Under isothermal (constant temperature) and isobaric (constant pressure) conditions, an equal volume of two different gases contains an equal number of particles.
Mathematically expressed as:
V α n
Where:
V= volume occupied by the gas, and n = number of moles of gas or the number of gas particles.
Thus, volume is directly proportional to the amount of gas in a container.
Converting proportionality into equality gives us:
V = k n
Where k is a constant, in this case, the temperature and pressure of the system stay unchanged.
If the amount of gas is doubled, the volume occupied by the gas will also increase two times. An example includes a balloon; the balloon size (volume) increases as you fill more gas in it.
What does V1/n1 = V2/n2 represent?
The expression V1/n1 = V2/n2 comes into play when we compare two different systems (system A and system B) filled with the same or different gases in varying amounts.
Systems A and B may also refer to the initial and final state of a container after changing the amount of gas present in it.
V1 = Volume occupied by gas in system A
n1 = Number of moles of gas in system A
V2 = Volume occupied by gas in the system B
n2 = Number of moles of gas in system B
As per Avogadro’s law:
V1 = kn1… Equation (i)
V2 = kn2… Equation (ii)
Making k the subject of the formula from both equations (i) and (ii) gives us:
k = V1/n1….Equation (iii)
k = V2/n2…..Equation (iv)
The value of k stays constant in either case, thus equating equations (iii) and (iv) gives us the final formula, i.e.,
V1/n1 = V2/n2
An alternative way of writing the above equation is:
V1 n2 = V2 n1
Units of V1/n1 = V2/n2
The volumes (V1 and V2) are usually measured in liters (L), while the number of moles of gas (n1 and n2) is given in mol on either side of the formula.
However, the units of the ratio (V/n), i.e., L/mol, get canceled on either side of the equation V1/n1 = V2/n2.
Where and how to use V1/n1 = V2/n2? -Examples
The formula V1/n1 = V2/n2 is quite simple to use.
It can be used to find the initial or final volume as well as the initial or final number of moles of gas in a system if three variables are known (or can be easily determined) while the fourth is unknown.
You will understand this concept better with the help of the examples given below. So, let’s start!
For example, A cylinder with a movable piston contains 0.6 moles of argon (Ar). More argon gas was added to the cylinder such that the volume increased from 2.0 L to 2.5 L, keeping temperature and pressure constant.
How many moles of argon gas were added to the cylinder leading to this volume change?
Solution
As per the question statement:
V1 = 2.0 L
n1 = 0.6 moles
V2 = 2.5 L
n2 =?
So let us find n2 by applying the formula (V1/n1 = V2/n2):
⇒ 2.0/0.6 = 2.5/n2
Solve for n2 by cross multiplication:
2.0(n2) = 2.5(0.6)
n2 = 1.5/2.0
∴ n2 = 0.75 moles
Result: The number of moles of argon gas added equals n2 –n1 = 0.75 – 0.6 = 0.15 moles.
Another example is- A balloon filled using 0.762 mol helium (He) gas occupies a volume of 1.90 L. If 0.035 mol of additional helium gas is added to this balloon while keeping the temperature and pressure constant, how much volume will the balloon now occupy?
Solution
In this example, system A is the balloon initially filled with 0.762 mol of helium gas. In contrast, system B is the balloon filled with an additional amount of 0.035 mol helium. This gives us:
n1 = 0.762 mol
n2 = 0.762 + 0.035 = 0.797 mol
V1 = initial volume occupied by the balloon = 1.90 L
V2 = final volume occupied by the balloon =? = x
Now let us substitute all the above data into the equation:
Solving for x via cross multiplication gives us:
(1.90)(0.797) = x (0.762)
1.5143 = 0.762 x
∴ x = 1.5143/0.762 = 1.99 L
Result: The final volume occupied by helium gas in the balloon is 1.99 L.
More examples on V1/n1 = V2/n2
1.79 moles of nitrogen (N2) gas occupies a volume of 30 L in a container. If 15 g of N2 is further added to the container with no change in temperature or pressure, what will be the new volume of the container?
As per the question statement:
n1 =1.79 moles
15 grams of nitrogen gas is added further, which implies:
Moles of nitrogen added = mass/molar mass of nitrogen = 15/14 = 1.071 moles
n2 = 1.79 + 1.071 = 2.861 moles
V1 = 30 L
V2 =?
⇒ V1/n1 = V2/n2
(30) (2.861) = 1.79 V2
∴ V2 = 85.83/1.79 = 47.95 L
Result: The final volume of nitrogen in the container is 47.95 L
When 0.15 moles of gas X are added to a piston containing 0.82 moles of another gas, by what percent does the total volume increase assuming isothermal and isobaric conditions?
As per the question statement:
n1 = 0.82 moles
n2 = 0.82 + 0.15 = 0.97 moles
Both V1 and V2 are unknown in this case. So is solving this question impossible? Not really; if you read carefully, the question statement asks by what percent does the total volume increase?
Let us suppose:
V1 = x L
Now let’s find V2 in terms of x by applying the formula V1/n1 = V2/n2 :
Result: Total volume increased by 18.29 %.
FAQ
What is Avogadro’s law and its equation?
Avogadro’s law explains the direct proportional relationship between the amount of a gas and the volume occupied by the gas .
An equal number of moles of two gas systems occupy the same volume (V1/n1 = V2/n2) under constant temperature and pressure conditions.
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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