Go to a Combined Gas Law Worksheet

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

Step 1: Write the problem-solving form of Boyle's Law:

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

Step 2: Multiply by the problem-solving form of Charles Law:

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

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

Step 3: Multiply by the problem-solving form of Gay-Lussac's Law:

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

The above is often how the combined gas law is written on the Internet. You may also see it typeset like this:

P _{1}V_{1}P _{1}V_{2}––––– = ––––– T _{1}T _{2}

In solving combined gas law problems, there is a lot of cross-multiplying involved. I have found using the formulation just above to be helpful in visualizing what to cross-multiply.

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 (mostly) ignore it in future discussions and use (mostly) the law given in step 4 above.

A different way to "derive" the combined gas law is discussed in example #5 below.

**Example #1:** 2.00 L of a gas is collected at 25.0 °C and 745.0 mmHg. What is the volume at STP?

**Solution:**

1) 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.

2) When problems like this were solved in the ChemTeam classroom (the ChemTeam is now retired from the classroom), I would 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.

3) 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

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

4) 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

5) Write the combined gas law equation:

P _{1}V_{1}P _{1}V_{2}––––– = ––––– T _{1}T _{2}

6) Solve for V_{2} by first cross-multiplying:

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

7) Then dividing both sides by P_{2}T_{1}:

P _{1}V_{1}T_{2}V _{2}= ––––– P _{2}T_{1}or:

V

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

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

(745.0 mmHg) (2.00 L) (273 K) x = ––––––––––––––––––––––––– (760.0 mmHg) (298 K) or:

x = [(745.0 mmHg) (2.00 L) (273 K)] / [(760.0 mmHg) (298 K)]

x = 1.796 L

to three significant figures, the answer is 1.80 L

**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.

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}

Notice how P_{2} is represented as being half of P_{1}. Notice how T_{2} is represented as being twice that of T_{1}.

2) Write, then 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 assign placeholder values that fit the requirements of the problem, as follows:

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

Note that the assigned values for pressure decrease by one-half and the assigned values for temperature double, 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

The 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.

**Example #3:** 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 common phrase used 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 terms "over water" and "wet gas" are equivalent; they means the same thing, that being that the gas is saturated with water vapor.

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 put the values into the Dalton's Law equation:

P_{gas}+ 2.6447 kPa = 98.0 kPa

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:

V_{2}= (P_{1}V_{1}T_{2}) / (P_{2}T_{1})x = [(95.3553 kPa) (1.85 L) (273 K)] / [(101.325 kPa) (295 K)

x = 1.61 L (to three sig figs)

Comment: a very common student mistake is to not realize that Dalton's Law must be used first when a gas is collected over water. There is a very common experiment in which some hydrogen gas is collected over water and the molar volume is determined. Dalton's Law will be used in the calculations associated with that lab.

**Example #4:** If the volume of an ideal gas is doubled while its temperature is quadrupled, does the pressure (a) reman the same, (b) decrease by a factor of 2, (c) decrease by a factor of 4, (d) increase by a factor of 2, or (e) increase by a factor of 4?

**Solution:**

1) Write the combined gas law:

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

2) I will assign a value of 1 to V_{1} and allow it to double. I will assign a value of 1 to T_{1} and allow its value to quadruple.

[(P_{1})(1)] / 1 = [(P_{2})(2)] / 4P

_{1}= P_{2}/ 22P

_{1}= P_{2}the answer is (d) increase by a factor of 2

By the way, any volume unit is fine for V_{1}, but the temperature unit must be understood to be Kelvin. In other words, do not select 1 °C, allow it to change to 4 °C and then convert those values to K.

**Example #5:**

The product of the pressure and volume of a gas, divided by the temperature, is a constant. This is represented by the formula:

PV/T = k

(x) If the pressure and volume of a gas both increase, will the temperature increase or decrease? Explain your answer.

(y) If the pressure is doubled and the volume is tripled, by what factor must the temperature increase or decrease? Show your work.

(z) If the pressure of the gas is decreased by removing some of the gas, is it possible to use the above formula to predict the change in volume and temperature? Why or why not?

**Solution:**

1) I would like to explain how PV/T = k comes about:

(a) write Boyle's Law (use k_{1}for the constant):PV = k_{1}(b) multiply by Charles' Law (use k

_{2}for the constant):PV^{2}/ T = k_{1}k_{2}(c) multiply by Gay-Lussac's Law (use k

_{3}for the constant):P^{2}V^{2}/ T^{2}= k_{1}k_{2}k_{3}(d) take the square root of both sides:

PV/T = kwhere k is the square root of k

_{1}k_{2}k_{3}

2) Answering (x):

(a) we know that PV/T = k(b) therefore for two different sets of conditions, we can write

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

P_{2}V_{2}/ T_{2}= k(c) since k = k, we can write the combined gas law:

P_{1}V_{1}/ T_{1}= P_{2}V_{2}/ T_{2}(d) isolate T

_{2}:T_{2}= T_{1}x (P_{2}/ P_{1}) x (V_{2}/ V_{1})(e) the rationale for answering that the temperature increases:

if P_{2}> P_{1}and V_{2}> V_{1}, then T_{2}must be > T_{1}

3) Answering (y):

(a) start here:T_{2}= T_{1}x (P_{2}/ P_{1}) x (V_{2}/ V_{1})(b) given P

_{2}= 2P_{1}and V_{2}= 3V_{1}T_{2}= T_{1}x (2P_{1}/ P_{1}) x (3V_{1}/ V_{1})T

_{2}= T_{1}x 2 x 3T

_{2}= 6T_{1}

4) Answering (z): The answer is no. Here's the rationale:

(a) start with the ideal gas law:PV = nRT(b) and rearrange

PV / T = nR(c) we get the original equation

PV/T = kONLY if nR is a constant

we know that R is a constant

so for nR to be a constant, n must be a constant also.

removing some gas makes n change, so that PV/T = k won't work