[Chemistry Class Notes] on Gay Lussacs Law Formula Pdf for Exam

States of the matter are one of the most asked about chapters in the student community. This is because the chapter has a lot of laws that need to be studied well. There are laws like the Charles Law, Boyle’s law, Gay Lussacs Law, etc. All of these laws are extremely important to be studied and require a good amount of focus and attention that can help them out in making sure that they are getting the most marks from this chapter. 

States of Matter is not the simplest chapter. It does require a good amount of effort and a lot of practice to master, however, the difficulty level of the questions that are asked from this chapter is not that high. The chapter needs to be studied thoroughly and once that is done, students can easily score well in this chapter.

This article is a brilliant resource that is going to focus majorly on the Gay Lussacs Law Formula. Through this article, we plan on giving you a better look at the problems that can be solved using the Gay Lussacs Law Formula and that is why we suggest that you already have a little knowledge regarding the law before reading this article. This article will mostly be aimed towards making sure that the application of the formula is taught to you carefully. 

recommends that you keep solving the examples in the article before you look at the solution so that you can learn as you go ahead. With all of these things being said, we hope that this article helps you in understanding this law better.

Gay Lussacs Law Formulas – Equation and Problem Solved with Example

The law states that at a constant volume, the pressure (in atm) of a given amount of gas is directly proportional to the temperature (in Kelvin).

Mathematically,

P∝T at constant volume,V

P/T = k where k is a constant

Which implies,

P1 / T1 = P2 / T2

Or graphically,

When the pressure of the gas is plotted against temperature at constant volume, we would obtain a straight line passing through the origin. At any point on the graph, the ratio of P: T will always be constant.

Example: 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:

From Gay Lussac’s Law we know that at constant volume (30 L in this question), 

P1 / T1 = P2 / T2

Convert given temperature in degree Celsius to Kelvin

T1 = 20 + 273

T1 = 293 K

T2 = 50 + 273

T2 = 323 K

Let, P1 = 3.00 atm and P2 = x;

Then, 

3 / 293 = x / 323

Which implies, 

x = 3.31 atm 

 

Question: A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure

Options:

(a) 310 K

(b) 380 K

(c) 400 K

(d) 340 K

Answer: (d)

Solution: 

P1 / T1 = P2 / T2

699.0 / 313 = 760 / x 

x = 340 K 

(notice how pressure is not converted to atm, it is because the conversion factor gets cancelled from both sides of the equation)

Conclusion 

Through this article, we hope that students were able to understand the applications of how Gay Lussac’s law works and how exactly is the formula applied on the numericals that will be given to you in your tests. The article included some of the most important examples that were very easy to understand and take references from to ensure that students get to the core of the concept very easily. In exams like JEE and NEET, students will have to solve these numericals based on laws like Gay Lussac’s Law. recommends that students practice a lot before taking any mock test so that they can appear confidently in the future for the final test. 

The only way for you to make the mathematical sums in chemistry easy for you is when you practice them over and over again. Repetition is the mother of all skills and truly believes in its power. We suggest that students practice as many problems as possible to fully understand the real application of Gay Lussac’s law.