Go to a Combined Gas Law Worksheet

Here is one way to "derive" the Combined Gas Law:

Step 1: Write Boyle's Law:

P_{1}V_{1}= P_{2}V_{2}

Step 2: Multiply by Charles Law:

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

Step 3: Multiply by Gay-Lussac's Law:

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

Step 4: Take the square root to get the combined gas law:

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

If all six gas laws are included (the three above as well as Avogadro, Diver, and "no-name"), we would get the following:

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

However, this more complete combined gas law is rarely discussed. Consequently, we will ignore it in future discussions and use only the law given in step 4 above.

**Example #1:** This type of combined gas law problem (where everything goes to STP) is VERY common:

2.00 L of a gas is collected at 25.0 °C and 745.0 mmHg. What is the volume at STP?

STP is a common abbreviation for "standard temperature and pressure."

You have to recognize that five (of six possible) values are given in the problem and the sixth is an x. Also, remember to change the Celsius temperatures to Kelvin.

When problems like this are solved in the ChemTeam classroom, I write a solution matrix, like this:

P _{1}=P _{2}=V _{1}=V _{2}=T _{1}=T _{2}=

and fill it in with data from the problem.

Here is the right-hand side filled in with the STP values:

P _{1}=P _{2}= 760.0 mmHgV _{1}=V _{2}= xT _{1}=T _{2}= 273 K

You can be pretty sure that the term "STP" will appear in the wording of at least one test question in your classroom. The ChemTeam recommends you memorize the various standards conditions. If your teacher allows a "cheat sheet" to be used on the test, MAKE CERTAIN those values are there.

Here's the solution matrix completely filled in:

P _{1}= 745.0 mmHgP _{2}= 760.0 mmHgV _{1}= 2.00 LV _{2}= xT _{1}= 298 KT _{2}= 273 K

Write the combined gas law equation:

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

and solve for V_{2} by first cross-multiplying:

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

then dividing:

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

Insert the five values in their proper places on the right-hand side of the above equation and carry out the necessary operations.

**Example #2:** The pressure of 8.40 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is the new volume?

**Solution:**

This is a combined gas law problem since you have three variables changing: pressure, temperature and volume. There will be six quantities: five values and one unknown.

1) Set up the six quantities:

P _{1}= P_{1}P _{2}= P_{1}/2V _{1}= 8.40 LV _{2}= xT _{1}= T_{1}T _{2}= 2T_{1}

2) Rearrange the Combined Gas Law:

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

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

3) Substitute into the rearranged gas law:

V_{2}= [(P_{1})(8.40 L)(2T_{1})] / [(T_{1}) (P_{1}/2) ]V

_{2}= 4(8.40 L) = 33.6 L

4) Another way to solve this is to use values, as follows:

P _{1}= 2P _{2}= 1V _{1}= 8.40 LV _{2}= xT _{1}= 1T _{2}= 2

Note that the pressure decreases by one-half and the temperature doubles, per the instructions in the problem.

5) Substitute into the rearranged gas law:

V_{2}= [(2)(8.40 L)(2)] / [(1) (1) ]V

_{2}= 4(8.40 L) = 33.6 L

**Example #3:** This next problem uses two gas laws in sequence. It involves using Dalton's Law of Partial Pressures first, then use of the Combined Gas Law. The explanation will assume you understand Dalton's Law. These two laws occuring together in a problem is VERY COMMON.

1.85 L of a gas is collected over water at 98.0 kPa and 22.0 °C. What is the volume of the dry gas at STP?

The key phrase is "over water." Another phrase to look for in this type of problem is "wet gas." This means the gas was collected by bubbling it into an inverted bottle filled with water which is sitting in a water bath. The gas bubbles in and is trapped. It displaces the water which flows out into the water bath.

The problem is that the trapped gas now has water vapor mixed in with it. This is a consequence of the technique and cannot be avoided. However, there is a calculation technique (Dalton's Law) that allows use to remove the effect of the water vapor and treat the gas as "dry." For this example, we write Dalton's Law like this:

P_{gas}+ P_{H2O}= P_{total}

We need to know the vapor pressure of water at 22.0 °C and to do this we must look it up in a reference source.

It is important to recognize the P_{total} is the 98.0 value. P_{total} is the combined pressure of the dry gas AND the water vapor. We want the water vapor's pressure OUT.

We solve the problem for P_{gas} and get 95.3553 kPa. Notice that it is not rounded off. The only rounding off done is at the FINAL answer, which this is not.

Placing all the values into the solution matrix yields this:

P _{1}= 95.3553 kPaP _{2}= 101.325 kPaV _{1}= 1.85 LV _{2}= xT _{1}= 295 KT _{2}= 273 K

Solve for x in the usual manner of cross-multiplying and dividing.