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AP®︎/College Chemistry

Course: ap®︎/college chemistry   >   unit 3.

  • The ideal gas law (PV = nRT)
  • Worked example: Using the ideal gas law to calculate number of moles
  • Worked example: Using the ideal gas law to calculate a change in volume
  • Gas mixtures and partial pressures
  • Worked example: Calculating partial pressures

Ideal gas law

  • (Choice A)   P 1 6 ‍   A P 1 6 ‍  
  • (Choice B)   P 1 2 ‍   B P 1 2 ‍  
  • (Choice C)   P 1 ‍   C P 1 ‍  
  • (Choice D)   2 P 1 ‍   D 2 P 1 ‍  

Chemistry Steps

Chemistry Steps

Gay-Lussacs law

General Chemistry

The ideal gas laws show the correlation of the temperature, pressure, volume, and the amount of a gas. Although, this is not how the laws were determined, I found my students grasping the concepts a lot easier using the following model.

We are going to take a pump filled with some gas and a freely moving plunger and by changing the gas parameters determine their correlation.

  • Boyle’s Law

Boyle’s Law shows the correlation of the pressure and the volume of a gas. To illustrate it, we change the volume of the gas by pushing down the plunger. This decreases the volume, and because the gas molecules have less space, the pressure is increasing:

practice problems gas laws answers

The observation is that the volume of a fixed quantity of gas at constant temperature is inversely proportional to its pressure.

practice problems gas laws answers

This relationship can be written as:

                                    

\[{\rm{V}} \sim \,\frac{{\rm{1}}}{{\rm{P}}}\]

                                                           

To bring in the equal sign, we introduce a constant:

                                                        PV = constant                               

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

                                                                 

Income ∼ number of cars

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income   = price x number of cars

So, for the gas pressure and volume, we are not interested too much in the constant, but rather in its linkage of pressure and volume at positions 1 and 2. Because the P x V product is constant, we can write that:

P 1 V 1 = constant = P 2 V 2

practice problems gas laws answers

This is the practical implication of the Boyle’s law that is used for solving gas problems.

For example ,

The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters.

First, write down what you gave and what needs to be determined. If nothing is mentioned about any parameter, for example, the moles and the temperature in this case, it is assumed that they are constant, so you don’t need to worry about them.

P 1 = 2.30 atm V 1 = 1.80 L V 2 = 1.20 L P 2 = ?

\[{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\]

Now, rearrange to calculate P 2 :

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{\rm{2}}{\rm{.30}}\;{\rm{atm}}\;{\rm{ \times 1}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{1}}{\rm{.20}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.45}}\;{\rm{atm}}\]

  • Charle’s Law

Let’s now consider what happens if we heat up the gas leaving the plunger to move freely. The gas is going to expand, and the correlation is that the volume of a gas increases with temperature:

practice problems gas laws answers

The volume is directly proportional to the temperature of the gas:

practice problems gas laws answers

Therefore, for different states of gas, we can write the Charle’s law as:

practice problems gas laws answers

For example,

What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200. °C?

Write down what is given and what needs to be determined first:

V 1 = 3.50 L T 1 = 20 o C T 2 = 200. o C V 2 = ?

Now, before doing anything else, remember to always convert the temperature to Kelvin when solving a gas problem:

practice problems gas laws answers

So, T 1 = 20 + 273 = 293 K ,  T 2 = 200 + 273 = 473 K  

The question studies the correlation between the volume and the temperature of a gas, so we need to use the Charle’s law.

Write it down and rearrange it to get an expression of V 2 .

\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{V}}_{\rm{2}}}\; = \;\frac{{{\rm{473}}\;\cancel{{\rm{K}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}}}{{{\rm{293}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.65}}\;{\rm{L}}\]

Gay-Lussac’s law

To study the relationship between the pressure and the temperature of a gas, the barrel is held at a fixed position to prevent changing the volume, and the sample is heated up:

practice problems gas laws answers

In this case, the pressure increases with the temperature and for different states of the gas, the Gay-Lussac’s law is written as:

practice problems gas laws answers

A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

P 1 = 1.40 atm T 1 = 23.0 °C P 2 = 400.0 K T 2 = ?

Convert the temperature to Kelvin right away!

T 1 = 23.0 + 273 = 296 K

Write down the gas law and rearrange it to get an expression of P 2 .

\[\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{\rm{1}}{\rm{.40}}\;{\rm{atm}}\;{\rm{ \times }}\;{\rm{400}}{\rm{.0}}\;\cancel{{\rm{K}}}}}{{{\rm{296}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.89}}\;{\rm{atm}}\]

  • Avogadro’s Law

The Avogadro’s law of ideal gases demonstrates that the volume of a gas is directly proportional to its number of moles. In other words, the more gas, the larger the volume which again is a very intuitive observation.

For this, the plunger is again left free, and pumping some gas into the system increases the volume that the gas occupies:

practice problems gas laws answers

This is summarized in the following formula:

practice problems gas laws answers

Talking of the volume and the moles of a gas, remember that at STP, one mole of any gas occupies a volume of 22.4 L called the molar volume , V o . 

This is because in ideal gases, the size of molecules is very small compared to the intermolecular distances. So, most of the volume is almost like an empty, and therefore, it does not matter what gas it is, the volume is determined to be 22.4 L at 0 o C and 1 atm.

  • The Ideal Gas Law

The examples above are great to demonstrate the individual gas law, however, notice that in all experiments, we assumed or set up the experiment, such that two parameters are constant, and we study the correlation of the other two. For example, in the Boyle’s law, we study a constant amount of gas at a constant temperature and find that the pressure increase as the volume is decreased.

To combine all the laws together and have the four variables (n, P, V, T) in one place, the Ideal Gas Law equation is obtained:

practice problems gas laws answers

The R is called the ideal gas constant . Although it has different values and units, you will mostly be using this:

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

The ideal gas law equation is used when you need to find P, V, T, or n , for a system where they do not change .

A sample of hydrogen gas is added into a 5.80 L container at 56.0 °C. How many moles of the gas is present in the container if the pressure is 6.70 atm?

Rearrange the ideal gas law to get an expression for the moles (n):

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\]

\[{\rm{n}}\;{\rm{ = }}\;\frac{{{\rm{6}}{\rm{.70}}\;\cancel{{{\rm{atm}}}}\; \times \;{\rm{5}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ }} \times \;{\rm{329}}\;\cancel{{\rm{K}}}}}\; = \;1.44\;{\rm{mol}}\]

Remember, to change the pressure to atm when the ideal gas law equation is used! This is because the units of R contain atm when the 0.08206 value is used. And this is what most problems in this chapter use.

practice problems gas laws answers

How do I know which gas law to use?

You are probably wondering about this question now that every gas law brings a new equation. For this, there is what is called the combined gas law and as long as you remember it, you do not need to remember all the gas laws to solve a problem.

Let’s keep it for another article because there is quite a lot of information in this one.

  • Gay-Lussac’s Law
  • Celsius or Kelvin
  • Ideal-Gas Laws
  • Combined Gas Law Equation
  • How to Know Which Gas Law Equation to Use
  • Molar Mass and Density of Gases
  • Graham’s Law of Effusion and Diffusion
  • Graham’s Law of Effusion Practice Problems
  • Dalton’s Law of Partial Pressures
  • Mole Fraction and Partial Pressure of the Gas
  • Gases in Chemical Reactions
  • Gases-Practice Problems

Boyle’s Law: The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 litters.

Boyle’s Law: After changing the pressure of a gas sample from 760.0 torr to 0.800 atm, it occupies 4.30 L volume. What was the initial volume of the gas?

Charles’s Law:  What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200 °C?

Charles’s Law:  The volume of a gas decreased from 2.40 L to 830. mL and the final temperature is set at 40.0 °C. Assuming a constant pressure, calculate the initial temperature of the gas in kelvins.

Gay-Lussac’s Law:  A sample of helium gas at 1.40 atm is heated from 23.0 °C to 400.0 K. How many atmospheres is the final pressure of the helium gas?

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Gas Laws Quiz

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This online quiz is intended to give you extra practice in performing a variety of gas laws calculations involving pressure, volume and temperature, as well as Ideal Gas Law problems.

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practice problems gas laws answers

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Chemistry: Gas Laws

V 1 / T 1 = V 2 / T 2
V 1 T 2 = V 2 T 1
(2.00 L) / 294.0 K) = (1.00 L) / (x) cross multiply to get: 2x = 293 x = 147.0 K Converting 147.0 K to Celsius, we find -126.0 °C, for a total decrease of 147.0 °C, from 21.0 °C to -126.0 °C.
(600.0 mL) / (293.0) = (x) / (333.0 K) x = 682 mL
(900.0 mL) / (300.0 K) = (x) / (405.0 K) x = 1215 mL
(60.0 mL) / (306.0 K) = (x) / (278.00 K) Cross multiply to get: 306x = 16680 x = 54.5 mL The volume decreases by 5.5 mL.
In cross-multiplied form, it is this: V 1 T 2 = V 2 T 1 V 2 = (V 1 T 2 ) / T 1 1 x = [(300.0 mL) (283.0 K)] / 290.0 K
In cross-multiplied form, it is this: V 1 T 2 = V 2 T 1 V 2 = (V 1 ) [T 2 / T 1 ] x = (1.00 L) [(606.0 K) / (273.0 K)] x = 2.22 L
(6.00 L) / (300.0 K) = (x) / (423.0 K) or (6.00 L) (423.0 K) = (x) (300.0 K)
V 2 = (V 1 ) [T 2 / T 1 ] x = (400.0 mL) [(400.0 K) / (498.0 K) x = 321 mL
(400.0 mL) / (498.0 K) = (x) / (400.0 K)
(8.00 L) / (483.0 K) = (x) / (250.0 K) Note how you can have a negative Celsius temperature, but not a negative Kelvin temperature.
V 1 / T 1 = V 2 / T 2 x / 588 K = 852 mL / 725 K (x) (725 K) = (852 mL) (588 K) x = 691 mL
2.05 L / 278 K = V 2 / 294 K Calculate V 2 . The volume that "escapes" is V 2 minus 2.05 L

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Chemistry LibreTexts

7.2: The Gas Laws (Problems)

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  • Page ID 98788

PROBLEM \(\PageIndex{1}\)

Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?

As temperature of a gas increases, pressure will also increase based on the ideal gas law. The volume of the tire can only expand so much before the rubber gives and releases the build up of pressure.

PROBLEM \(\PageIndex{2}\)

Explain how the volume of the bubbles exhausted by a scuba diver change as they rise to the surface, assuming that they remain intact.

As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.

PROBLEM \(\PageIndex{3}\)

One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.”

(a) What is the meaning of the term “inversely proportional?”

(b) What are the “other things” that must be equal?

The pressure of the gas increases as the volume decreases

amount of gas, temperature

PROBLEM \(\PageIndex{4}\)

An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.”

  • What is the meaning of the term “directly proportional?”
  • What are the “other things” that must be equal?

The number of particles in the gas increases as the volume increases

temperature, pressure

PROBLEM \(\PageIndex{5}\)

A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?

3.40 × 10 3 torr

PROBLEM \(\PageIndex{6}\)

What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?

PROBLEM \(\PageIndex{7}\)

A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.

PROBLEM \(\PageIndex{8}\)

A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?

PROBLEM \(\PageIndex{9}\)

A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?

PROBLEM \(\PageIndex{10}\)

How many grams of gas are present in each of the following cases?

  • 0.100 L of CO 2 at 307 torr and 26 °C
  • 8.75 L of C 2 H 4 , at 378.3 kPa and 483 K
  • 221 mL of Ar at 0.23 torr and –54 °C

7.24 × 10 –2 g

1.5 × 10 –4 g

PROBLEM \(\PageIndex{11}\)

A high altitude balloon is filled with 1.41 × 10 4 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?

1.2741 × 10 5 L or more correctly to 3 significant figures 1.27 × 10 5 L

PROBLEM \(\PageIndex{12}\)

While resting, the average 70-kg human male consumes 14 L of pure O 2 per hour at 25 °C and 100 kPa. How many moles of O 2 are consumed by a 70 kg man while resting for 1.0 h?

PROBLEM \(\PageIndex{13}\)

A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet?

PROBLEM \(\PageIndex{14}\)

If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?

Temperature and Pressure are directly proportional. Pressure will also have to increase (doubling).

PROBLEM \(\PageIndex{15}\)

If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?

Volume and pressure are inversely proportional. The pressure decreases by a factor of 3.

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).

  • Adelaide Clark, Oregon Institute of Technology

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    1. The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters. answer This content is available to registered users only. Click here to Register!

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    PROBLEM 7.2.1.2 7.2.1. 2. Explain how the volume of the bubbles exhausted by a scuba diver change as they rise to the surface, assuming that they remain intact. Answer. PROBLEM 7.2.1.3 7.2.1. 3. One way to state Boyle's law is "All other things being equal, the pressure of a gas is inversely proportional to its volume.".

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    You might need: Calculator A particular amount of ideal gas occupies 3 L at 27 o C . Calculate the decrease in volume observed if the gas is cooled down to 17 o C . Given: The pressure remains constant. L . Show Calculator Stuck? Use a hint. Report a problem Do 4 problems

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    1. A sample of gas has a pressure of 100.0 torr and 27.0 C. Calculate the pressure if the temperature is changed to 127 C while the volume remains constant. 2. A gas initially at STP is changed to 248 K. Calculate the final pressure of the gas. 3. A gas occupies a volume of 50.0 mL at 27 C and 630 mmHg.

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    Lesson 4: Ideal gas law. The ideal gas law (PV = nRT) Worked example: Using the ideal gas law to calculate number of moles. Worked example: Using the ideal gas law to calculate a change in volume. Gas mixtures and partial pressures. Worked example: Calculating partial pressures. Ideal gas law. Science >.

  7. 1B: Gas Laws

    Ideal Gas Law. For any sample of gas under ideal conditions, the relationship between the amount of gas in moles ( n) and its temperature, pressure, and volume is given by the relationship. PV = nRT P V = n R T. in which R is the gas constant, with a value of 0.08206 L × atm/K × mol.

  8. 7.3.1: Practice Problems- Applications of the Ideal Gas Law

    Answer. PROBLEM 7.3.1.10 7.3.1. 10. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN 3 ). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide. Answer.

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    The ideal gas laws show the correlation of the temperature, pressure, volume, and the amount of a gas. Although, this is not how the laws were determined, I found my students grasping the concepts a lot easier using the following model.

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