Ideal Gas Law Practice Worksheet Answers Explained
Table of Contents :
The Ideal Gas Law is a fundamental principle in chemistry and physics that describes the behavior of ideal gases. It combines several gas laws and establishes a relationship between pressure, volume, temperature, and the number of moles of a gas. Understanding and applying the Ideal Gas Law can be crucial for students and professionals in scientific fields. In this article, we will explore the Ideal Gas Law, provide a practice worksheet, and explain the answers to enhance comprehension.
Understanding the Ideal Gas Law
The Ideal Gas Law is expressed with the formula:
[ PV = nRT ]
Where:
- ( P ) = Pressure of the gas (in atmospheres, mmHg, or pascals)
- ( V ) = Volume of the gas (in liters)
- ( n ) = Number of moles of the gas
- ( R ) = Ideal gas constant (0.0821 L·atm/(K·mol) or 8.314 J/(K·mol))
- ( T ) = Temperature of the gas (in Kelvin)
Key Components of the Ideal Gas Law
-
Pressure (P): It is the force exerted by gas molecules colliding with the walls of their container. It can be measured in different units, including atm (atmospheres), torr, or pascals (Pa).
-
Volume (V): The amount of space that the gas occupies, typically measured in liters (L).
-
Temperature (T): The measure of the average kinetic energy of gas molecules, which must always be expressed in Kelvin for gas law calculations.
-
Moles (n): The quantity of substance measured in moles, which relates to the number of gas particles.
The Ideal Gas Constant (R)
The value of R is critical in calculations, and it can vary based on the units being used. Here is a quick reference:
<table> <tr> <th>Units</th> <th>Ideal Gas Constant (R)</th> </tr> <tr> <td>Liters, atm, K, mol</td> <td>0.0821 L·atm/(K·mol)</td> </tr> <tr> <td>Joules, K, mol</td> <td>8.314 J/(K·mol)</td> </tr> </table>
Importance of the Ideal Gas Law
The Ideal Gas Law is pivotal in predicting how gases behave under various conditions. It can be applied in numerous scenarios, from calculating the volume of gas needed in a chemical reaction to determining how changing temperature and pressure affects gas volume.
Practice Worksheet
Now that we have an understanding of the Ideal Gas Law, let's create a practice worksheet that includes several problems for students to solve.
Worksheet Problems
-
A gas occupies a volume of 5.00 L at a pressure of 2.00 atm and a temperature of 300 K. Calculate the number of moles of the gas.
-
If 1 mole of an ideal gas occupies 22.4 L at standard temperature and pressure (STP: 0°C or 273 K, 1 atm), what is the volume of 2.5 moles of the gas at STP?
-
A gas has a volume of 10.0 L and contains 0.50 moles at a temperature of 250 K. What is the pressure of the gas?
-
If the temperature of a gas is raised from 300 K to 600 K while the volume remains constant, how does the pressure change if the initial pressure was 1.00 atm?
-
A gas in a container has a volume of 15.0 L and a pressure of 3.00 atm at 350 K. How many moles of gas are present in the container?
Worksheet Answers Explained
Now let's go through the solutions to the practice worksheet step by step.
Problem 1 Solution
Given:
- ( P = 2.00 , \text{atm} )
- ( V = 5.00 , \text{L} )
- ( T = 300 , \text{K} )
Solution: Using the Ideal Gas Law formula:
[ n = \frac{PV}{RT} ]
Plugging in the values:
[ n = \frac{(2.00 , \text{atm})(5.00 , \text{L})}{(0.0821 , \text{L·atm/(K·mol)})(300 , \text{K})} ]
[ n = \frac{10.00}{24.63} \approx 0.406 , \text{moles} ]
Problem 2 Solution
Given:
- ( n = 2.5 , \text{moles} )
- ( T = 273 , \text{K} )
- ( P = 1 , \text{atm} )
Solution: Using the Ideal Gas Law:
[ V = \frac{nRT}{P} ]
Plugging in the values:
[ V = \frac{(2.5)(0.0821)(273)}{1} ]
[ V \approx 56.02 , \text{L} ]
Problem 3 Solution
Given:
- ( V = 10.0 , \text{L} )
- ( n = 0.50 , \text{moles} )
- ( T = 250 , \text{K} )
Solution:
[ P = \frac{nRT}{V} ]
Plugging in the values:
[ P = \frac{(0.50)(0.0821)(250)}{10.0} ]
[ P \approx 1.025 , \text{atm} ]
Problem 4 Solution
Given:
- ( P_1 = 1.00 , \text{atm} )
- ( T_1 = 300 , \text{K} )
- ( T_2 = 600 , \text{K} )
Solution: Using the relationship ( \frac{P_1}{T_1} = \frac{P_2}{T_2} ):
[ P_2 = P_1 \times \frac{T_2}{T_1} ]
[ P_2 = 1.00 , \text{atm} \times \frac{600}{300} = 2.00 , \text{atm} ]
Problem 5 Solution
Given:
- ( V = 15.0 , \text{L} )
- ( P = 3.00 , \text{atm} )
- ( T = 350 , \text{K} )
Solution:
[ n = \frac{PV}{RT} ]
Plugging in the values:
[ n = \frac{(3.00)(15.0)}{(0.0821)(350)} ]
[ n \approx 1.66 , \text{moles} ]
Understanding the Ideal Gas Law through practice is essential for mastering gas behavior and conducting various scientific calculations. Students are encouraged to solve similar problems and seek further clarification on concepts to strengthen their grasp of this crucial scientific principle.