value of V_{1} to make an equality. That means that the answer to the question is that the volume doubles.

4) This problem can also be solved by assigning a value to V_{1} and seeing what happens to V_{2}. Like this:

(1 L) (2 K) = V_{2}(1 K)V

_{2}= 2 LThe volume doubles.

**Example #9:** Under constant pressure, at what temperature (in K) will a balloon double in size when originally at 102.4 K

**Solution:**

The answer is 204.8 K.We know that Charles' Law is a direct relationship when the temperature is expressed in Kelvins. To double the volume (at constant pressure), the Kelvin temperature would have to double.

You can solve this problem using V

_{1}T_{2}= V_{2}T_{1}if you so desire. Use 1 and 2 for the volumes and 102.4 for T_{1}. Solve for T_{2}.

**Example #10:** Under constant pressure, at what temperature (in °C) will a balloon double in size when originally at 102.4 °C.

102.4 + 273.15 = 375.55 K375.55 K doubles to 751.10 K

751.10 K − 273.15 = 477.95 C

Rounded off, the answer is 478.0 °C

Notice that I used 273.15 rather than the usual 273. I did that so as to preserve four digits in the final answer. Using 273 results in an answer of 478 °C, so my choice amounted to not making much of a difference, except in number of significant figures in the answer.

**Bonus Example:** An ideal gas at 7.00 °C is in a spherical flexible container having a radius of 1.18 cm. The gas is heated at constant pressure to 88.0 °C. Determine the radius of the spherical container after the gas is heated. [Volume of a sphere = (^{4}⁄_{3})πr^{3}]

**Solution:**

1) Convert Celsius to Kelvin:

7.00 °C = 280.0 K

88.0 °C = 361.0 K

2) Set up Charles' law equation:

V_{1}/ T_{1}= V_{2}/ T_{2}[(

^{4}⁄_{3}) (3.14159) (1.18)^{3}] / 280 = [(^{4}⁄_{3}) (3.14159) (x^{3})] / 361(1.18)

^{3}/ 280 = x^{3}/ 361x = 1.28 cm

Note that the (^{4}⁄_{3})π factor drops out. This is because it remains constant during the entire problem, so it can be divided out. If the problem had asked for the new volume (as opposed to the new radius), we could have include it at the end of the problem. Like this:

V = (^{4}⁄_{3}) (3.14159) (1.28)^{3}

Also, note that the cube remains in step #2 above. This is because the volume change is proportional to the cube of the radius, not the radius itself.

See the Bonus Problem after problem #25 here. It also uses the formula for volume of a sphere.