Lesson 2: Gas Laws

Part c: Pressure and Volume

The Experiment

The previous two pages of Lesson 2 presented the pressure-temperature relationship and the volume-temperature relationship. On this page, we will look at the pressure-volume relationship and, like the previous page, we will begin with an experiment.
 
This experiment involves an air-filled syringe with an attached pressure sensor. The syringe is calibrated to measure the volume of air inside. The volume is read from the scale and the sensor is used for pressure readings. The plunger of the syringe can be depressed and held steady with volumes ranging between 8.0 mL and 20.0 mL. Pressure readings are obtained for each volume setting. The air temperature is constant. Because the syringe is sealed at each end, the number of moles of gas is also a constant. Sample data and a graph for this experiment is shown below.

 The collected data points fit a curved line. This is the pattern we might expect for two quantities that are inversely proportional to one another. One way to confirm an inversely proportional relationship is to plot Pressure vs. 1/Volume. If the plot is linear with a y-intercept of 0, then it is safe to assert that the pressure is inversely proportional to the volume. Such a plot is shown below. The red line is the best fit line for the collected data points The grey dashed line is an extension of the best fit line. The extension line passes through the origin. The y-intercept is 0.
 

 
The graph confirms our hypothesis that the pressure and the volume are inversely proportional to on another. Any two quantities that are inversely proportional are described by an equation of the form x•y = k where k is a constant. For the collected data in our experiment, we might assert that …

Pressure (atm) • Volume (L) = k

 The k value for our data is approximately 20.0 L•atm.
 

Boyle’s Law

This experiment leads us to another gas law. It is referred to as Boyle’s Law, in honor of Irish chemist Robert Boyle who is credited with its discovery. The law can be stated as follows:

P•V = k

where P is the pressure, V is the volume, and k is a proportionality constant.

The idea of an inversely proportional relationship indicates that the factor by which the pressure is changed is the reciprocal of the factor by which the volume changes. The sample data at the right illustrate an inversely proportional relationship. A doubling of the volume causes a halving of the pressure. That is, if the volume increases by a factor of two, the pressure decreases by a factor of two. We see this pattern in Rows 1 and 2. The volume is doubled from 1.0 L to 2.0 L and the pressure is halved as a result, from 24.0 atm to 12.0 atm. Similarly, a tripling of the volume causes the pressure to be one-third of its original value. This is observed in Rows 1 and 3. The volume is tripled from 1.0 L to 3.0 L and the volume changes to 8.0 atm, which is one-third of the row 1 value of 24.0 atm. Finally, a quadrupling of the volume causes the pressure to change to one-fourth of its original value. We see this pattern in Rows 1 and 4. The volume is quadrupled from 1.0 L to 4.0 L while the pressure changes to 6.0 atm, which is one-fourth of the original pressure of 24.0 atm.

 
The data table also show that the product of volume and pressure is a constant value. For any given row of the table, the product is 24.0 L•atm. We could describe this data set by the equation

P•V = 24.0 L•atm.

The State Equation

There has been a steady emphasis throughout the chapter on four state variables that describe any gas sample. They are pressure, temperature, volume, and the number of moles. The values of these variables will often change. In the experiment discussed above, the volume and pressure change in an inversely proportional manner while the temperature and the number of moles were held constant. It can be said that the gas sample experienced a change of state. The gas sample changed from one state with a set of P-T-V-n conditions to another state with a different set of a set of P-T-V-n conditions.
 
The equation P•V = k describes the inversely proportional between the two quantities. But it is often more useful to express the relationship as a state equation. A state equation relates the variable values at one state to the variable values for another state after a change has occurred. Through a collection of sequential and logical steps, the P•V = k equation can be converted to a state equation. This is shown below:
 

 
In the last step, a state equation is generated that relates the volume and pressure in one state (V₁ and P₁) to the volume and pressure of a second state (V₂ and P₂). This is one of the forms in which Boyle’s law is commonly stated.

 
This state equation is often used in the solution of gas law problems. Here are two examples with accompanying solutions. Additional examples are provided in the Check Your Understanding section.
 

Example 1

Perry O'Dictable, a close friend of Irishman Robert Boyle, measures the volume of air inside a canister to be 3.54 L when 1.38 atm of pressure is applied. If the pressure is increased to 3.99 atm, the new volume will be _____ L.
 

Solution:

Example 2

A 55.0-L oxygen tank is filled with oxygen gas at a pressure of 7.14 atm. The tank is connected by some tubing and valves to a smaller empty tank having a volume of 35.0 L. The valves are opened, allowing some oxygen to expand into the second tank. Once the expansion has ceased, what is the final pressure? Assume a constant temperature.
 

Solution:

Conclusion

We now have covered three gas laws in Lesson 2 - one for pressure and temperature, one for volume and temperature, and now one for pressure and volume. Next up is the volume and the number of moles. But before you move onward, please check out our learning suggestions below.
 
 
 
 
 

Before You Leave

• Download our Study Card on Pressure and Volume. Save it to a safe location and use it as a review tool. (Coming Soon)
• Try our Pressure and Volume Concept Builder with great conceptual practice on the inversely proportional relationship between these two quantities.
• Problem Set GL5 in our Calculator Pad section provides six awesome practice problems. Visit Pressure and Volume.
• The Check Your Understanding section below include questions with answers and explanations. It provides a great chance to self-assess your understanding.

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.
 
1. Consider the table at the right in answering the questions.

1. Identify at least two sets of two rows that illustrate that a doubling of the volume will halve the pressure. 
   
   Check Answer
   Answer: Rows 1 and 2 and Rows 2 and 4 and Rows 3 and 5
   
   In each pair of rows, the volume of the second listed row is twice the volume of the first listed row. And the pressure of the second row of the pair is one-half the pressure of the first row of the pair.
   
    
   
    
2. Complete the sentence: Rows 1 and 4 illustrate that increasing the volume by a factor of ______ will cause the pressure to __________ (increase, decrease) by a factor of _____. 
   
   Check Answer
   Answer:
   Rows 1 and 4 illustrate that increasing the volume by a factor of 4 will cause the pressure to decrease by a factor of 4.
   
    
   
    
3. Predict the value of the pressure when the volume is 16.0 L. 
   
   Check Answer
   Answer: 4.5 atm
   
   The volume is 8.0 L in row 4; the corresponding pressure is 9.0 atm. A volume of 16.0 L is twice as big as the row 4 volume. So the coresponding pressure will be one-half of the row 4 pressure. One-half of 9.0 atm is 4.5 atm.
   
    
   
    
4. Predict the value of the pressure when the volume is 24.0 L. 
   
   Check Answer
   Answer: 3.0 atm
   
   The volume is 12.0 L in row 5; the corresponding pressure is 6.0 atm. A volume of 24.0 L is twice as big as the row 5 volume. So the coresponding pressure will be one-half of the row 5 pressure. One-half of 6.0 atm is 3.0 atm.
   
    
   
    
5. Predict the value of the pressure when the volume is 20.0 L. 
   
   Check Answer
   Answer: 3.6 atm
   
   The volume is 2.0 L in row 1; the corresponding pressure is 36.0 atm. A volume of 20.0 L is ten times as big as the row 1 volume. So the coresponding pressure will be one-tenth of the row 1 pressure. One-tenth of 36.0 atm is 3.6 atm.
   
   Alternatively, the product of pressure•volume is 72 L•atm in every row. So with P•V = 72 L•atm, we know that P = 72/V. Substitute 20.0 L for V and solve for P.
   
    
   
    

1. Increase the volume of the gas.
2. Decrease the volume of the gas.