Ten Examples

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 _{2}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. I put a four-variable problem as #11 in the Probs 1-10 file. (That 11 is not a typo.)

A different way to "derive" the most common three-equation combined gas law is discussed in example #5 below. In it, I use three laws: Boyle, Charles and Gay-Lussac.

Please follow this link, for getting the same three-equation combined gas law from just Boyle's and Charles' Laws.

**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 _{2}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 example 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

**Example #6:** At constant temperature, if the gas amount in the sample is doubled while the pressure is halved, what will happen to the volume of the gas sample?

**Solution:**

1) You can determine this by assigning values to use in a combined gas law problem. I'll start from the less common form that has all 4 variables.

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

2) Since the T is constant, let us drop it:

P_{1}V_{1}/ n_{1}= P_{2}V_{2}/ n_{2}<--- another seldom seen form of the combined gas law (one with three variables)

3) The amount of the gas is doubled:

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

4) The pressure is halved:

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

5) I will assign a volume of 1 to V_{1} and see what V_{2} will come to be:

(2 * 1) / 1 = (1 * V_{2}) / 2V

_{2}= 4The volume of the gas sample increases by a factor of 4.

5) Speaking of seldom seen combined gas law formulations, please go here for another example. Scroll down to the Bonus Problem at the end of the file. I decided to start from the Ideal Gas Law in my solution to that problem and I wind up with this:

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

**Example #7:** Using the Combined Gas Law, write each of the six symbolic values in terms of the other five symbolic values.

**Solution:**

1) Here is the combined gas law most likely assumed by the question writer:

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

2) Cross multiply:

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

3) To obtain P_{1} by itself, divide both sides by V_{1}T_{2}:

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

4) To obtain V_{2} by itself, divide both sides of the cross-multiplied equation in step 2 by P_{2}T_{1}:

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

5) The other four are left to the reader. Indeed, you may want to try your hand at the four-variable form of the combined gas law. Here's a bit of the start:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– n _{1}T_{1}n _{2}T_{2}cross multiply:

P

_{1}V_{1}n_{2}T_{2}= P_{2}V_{2}n_{1}T_{1}You may proceed from there.

**Example #8:** 35.4 mL of hydrogen gas is collected over water at 24.0 °C and a total pressure of 745.0 mmHg. What is the volume of the gas at STP?

**Solution:**

I decided to not use the word dry in front of gas in the last sentence. Often, a teacher or question writer will assume that dry gas is the item desired in this type of problem. That's because the water vapor is just in the way of doing more calculations focused on the hydrogen. The assumption is made that the reader (you!) simply understands this and that there is no need to spell out that dry gas is the desired quantity.

I did decide to use the phrase ""a total pressure of." Sometimes, it is not made explicit that the pressure given is a total pressure and is composed of two gases. Once again, the question writer is assuming you know this by context and from experience.

1) Use Dalton's Law to remove the pressure of the water vapor:

From the reference source, we determine that the vapor pressure of water at 24.0 °C is 2.985 kPa.Let us convert to mmHg:

(2.985 kPa) (760.0 mmHg / 101.325 kPa) = 22.39 mmHgNow, use Dalton's Law:

P_{tot}= P_{H2}+ P_{H2O}745.0 = x + 22.39

x = 722.61 mmHg

2) Set up the data for the problem:

P _{1}= 722.61 mmHgP _{2}= 760.0 mmHgV _{1}= 35.4 mLV _{2}= xT _{1}= 297.0 KT _{2}= 273.0 K

3) Use the combined gas law:

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

(722.61 mmHg) (35.4 mL) (760.0 mmHg) (x) ––––––––––––––––––––– = ––––––––––––––– 297.0 K 273.0 K x = 30.9 mL (to three sig figs)

**Example #9:** A tire has 25 air particles and a volume of 205 mL with a pressure of 0.950 atm. If 10 air particles are added the tire, the volume is 215 mL. What is the new tire pressure?

**Solution:**

Before starting the solution, you have to recognize that the word 'moles' can be substituted for the word 'particles.' In other words, there is a 25 to 10 ratio of particles. If expressed in moles, the ratio is still 25 to 10.

1) Let's start with the four-variable form of the combined gas law:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– n _{1}T_{1}n _{2}T_{2}

2) Since temperature is never mentioned, we assume it is constant. So, T_{1} = T_{2}, which means T will drop out. This results in an unusual formulation of the combined gas law.

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

5) Substituting values:

25 and 35 (from 25 + 10) are our moles.[(0.950 atm) (205 mL)] / 25 = [(x) (215 mL)] / 35

x = 1.27 atm

**Example #10:** At constant temp, if the amount of gas in the sample is doubled while the pressure is halved, what will happen to the volume of the gas sample?

**Solution:**

1) The answer can determined by assigning values to use in a combined gas law problem. I'll start with the less common form that has all 4 variables:

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

2) Since the T is constant, let us drop it:

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

3) The amount of the gas is doubled:

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

4) The pressure is halved:

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

5) I will assign a volume of 1 to V_{1} and see what V_{2} will come to be:

(2 * 1) / 1 = (1 * V_{2}) / 2V

_{2}= 4The volume of the gas sample increases by a factor of 4.

**Bonus Example:** When the pressure exerted on 1.00 L of an ideal gas is tripled, and the absolute temperature is doubled, the volume becomes what value?

**Solution #1:**

1) Use the combined gas law:

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

2) Assign values as follows:

P _{1}= 1.00 atmP _{2}= 3.00 atmV _{1}= 1.00 LV _{2}= xT _{1}= 1.00 KT _{2}= 2.00 K

3) Insert values into the equation and solve for x:

[(1.00 atm) (1.00 L)] / 1.00 K = [(3.00 atm ) (x)] / 2.00 Kx = 2/3 L = 0.667 L

**Solution #2:**

Tripling the pressure on a gas will divide its volume by 3 (Boyle's Law). Therefore, after the increase in pressure, the volume will be 1/3 L.However, doubling the absolute temperature of a gas will also double its volume (Charles' Law). Multiply the previous answer by 2: 1/3 L x 2 = 2/3 L

**Solution #3:**

use PV = nRTlet V

_{initial}be the initial volume of your gasso by rearranging the equation you get V

_{initial}= nRT/Pthe question says that later the pressure is tripled and the temperature is doubled, so now you have

V

_{new}= nR times (2T)/(3P)V

_{new}= (2/3) times (nRT/P)by comparing V

_{new}with V_{initial}, you can see thatV

_{new}= 2/3 times V_{initial}you know V

_{initial}is 1L, so your V_{new}has to be 2/3 L