Combined Gas Law Example With Answer. A series of free High School Chemistry Video Lessons. In this lesson, we will learn and apply Boyle's Law Charles' Law Gay-Lussac's Law Combined Gas Law Ideal Gas Law The following table gives the Gas Law Formulas. Scroll down the page for
more examples and solutions on how to use the Boyle's Law, Charles'Law, Gay-Lussac's Law, Combined Gas Law and Ideal Gas Law. Boyle's Law Boyle's Law states that volume of a given amount of gas held at a constant temperature varies inversely the with pressure. The relationship between
pressure and volume of Boyle's Law is expressed in mathematical terms as P1V1= P2V2. An introduction to the relationship between pressure and volume, and an explanation of how to solve gas problems with Boyle's Law Example: At 1.70 atm, a sample of gas takes up 4.25L. If the pressure in
the gas is increased to 2.40 atm, what will the new volume be? Show Step-by-step Solutions Understanding and applying Boyle's Law Example: A sample of Ne gas occupies 0.220L at 0.86 atm. What will be its volume at 29.4kPa? Show Step-by-step Solutions Charles' Law states that the volume
of a given mass of a gas is directly proportional to its Kelvin temperature at constant pressure. In mathematical terms, the relationship between temperature and volume is expressed as V1/T1=V2/T2. What is the relationship between volume and temperature of a gas and how to solve problems
using Charles' Law? Example: A balloon takes up 625L at 0°C. If it is heated to 80°C, what will its new volume be? Show Step-by-step Solutions Understanding and applying Charles' Law Example: A gas at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the
new volume be if the pressure is constant? Show Step-by-step Solutions Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant. Gay-Lussac's Law is expressed in a formula form as P1/T1 = P2/T2. When
dealing with Gay-Lussac's Law, the unit of the temperature should always be in Kelvin. Using Gay-Lussac's Law to understand the relationship between a gas' pressure and temperature Example: If the pressure in a car tire is 1.88 atm at 25°C, what will be the pressure if the temperature warms to
37°C? Show Step-by-step Solutions How to solve word problems that show how to use Gay-Lussac's Law? Examples: 1. The pressure in a sealed can of gas is 235kPA when it sits at room temperature (20°). If the can is warmed to 48°C, what will the new pressure inside the can be? 2. A car tire
has a pressure of 2.38 atm at 15.2°C. If the pressure inside reached 4.08 atm, the tire will explode. How hot would the tire have to get for this to happen? Report the temperature in degrees Celsius. Show Step-by-step Solutions Practice Problem to show how to use Gay-Lussac's Law Example: In
the morning, a paintball pressure tank is at 306 atm. The weather heats up over the course of the day, and by 3 PM, the outside temperature is roasting at 38.5°C, and the pressure inside the tank is 324 atm. What was the temperature (in degree Celsius) in the morning? Show Step-by-step
Solutions The Combined Gas Law combines Charles' Law, Boyle's Law and Gay Lussac's Law. The Combined Gas Law states that a gas' (pressure × volume)/temperature = constant. The combined law for gases. Example: A gas at 110kPa at 30.0°C fills a flexible container with an initial volume
of 2.00L. If the temperature is raised to 80,0°C and the pressure increases to 440Kpa, what is the new volume? Show Step-by-step Solutions How to solve problems with the Combined Gas Equation? Example: A 40.0L balloon is filled with air at sea level (1.00 atm, 25.0°C). It is tied to a riavk and
thrown in a cold body of water, and it sinks to the point where the temperature is 4.0°C and the pressure is 11.0 atm. What will its new volume be? Show Step-by-step Solutions The Ideal Gas Law mathematically relates the pressure, volume, amount and temperature of a gas with the equation:
pressure × volume = moles × ideal gas constant × temperature; PV = nRT. The Ideal Gas Law is ideal because it ignores interactions between the gas particles in order to simplify the equation. There is also a Real Gas Law which is much more complicated and produces a result which, under most
circumstances, is almost identical to that predicted by the Ideal Gas Law. Understanding and applying the ideal gas law Example: What is the pressure in atm of a 0.108 mol sample of the gas at a temperature of 20.0°C if its volume is 0.505L? Show Step-by-step Solutions Sample problems for
using the Ideal Gas Law, PV = nRT Examples: 1) 2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is the volume of the gas? 2) At a certain temperature, 3.24 moles of CO2 gas at 2.15 atm take up a colume of 35.28L. What is this temperature (in
Celsius)? Show Step-by-step Solutions Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to
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[?] Subscribe To This Site The combined gas law makes use of the relationships shared by pressure, volume, and temperature: the variables found in other gas laws, such as Boyle's law, Charles' law and Gay-Lussac's law. Let's review the basic principles of these three laws. Imagine you are a
diver, and you begin your dive with lungs full of air. As you go deeper under water, the pressure you experience in your lungs increases. When this happens, the air inside your lungs gets squished, so the volume decreases. This is an example of Boyle's law in action, which states that the higher
the pressure (P), the lower the volume (V), as shown in this image. Here, k is any constant number. Have you ever tried putting a balloon in the refrigerator and notice that it shrinks? As the temperature of the refrigerated balloon decreases, the volume of the gas inside the balloon also decreases.
When you take the balloon out of the refrigerator, it reverts to its original size, so the opposite is also true; when the temperature increases, the volume also increases. The shrinking balloon serves as a demonstration of Charles' law, which states that the higher the temperature (T), the higher the
volume (V). Imagine yourself driving down a road, which can cause the temperature to increase within your tires. As a result, the air inside the tires expands, and the pressure increases. This is an example of Gay-Lussac's law, which shows the relationship between pressure (P) and temperature
(T) when the volume remains constant; as the temperature increases, the pressure also increases. When we put Boyle's law, Charles' law, and Gay-Lussac's law together, we come up with the combined gas law, which shows that: Pressure is inversely proportional to volume, or higher volume
equals lower pressure. Pressure is directly proportional to temperature, or higher temperature equals higher pressure. Volume is directly proportional to temperature, or higher temperature equals higher volume. The FormulaLet's take a look at the formula for the combined gas law. Here, PV / T =
k shows how pressure, volume and temperature relate to each other, where k is a constant number. The formula for the combined gas law can be adjusted to compare two sets of conditions in one substance. In the equation, the figures for pressure (P), volume (V), and temperature (T) with
subscripts of one represent the initial condition, and those with the subscripts of two represent the final condition. P1V1 / T1 = P2V2 / T2 It is important to note that the temperature should always be in Kelvin, so if the given units are in Celsius, then those should be converted to Kelvin by adding
273. We will demonstrate how this is done in the next section. ExamplesHow do we use the combined gas law? Let's go over a few sample problems. Example One: 450 mL of a gas occupies a container that has a temperature of 28°C and a pressure of 788 mmHg. What is the temperature if the
volume is reduced to 50 mL at 760 mmHg? As the unit of measurement is in Celsius, remember to convert it to Kelvin. Examples and Problems only Probs 1-15 KMT & Gas Laws Menu 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 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: 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 mmHgP2 = 760.0 mmHg V1 = 2.00 LV2 = x T1 = 298 KT2 = 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: 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 = P1P2 = P1/2 V1 = 8.40 L V2 = x T1 = T1T2 = 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 = 2P2 = 1 V1 = 8.40 L V2 = x T1 = 1T2 = 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 kPaP2 = 101.325 kPa V1 = 1.85 LV2 = x T1 = 295 KT2 = 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 2)
Answering (x): (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 3) Answering (y): (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 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 = 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 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: 4) To obtain V2 by itself, divide both sides of the cross-multiplied equation in step 2 by 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 mmHgP2 = 760.0 mmHg V1 = 35.4 mLV2 = x T1 = 297.0 KT2 = 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 atmP2 = 3.00 atm V1 = 1.00 LV2 = x T1 = 1.00 KT2 = 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 x 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 times (2T)/(3P) Vnew = (2/3) times (nRT/P) by comparing Vnew with Vinitial, you can see that Vnew = 2/3 times Vinitial you know Vinitial is 1L, so your Vnew has to be 2/3 L Examples and Problems only Probs 1-15 KMT & Gas Laws Menu The combined gas law combines the three
gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law.It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant. When Avogadro's law is added to the combined gas law, the ideal gas law results. Unlike the named gas laws, the
combined gas law doesn't have an official discoverer. 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. The combined gas law is the combination of Boyle's law, Charles' law and Gay-Lussac's law and shows the relationship shared by pressure, temperature and volume. By combining the formulas, the ... Combined Gas Law The Combined Gas Law combines
Charles' Law, Boyle's Law and Gay Lussac's Law. The Combined Gas Law states that a gas' (pressure × volume)/temperature = constant. The combined law for gases. Example: A gas at 110kPa at 30.0°C fills a flexible container with an initial volume of 2.00L. 8/15/2019 · The combined gas law
has practical applications in situations where pressure, volume, or temperature can change. It is used in engineering, thermodynamics, fluid mechanics, and meteorology. For example, it can be used to predict cloud formation and the behavior of refrigerants in air conditioners and refrigerators.
combined gas law example problems with answers
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