Gay lussac worksheet

Gay-Lussac's Law
Problems #1 - 10

Ten Examples

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Problem #1: A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 °C?

Solution:

P1  P2
––– = –––
T1  T2

3.00   x
––– = –––
293   323

Solution technique: cross-multiply and divide.

x = 3.31 atm (to three sig figs)

Note: you will see set ups (especially in gas laws) that simply omit all the units in the answer. If you do that on a homework difficulty or test, you may get a deduction. It's not laziness on the part of the person writing the solution, it's simply assuming the reader knows what the units are and how they cancel out to depart the final unit.

Many times, you (as the student) are not allowed that luxury.


Problem #2: Determine the pressure alter when a constant volume of gas at 1.00 atm is heated

NAME_________________________________ CLASS___________________ DATE____________ 

1. Which variable is plotted on the graph's vertical axis? __________________

2. Which variable is plotted on the graph's horizontal axis? ________________

3. What is the pressure in kilopascals? _________________

4. The red plunger is used to exert pressure on the gas molecules in which colored area? _____________.

5. Finalize the table below as you view the animated gas lab.

6. What complete you predict the temperature will be when the volume is 5.5? ______________________

7. Sketch the completed temperature-volume graph to the right:

 

 

8. Click on "Effect of changing temperature on volume." Describe what is used to increase the temperature._________________

9. Sketch the completed volume-temperature graph to the right:

 

 

10. Compose the formula equation for Charles and Gay-Lussac's Law. ___________________________

11. Write the equation for Charles and Gay-Lussac's Law in words. _____________________________

12. In the animated gas lab, the unit of temperature is _________________.

13. What are the units of volume used in th

Gay-Lussac's Law

Gay-Lussac's Law, also known as the pressure-temperature law, is one of the gas laws discovered by the French chemist Joseph Louis Gay-Lussac in 1802. This law describes the relationship between the pressure and temperature of a gas at constant volume. The law can be stated as:

Pf / Tf = Pi / Ti

Where Pf and Tf are the final pressure and temperature of the gas, and Pi and Ti are the initial pressure and temperature of the gas, respectively.

Explanation of Gay-Lussac's Law

Gay-Lussac's Law states that the pressure of a given amount of gas held at unchanging volume is directly proportional to the Kelvintemperature. This means that as the temperature of a gas increases, so does its pressure, and vice versa. Mathematically, this can be expressed as P / T = k, where P is the pressure, T is the temperature in Kelvin, and k is a constant.

Application of Gay-Lussac's Law

This rule is particularly important in comprehending the behavior of gases in various real-world scenarios. For instance, it helps in predicting how changes in temperature will influence the pressure of a gas in a sealed container, such as a gas cylinder or a balloon. It also forms th

Gay-Lussac's Law Worksheet: Train Problems

Name: ____________________ Lussac’s Law Worksheet 1. Write the equation for Gay Lussac’s Law. Define the symbols used. 2. What gas commandment variable is steady in Gay-Lussac’s Law? 3. A rigid container has an initial pressure of 1.50 atm at 294 K. What will the pressure be if the temperature is increased to 394 K? 4. The pressure inside a container is 1.10 atm at a temperature of 330 K. What would the pressure be at 348 K? 5. A rigid container is at a temperature of 65 oC. When heated to 224 oC, the pressure was 2.20 atm. What was the initial pressure? 6. If a gas in a closed container is pressurized from 1520 kPa to 1621 kPa and its original temperature was 298 K, what would the final temperature of the gas be? 7. At -23 oC a gas had a pressure of 0.853 atm. At what temperature would it be at 1.32 atm? 8. The pressure in an automobile tire is 2.00 atm at 300 K. At the end of a road trip the pressure has risen to 2.20 atm. What is the temperature of the air in the tire, assuming the volume has not changed? 9. A balloon is filled with helium gas to a pressure of 1140 mm Hg when the temperature is 295 K. If the temperatu