### Combined Gas LawTen Examples

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

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

P1V1 = P2V2

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

(P1V1) (V1 / T1) = (P2V2) (V2 / T2)

P1V12 / T1 = P2V22 / T2

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

(P1V12 / T1) (P1 / T1) = (P2V22 / T2) (P2 / T2)

P12V12 / T12 = P22V22 / T22

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

P1V1 / T1 = P2V2 / T2

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

 P1V1 P2V2 ––––– = ––––– T1 T2

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:

P1V1 / n1T1 = P2V2 / n2T2

Or, like this:

 P1V1 P2V2 ––––– = ––––– n1T1 n2T2

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. Comment: I have seen some online material that refers to the Combined Gas Law as the General Gas Law. I think it is unwise to attempt to rename the Combined Gas Law, but I can't stop the attempt from being made. You, as the student, are going to have to be aware of this and, as a consequence, learn both terms.

Example #0: What gas law would you use if the pressure, volume, and moles changed while the temperature remained constant?

Solution: You would use an unusual form of the combined gas law.

1) Write the full Combined Gas Law:

 P1V1 P2V2 ––––– = ––––– n1T1 n2T2

2) The temperature, being constant, drops out:

 P1V1 P2V2 ––––– = ––––– n1 n2

This is the form of the Combined Gas Law that would be used.

3) Other forms of this question can easily be constructed: What gas law would you use if the volume, moles, and temperature changed while the pressure remained constant? The answer: another form of the Combined Gas Law not commonly encountered.

 V1 V2 ––––– = ––––– n1T1 n2T2

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:

 P1 = P2 = V1 = V2 = T1 = T2 =

and fill it in with data from the problem.

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

 P1 = P2 = 760.0 mmHg V1 = V2 = x T1 = T2 = 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:

 P1 = 745.0 mmHg P2 = 760.0 mmHg V1 = 2.00 L V2 = x T1 = 298 K T2 = 273 K

5) Write the combined gas law equation:

 P1V1 P2V2 ––––– = ––––– T1 T2

6) Solve for V2 by first cross-multiplying:

P1V1T2 = P2V2T1

7) Then dividing both sides by P2T1:

 P1V1T2 V2 = ––––– P2T1

or:

V2 = (P1V1T2) / (P2T1)

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:

 P1 = P1 P2 = P1/2 V1 = 8.40 L V2 = x T1 = T1 T2 = 2T1

Notice how P2 is represented as being half of P1. Notice how T2 is represented as being twice that of T1.

2) Write, then rearrange the Combined Gas Law:

P1V1 / T1 = P2V2 / T2

V2 = P1V1T2 / T1P2

3) Substitute into the rearranged gas law:

V2 = [(P1)(8.40 L)(2T1)] / [(T1) (P1/2) ]

V2 = 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:

 P1 = 2 P2 = 1 V1 = 8.40 L V2 = x T1 = 1 T2 = 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:

V2 = [(2)(8.40 L)(2)] / [(1) (1) ]

V2 = 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:

Pgas + PH2O = Ptotal

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 Ptotal is the 98.0 value. Ptotal 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:

Pgas + 2.6447 kPa = 98.0 kPa

We solve the problem for Pgas 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:

 P1 = 95.3553 kPa P2 = 101.325 kPa V1 = 1.85 L V2 = x T1 = 295 K T2 = 273 K

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

V2 = (P1V1T2) / (P2T1)

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:

P1V1 / T1 = P2V2 / T2

2) I will assign a value of 1 to V1 and allow it to double. I will assign a value of 1 to T1 and allow its value to quadruple.

[(P1)(1)] / 1 = [(P2)(2)] / 4

P1 = P2 / 2

2P1 = P2

the answer is (d) increase by a factor of 2

By the way, any volume unit is fine for V1, 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 k1 for the constant):
PV = k1

(b) multiply by Charles' Law (use k2 for the constant):

PV2 / T = k1k2

(c) multiply by Gay-Lussac's Law (use k3 for the constant):

P2V2 / T2 = k1k2k3

(d) take the square root of both sides:

PV/T = k

where k is the square root of k1k2k3

(a) we know that PV/T = k

(b) therefore for two different sets of conditions, we can write

P1V1 / T1 = k
P2V2 / T2 = k

(c) since k = k, we can write the combined gas law:

P1V1 / T1 = P2V2 / T2

(d) isolate T2:

T2 = T1 x (P2 / P1) x (V2 / V1)

(e) the rationale for answering that the temperature increases:

if P2 > P1 and V2 > V1, then T2 must be > T1

(a) start here:
T2 = T1 x (P2 / P1) x (V2 / V1)

(b) given P2 = 2P1 and V2 = 3V1

T2 = T1 x (2P1 / P1) x (3V1 / V1)

T2 = T1 x 2 x 3

T2 = 6T1

PV = nRT

(b) and rearrange

PV / T = nR

(c) we get the original equation

PV/T = k

ONLY 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.

P1V1 / n1T1 = P2V2 / n2T2

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

P1V1 / n1 = P2V2 / n2 <--- another seldom seen form of the combined gas law (one with three variables)

3) The amount of the gas is doubled:

P1V1 / 1 = P2V2 / 2

4) The pressure is halved:

2V1 / 1 = 1V2 / 2

5) I will assign a volume of 1 to V1 and see what V2 will come to be:

(2 * 1) / 1 = (1 * V2) / 2

V2 = 4

The 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:

P1 / n1T1 = P2 / n2T2

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:

 P1V1 P2V2 ––––– = ––––– T1 T2

2) Cross multiply:

P1V1T2 = P2V2T1

3) To obtain P1 by itself, divide both sides by V1T2:

 P2V2T1 P1 = ––––– V1T2

4) To obtain V2 by itself, divide both sides of the cross-multiplied equation in step 2 by P2T1:

 P1V1T2 V2 = ––––– P2T1

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:

 P1V1 P2V2 ––––– = ––––– n1T1 n2T2

cross multiply:

P1V1n2T2 = P2V2n1T1

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 mmHg

Now, use Dalton's Law:

Ptot = PH2 + PH2O

745.0 = x + 22.39

x = 722.61 mmHg

2) Set up the data for the problem:

 P1 = 722.61 mmHg P2 = 760.0 mmHg V1 = 35.4 mL V2 = x T1 = 297.0 K T2 = 273.0 K

3) Use the combined gas law:

 P1V1 P2V2 ––––– = ––––– T1 T2

 (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:

 P1V1 P2V2 ––––– = ––––– n1T1 n2T2

2) Since temperature is never mentioned, we assume it is constant. So, T1 = T2, which means T will drop out. This results in an unusual formulation of the combined gas law.

P1V1 / n1 = P2V2 / n2

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:

P1V1 / n1T1 = P2V2 / n2T2

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

P1V1 / n1 = P2V2 / n2

3) The amount of the gas is doubled:

P1V1 / 1 = P2V2 / 2

4) The pressure is halved:

(2 * V1) / 1 = (1 * V2) / 2

5) I will assign a volume of 1 to V1 and see what V2 will come to be:

(2 * 1) / 1 = (1 * V2) / 2

V2 = 4

The 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:

P1V1 / T1 = P2V2 / T2

2) Assign values as follows:

 P1 = 1.00 atm P2 = 3.00 atm V1 = 1.00 L V2 = x T1 = 1.00 K T2 = 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 K

x = 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) (2) = 2/3 L

Solution #3:

use PV = nRT

let Vinitial be the initial volume of your gas

so by rearranging the equation you get Vinitial = nRT/P

the question says that later the pressure is tripled and the temperature is doubled, so now you have

Vnew = (nR) (2T / 3P)

Vnew = (2/3) (nRT/P)

by comparing Vnew with Vinitial, you can see that

Vnew = 2/3 times Vinitial

you know Vinitial is 1 L, so your Vnew has to be 2/3 L