Gay-Lussac’s law; what is it, where to use P1/T1 = P2/T2
The formula P1/T1 = P2/T2 represents Gay-Lussac’s law, aka the pressure-temperature relationship.
It is important to understand the effect of varying pressure and temperature conditions on the chemical behavior of a fixed amount of gas that occupies a constant volume.
As per the kinetic theory of gases, in an ideal gas sample, the individual gas molecules are continuously moving, in random directions, occupying maximum space.
At a fixed volume and concentration, if the temperature of the gas is increased, the kinetic energy possessed by these gas molecules will also increase which in turn leads to a greater force per unit area i.e., a higher pressure exerted by the gas on the walls of the container.
Doubling the temperature increases the pressure exerted by the gas two times and vice versa.
This is known as Gay-Lussac’s law. Mathematically represented as follows:
P α T
Converting proportionality into equality gives us:
P = k T
k = P/T
Where P = pressure, T = temperature and k = proportionality constant
If the temperature of a gaseous system is changed from T1 to T2, its pressure changes from P1 to P2 and the relationship between the two variables can be expressed as per Gay-Lussac’s law:
P1/T1 = k……Equation (i)
P2/T2 = k ….Equation (ii)
As the value of k stays unchanged so, the combined equation becomes:
P1/T1 = P2/T2
What do the components of P1/T1 = P2/T2 represent?
For a gaseous system undergoing some change in temperature and pressure at a constant volume and concentration, P1/T1 = P2/T2 represents:
P1 = Initial pressure of the gaseous system (Units = Pa, atm or mm Hg)
T1= Initial temperature (Units = K)
P2 = Final pressure of the gaseous system (Units = Pa, atm or mm Hg)
T2 = Final temperature (Units = K)
Where and how to use P1/T1 = P2/T2? – Examples
The formula P1/T1 = P2/T2 can be used to find the initial or final pressure or temperature of a gas by rearranging this equation into the following four different ways, depending upon which variable is unknown.
Another important point is that consistency in units is very important. The units of pressure (P1 and P2) should be the same on either side of P1/T1 = P2/T2. However, temperature (T1 and T2) should always be given Kelvin.
The temperature in °C can be converted into K as follows:
For example, A 150 mL chamber with a fixed amount of oxygen (O2) gas has a pressure of 172 atm and a temperature of 60°C. If the temperature is raised to 80°C, what is the new pressure of the chamber?
Solution
The following data is obtained from the question statement given above:
P1 = 172 atm
T1 = 60°C
P2 = ?
T2 = 80°C
As per the question statement, both the volume and the amount of oxygen gas are kept constant. Therefore, we can apply Gay-Lussac’s law to find P2. But before that, we need to convert both T1 and T2 into Kelvin (K).
1 K = 1°C + 273
T1 = 60 + 273 = 333 K
T2 = 80 + 273 = 353 K
Cross multiply to make P2 the subject of the formula:
172 x 353 = P2 x 333
60716 = 333 P2
P2 = 60716/333
∴ P2 = 182.3 atm
Result: The final pressure in the gas chamber is 182.3 atm.
Another example is – The final pressure of a gas inside a tank is 12 Pa at 200 K. Find the initial pressure if the initial temperature of the gas was 50 K.
Solution
As per the question statement;
P1 = ?
T1 = 50 K
P2 = 12 Pa
T2 = 200 K
Substitute the above data into the equation and make P1 the subject of the formula as shown below:
200 P1 = 50 x 12
P1 = 600/200
∴ P1 = 3 Pa
Result: The initial pressure in the gas tank equals 3 Pa.
Note: As the final pressure is given in Pascals (Pa) so the unit used for initial pressure should also be Pa.
More examples on P1/T1 = P2/T2
A chemist has a 140 mL chamber with a fixed amount of carbon dioxide (CO2) gas. The chemist noted the initial pressure to be 300 atm but forgot to record the initial temperature. If the temperature of the chamber is eventually raised to 50 K, its pressure increases to 550 atm. Use this information and the formula P1/T1 = P2/T2 to find the initial temperature inside the gas chamber.
As per the question statement:
P1 = 300 atm
T1 = ?
P2 = 550 atm
T2 = 50 K
300 (50) = 550 (T1)
T1 = 15000/550
∴ T1 = 27.3 K
Result: The initial temperature in the chamber is 27.3 K.
The pressure of a gas at 267 K is 3.5 atm. If the volume and amount of gas are kept fixed while the pressure is reduced to 2.6 atm, find the temperature of the gas.
As per the question statement:
P1 = 3.5 atm
T1 = 267 K
P2 = 2.6 atm
T2 = ?
3.5 (T2) = 267 (2.6)
T2 = 694.2/3.5
∴ T2 = 198.3 K
Result: The final temperature of the gas is 198.3 K.
Further insight: You may note that the decrease in temperature also decreased the pressure of the gas in this example, which is thus proof of the direct proportional relationship between P and T as per Gay-Lussac’s law.
FAQ
What is Gay Lussac’s law?
As per Gay-Lussac’s law, the pressure (P) of a gas is directly proportional to its temperature (T) if the number of moles of gas (n) and its volume (V) are kept constant.
P α T
P = k T
Where P = pressure, T = temperature, and k = proportionality constant.
What does the terms P1, T1, P2 and T2 represent in P1/T1 = P2/T2?
At a constant volume, if the temperature of a gas changes from T1 to T2 while the pressure changes from P1 to P2,then the formula P1/T1 = P2/T2 represents:
P1 = Initial pressure of the gas
T1 = Initial temperature of the gas
P2 = Final pressure of the gas
T2= Final temperature of the gas
In either case, pressure is measured in atm, Pa, or mm Hg, while temperature must be recorded in Kelvin (K).
Two different pressure versus temperature graphs are shown below. Why does graph A pass through the origin (0, 0) while graph B does not?
As per Gay-Lussac’s law, the pressure (P) of a gas is directly proportional to Kelvin’s temperature (T). Graph A is a plot of pressure versus temperature (in K).
At 0 K, i.e., absolute zero, the pressure of the gas equals zero.
P = k T
P = k(0) = 0
Therefore, this graph passes through the origin.
Contrarily, Graph B is a plot of pressure versus temperature (in °C). At 0°C, the temperature in Kelvins is 273 K.
P= k (273) ≠ 0. Thus, the graph does not pass through the origin.
What is the difference between Gay-Lussac’s law and Charles’ law?
Gay-Lussac’s law defines the direct proportional relationship between the pressure exerted by a gas and the temperature provided to it at constant volume and concentration.
P =kT (pressure-temperature relationship)
Contrarily, Charles’ law states that the volume of a gas is directly proportional to its temperature at constant pressure and concentration.
V = kT (volume-temperature relationship)
Which of the following is a correct formula for finding P1 using P1/T1 = P2/T2?
A) P1 = T1T2/P2
B) P1 = P2T1/T2
C) P1 = P2T2/T1
D) P1 = P2/T1T2
Option B is the correct answer.
The equation P1/T1 = P2 /T2 can be rearranged to make P1 the subject of the formula, as shown below:
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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