The gas laws are a group of laws that govern the behaviour of gases by providing relationships between the following the volume occupied by a gas, pressure exerted by a gas on the walls of its container, the absolute temperature of the gas, and the amount of gaseous substance (or) the number of moles of gas.
All gases generally show similar behaviour when the conditions are normal. But with a slight change in physical conditions like pressure, temperature or volume these show a deviation. Gas laws are an analysis of this behaviour of gases. The variables of state like the Pressure, Volume and Temperature of a gas depict its true nature. Hence gas laws are relations between these variables.
You will be expected to know each of the gas laws presented here, the conditions under which they work and do not work. You will also be expected to be able to use these ideas and the equations within to solve problems.
In this article, you will Study all the gas laws below, be able to understand the derivative of the laws, and their formulae.
- Boyle’s Law
- Charle’s Law
- Gay-Lussac Law
- General Gas Law
- Avogadro’s Law
- Ideal Gas Law
- Dalton Law of Partial Pressure
Boyle’s Law
Boyle’s law states that at constant temperature, the volume of a gas is inversely proportional to its pressure in a closed system. Basically, this law describes the relationship between the pressure and volume of a gas.
Mathematically, Boyle’s law can be expressed as;
V ∝ 1/P
Or
P ∝ 1/V
Or
PV = k
Where V is the volume of the gas, P is the pressure of the gas and Kis the constant. Boyle’s Law can be used to determine the current pressure or volume of gas and can be represented as;
P1V1 = P2V2
Boyle measured the relationship between the pressure and the volume of a given sample of gas at fixed temperature. He found that a sample of gas compresses if the external pressure applied to it increases and that the product PV is constant.
Diagrammatic Representation of Boyle’s Law
Boyle’s law, stated in mathematical terms for a gas whose pressure and volume is measured at two different pressure/volume states at a constant temperature is then,
P1V1 = P2V2
Note: This law is only truly valid at infinitely low pressures. Standard pressures, however lead to a reasonable approximation for most gases.
Graphs of Boyle’s Law
A gas that obeys both Boyle’s Law and Avogadro’s law is an Ideal gas. The atoms (molecules) of an ideal gas are infinitely small such that they never collide with each other. They never even interact with each other in any way.
Thus, the atoms/molecules of an ideal gas each behaves as if they were the only atom in the container. Simple experience tells us that this is not exactly true, even at room temperature (gas-phase reactions occur). It is a reasonably good assumption for most gases at or near STP.
The assumption of ideality breaks down as the pressure is increased or as the temperature is lowered. It also breaks down sooner for particularly large or massive or polar gas molecules.
Example Questions of Boyle’s Laws
Example 1
A gas occupies 11.2 cm3 at 0.860 mmHg. What is the pressure if the volume becomes 15.0 cm3?
Solution
P1V1 = P2V2
P1 = 0.860 mmHg, V1 = 11.2 cm3 V2 = 15.0 cm3 P2 = ?
0.860 mmHg X 11.2 cm3 = P2 X 15.0 cm3
P2 = 0.642 mmHg
Example 2
A helium balloon has a volume of 735 dm3 when it is at ground level. The balloon is transported to an elevation of 5km, where the pressure is only 0.8atm. At this altitude the gas occupies a volume of 1286 dm3. Assuming the temperature has remained the same, what was the ground level pressure?
Solution
P1V1 = P2V2
We are given the final pressure and volume, along with the initial volume. Using these values, we can calculate the initial pressure.
P1 = ?, V1 = 735mL V2 = 1286mL P2 = 0.8atm
P1 X 735mL = 0.8atm X 1286mL
P1= (0.8atm)
1286mL X 735mL
P1 = 1.4atm
Note that the pressure at sea level is equal to 1.0atm. A pressure greater than 1.0atm simply indicates that the ground level is below sea level at this point.
Charle’s Law
Charle’s law states that at constant pressure, the volume of a gas is directly proportional to the temperature (in Kelvin) in a closed system. Basically, this law describes the relationship between the temperature and volume of the gas.
Mathematically, Charle’s law can be expressed as;
V ∝ T
V / T = K
Where V is the volume of the gas, T is the temperature of the gas in Kelvin and Kis the constant. Charle’s Law can be used to determine relationship between the temperature and volume of a gas and can be represented as;
Charles’ Law- gives the relationship between volume and temperature if the pressure and the amount of gas are held constant, the following holds:
1) If the Kelvin temperature of a gas is increased, the volume of the gas increases. (P, n Constant)
2) If the Kelvin temperature of a gas is decreased, the volume of the gas decreases. (P, n Constant)
This means that the volume of a gas is directly proportional to its Kelvin temperature.
Diagrammatic Representation of Chale’s Law
The volume of any gas increases linearly with increasing temperature at constant pressure.
If we plot a graph of the volume of a sample of gas versus the temperature we get something that looks like the following:
Graphs of Charle’s Law
This graph provides us with another way of defining absolute zero on the temperature scale. Absolute zero is the temperature at which the volume of a gas becomes zero when the a plot of the volume versus temperature for a gas are extrapolated.
As expected, the value of absolute zero obtained by extrapolating the data is essentially the same as the value obtained from the graph of pressure versus temperature in the preceding section. Absolute zero can therefore be more accurately defined as the temperature at which the pressure and the volume of a gas extrapolate to zero.
A plot of the volume versus the temperature of a gas (when the temperatures obtained are converted from Celsius to the Kelvin scale) becomes a straight line that passes through the origin. Any two points along this line can therefore be used to construct the following equation, which is known as Charles’ law.
Before using this equation, it is important to remember that temperatures must be converted from oC to K. (K = oC + 273)
Example Questions of Charle’s Laws
Example 1
A gas occupies 900.0 dm3 at a temperature of 27.0 °C. What is the volume at 132.0 °C?
Solution
T1 = (27+273)K = 300K, V1 = 900.00 dm3 V2 = ? T2 = (132+273) = 405K
(900dm3) / (300K) = V2 / (405K)
V2 = 1215 dm3
Example 1
When the volume of a gas is changed from ___ mL to 852 mL, the temperature will change from 315 °C to 452 °C. What is the starting volume?
Solution
Write Charles Law and substitute values in:
V1 / T1 = V2 / T2
V1 / 588 K = 852 mL / 725 K
V1 (725 K) = (852 mL) (588 K)
V1 = 691 mL
Note the large °C values, trying to get you to forget to add 273. Remember, only Kelvin temperatures are allowed in the calculations.
Gay Lussac’s Law
Gay Lussac’s Law – states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
This mathematically can be written as: p ∝ T
⇒ p / T = constant= K
The temperature here is measured on the Kelvin scale. The graph for the Gay- Lussac’s Law is called as an isochore because the volume here is constant.
Graphs of Gay Lussac’s Law
To calculate a change in pressure or temperature using Gay Lussac’s Law the equation looks like this:
For an enclosed ideal gas at the constant volume;
Combining both equations,
From the above formula, we can calculate the pressure or the temperature at any unknown condition if the pressure and the temperature are known at any one condition.
Let take examples to better understand it.
Example Questions of Gay Lussac’s Laws
Example 1
An ideal gas is contained in a 2.5L container at a pressure of 4.0atm. The container is at a temperature of 25∘C. What will be the final pressure if the temperature is increased to 50∘C?
Solution
Since the only variable that has changed is temperature, we can use Gay-Lussac’s law in order to compare pressure to temperature. Because temperature and pressure are on opposite sides of the ideal gas law, they are directly proportional to one another. In other words, as one increases, the other increases as well.
Gay-Lussac’s law is written as follows:
P1 / T1=P2 / T2
To use this equation, we must convert the temperature to Kelvin.
25oC+273=298K
50oC+273=323K
Now, use these temperatures and the initial pressure to solve for the final pressure.
4atm / 298K=P2 / 323K
P2 = 4.3atm
Example 2
A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure?
Solution
| 699.0 mmHg | 760.0 mmHg | |
| –––––––– | = | –––––––––– |
| 313 K | x |
x = 340. K (or 67.0 °C. to three sig figs)
Note that the problem did not specify what temperature unit to use. Usually, the expectation is that the answer be given in the same unit as used in the problem (in this case, degrees Celsius). If you notice this done on a test question, consider asking the teacher what he/she wants done (or just supply both values).
General Gas Law
The general gas law is the combination of the Boyle’s, Charles and Gay-Lussac’s/Pressure Law. The law shows the relationship between temperature, volume and pressure for a fixed quantity of gas.
The general equation of combined/general gas law is given as;
PV / T = k
If we want to compare the same gas in different cases, the general gas law can be represented as;
P1V1 / T1 = P2V2 / T2
The combined gas (general gas law) law allows you to derive any of the relationships needed by combining all of the changeable peices in the ideal gas law: namely pressure, temperature and volume.
R and the number of moles do not appear in the equation as they are generally constant and therefore cancel since they appear in equal amounts on both sides of the equation.
Questions on General Gas Law
Example 1
What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from standard pressure to 360.0 mmHg?
Solution
We are looking to determine V2 in this problem. Here’s the set up:
| P1 = 760.0 mmHg | P2 = 360.0 mmHg |
| V1 = 400.0 L | V2 = x |
| T1 = 295 K | T2 = 303 K |
V2 = [(760 mmHg) (400 mL) (303 K)] / [(295 K) (360 mmHg)]
V2 = 867 mL (to three sig figs)
Example 2
What is the density, in g/L, for a gaseous compound at STP if the gas in a 1.00 L bulb weighs 0.672 g at 25.0 °C and 733.4 mm Hg?
Solution:
1) We need to know the volume of the gas at STP. For this, we use the combined gas law. Here’s the data:
| P1 = 733.4 mmHg | P2 = 760.0 mmHg |
| V1 = 1.00 L | V2 = x |
| T1 = 298 K | T2 = 273 K |
2) Here’s the combined gas law with the data filled in:
| (733.4 mmHg) (1.00 L) | (760.0 mmHg) (x) | |
| ––––––––––––––––––– | = | ––––––––––––––– |
| 298 K | 273 K |
After a bit of math, we find:
x = 0.884 L
3) We are now ready to determine the density:
0.672 g / 0.884 L = 0.760 g/L
Avogadro’s Law
Avogadro’s law states that if the gas is an ideal gas, the same number of molecules exists in the system. The law also states that if the volume of gases is equal it means that the number of the molecule will be the same as the ideal gas only when it has equal volume. This above statement can be mathematically expressed as;
V / n = constant
Or
V1 / n1 = V2 / n2
Where V is the volume of an ideal gas and n in the above equation represent the number of gas molecules. n is the number of moles of the gas. Hence, V= kn
The number of molecules in a mole of any gas is known as the Avogadro’s constant and is calculated to be 6.022 X 1023. The values for temperature and pressure here are the standard values.
For temperature, we take it to be 273.15 K while for the pressure it equals 1 bar or 105 pascals. At these Standard Temperature Pressure (STP) values, one mole of a gas is supposed to have the same volume.
Avagadro’s Law- Gives the relationship between volume and amount of gas in moles when pressure and temperature are held constant.
If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases. This is assuming of course that the container has expandible walls.
The relationship is again directly proportional so the equation for calculations is
The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules.
Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.
In equation form, this is written as:
V∝n
or
V=k×n
or
V1 / n1 = V2 / n2
Avogadro’s law assumes all atoms (molecules) are the same size. This assumption is only true at extremely small pressures.
Standard Temperatures and Pressures STP are defined to be
0 ℃ and 100 kPa (=0.98692 atm)
One mole of Gas at STP occupies 22.7 L
There are slight variances in real gases at normal conditions but not much.
Questions on Avogadro’s Gas Law
Example 1
A 6.0 L sample at 25°C and 2.00 atm of pressure contains 0.5 mole of a gas. If an additional 0.25 mole of gas at the same pressure and temperature are added, what is the final total volume of the gas?
Solution:
First, express Avogadro’s law by its formula:
Vi/ni = Vf/nf
where
Vi = initial volume
ni = initial number of moles
Vf = final volume
nf = final number of moles
For this example, Vi = 6.0 L and ni = 0.5 mole. When 0.25 mole is added:
nf = ni + 0.25 mole
nf = 0.5 mole = 0.25 mole
nf = 0.75 mole
The only variable remaining is the final volume.
Vi/ni = Vf/nf
Solve for Vf
Vf = Vinf/ni
Vf = (6.0 L x 0.75 mole)/0.5 mole
Vf = 4.5 L/0.5 Vf = 9 L
Check to see if the answer makes sense. You would expect the volume to increase if more gas is added. Is the final volume greater than the initial volume?
Yes. Doing this check is useful because it is easy to put the initial number of moles in the numerator and the final number of moles in the denominator.
If this had happened, the final volume answer would have been smaller than the initial volume.
Thus, the final volume of the gas is 9.0
Example 2
At a certain temperature and pressure, one mole of a diatomic H2 gas occupies a volume of 20 L. What would be the volume of one mole of H atoms under those same conditions?
Solution:
One mole of H2 molecules has 6.022 x 1023 H2 molecules.
One mole of H atoms has 6.022 x 1023 H atoms.
The number of independent “particles” in each sample is the same.
Therefore, the volumes occupied by the two samples are the same. The volume of the H atoms sample is 20 L.
By the way, I agree that one mole of H2 has twice as many atoms as one mole of H atoms. However, the atoms in H2 are bound up into one mole of molecules, which means that one molecule of H2 (with two atoms) counts as one independent “particle” when considering gas behavior.
Example 3
A flexible container at an initial volume of 6.13 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 15.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Solution:
1) State Avogadro’s Law in problem-solving form:
| V1 | V2 | |
| ––– | = | –––– |
| n1 | n2 |
2) Substitute values into equation and solve:
| 6.13 L | 15.5 L | |
| ––––––– | = | –––––– |
| 8.51 mol | x |
x = 21.5 mol
3) Determine moles of gas added:
21.5 mol − 8.51 mol = 13.0 mol (when properly rounded off)
Example 4
Three balloons are filled with different amounts of an ideal gas. One balloon is filled with 3 moles of the ideal gas, filling the balloon to 30 L.
a) One balloon contains 2 moles of gas. What is the volume of the balloon?
b) One balloon encloses a volume of 45 L. How many moles of gas are in the balloon?
Solution:
Avogadro’s law says the volume (V) is directly proportional to the number of molecules of gas (n) at the same temperature.
n ∝ V
This means the ratio of n to V is equal to a constant value.
Since this constant never changes, the ratio will always be true for different amounts of gas and volumes.
where
ni = initial number of molecules
Vi = initial volume
nf = final number of molecules
Vf = final volume.
Part a) One balloon has 3 moles of gas in 30 L. The other has 2 moles in an unknown volume. Plug these values into the above ratio:
Solve for Vf
(3 mol)Vf = (30 L)(2 mol)
(3 mol)Vf = 60 L⋅mol
Vf = 20 L
You would expect less gas to take up a smaller volume. In this case, 2 moles of gas only took up 20 L.
Part b) This time, the other balloon has a known volume of 45 L and an unknown number of moles. Start with the same ratio as before:
Use the same known values as in part a, but use 45 L for Vf.
Solve for nf
(3 mol)(45 L) = (30L)nf
135 mol⋅L = (30L)nf
nf = 4.5 moles
The larger volume means there is more gas in the balloon. In this case, there are 4.5 moles of the ideal gas in the larger balloon.
An alternative method would be to use the ratio of the known values. In part a, the known values were the number of moles. There was second balloon had 2⁄3 the number of moles so it should have 2⁄3 of the volume and our final answer is 2⁄3 the known volume. The same is true of part b. The final volume is 1.5 times larger so it should have 1.5 times as many molecules. 1.5 x 3 = 4.5 which matches our answer. This is a great way to check your work.
Ideal Gas Law
The Ideal Gas Laws states that; The volume of a specified amount of gas is inversely proportional to pressure and directly proportional to volume and temperature.
To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas: Boyle’s law, Gay Lussac’s law, Charles’s law, and Avogadro’s law
Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:
PV=nRT
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant.
The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol–1 K–1 and 8.314 kPa L mol–1 K–1.
This law relates four different variables which are pressure, volume, no of moles or molecules and temperature. Basically, the ideal gas law gives the relationship between these above four different variables.
Mathematically Ideal gas law is expressed as;
PV=nRT
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant.
We can also use an equivalent equation given below.
PV = kNT
Where, k = Boltzman constant and N = number of gas molecules.
When using the Ideal Gas Law to calculate any property of a gas, you must match the units to the gas constant you choose to use and you always must place your temperature into Kelvin.
To use the equation, you simply need to be able to identify what is missing from the question and rearrange the equation to solve for it.
Because the units of the gas constant are given using atmospheres, moles, and Kelvin, it’s important to make sure you convert values given in other temperature or pressure scales.
Value of R
R is the gas constant or proportionality constant in the Ideal Gas Equation. Mathematically, if you need to find the value of any variable, then you can do so if you have the other values. In the Ideal gas law equation pV = nRT, we can write R = pV/ nT. Now at Standard Temperature Pressure conditions i.e. when T = 273.15 K and p = 1 bar the volume is 22.710 L mol-1 and number of moles(n) is 1, then R = (105 Pa)(22.71×10-3 m3) / (1 mol)(273.15K)
= 8.314 Pa m3K-1 mol-1
= 8.314 JK-1 mol-1
Ideal Gas Law – More Details
Gases can described in terms of four variables: pressure (P), volume (V), temperature (T), and the amount of gas (n). There are five relationships between pairs of these variables in which two of the variables were allowed to cahnge while the other two were held constant.
| P |  | n | (T and V constant) | ||||
| Boyle’s law: | P | | 1/V | (T and n constant) | |||
| Amontons’ law: | P | | T | (V and n constant) | |||
| Charles’ law: | V | | T | (P and n constant) | |||
| Avogadro’s hypothesis: | V | | n | (P and T constant) |
Each of these relationships is a special case of a more general relationship known as the ideal gas equation.
PV = nRT
In this equation, R is a proportionality constant known as the ideal gas constant and T is the absolute temperature. The value of R depends on the units used to express the four variables P, V, n, and T. By convention, most chemists use the following set of units.
Ideal Gas Behaviour
Many properties of Ideal gases are very closely similar to real gases. Some of these properties specific to Ideal gases are:
- Ideal gas constitutes a large number of molecules that are identical to each other.
- These molecules move in random motion and obey Newton’s Laws of Motion.
- The volume occupied by the gas is larger as compared to that of the volume occupied by the molecules.
It is only during a collision that molecules experience external forces. These collisions are elastic in nature and end in very short intervals of time.
Questions on Ideal Gas Law
Example 1
Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.
Solution:
1) Rearrange PV = nRT to this:
V = nRT / P
2) Substitute:
V = [(2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K)] / 1.00 atm
V = 1.19 L (to three significant figures)
Example 2
At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?
Solution:
1) Rearrange PV = nRT to this:
T = PV / nR
2) Substitute:
T = [(1.95 atm) (12.30 L)] / [(0.654 mol) (0.08206 L atm mol¯1 K¯1)]
T = 447 K
Dalton Law of Partial Pressure
State that; the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.
Therefore, the pressure of a mixture of gases is equal to the sum of the partial pressure that each individual gas would exert by itself at the same volume and temperature.
Pt = P1 + P2 + P3
where Pt is total pressure of the mixture of gases and P1, P2, P3 are the partial pressures of the individual gases.
Explanation
Based on the kinetic theory of gases, a gas will diffuse in a container to fill up the space it is in and does not have any forces of attraction between the molecules. In other words, the different molecules in a mixture of gases are so far apart that they act independently; they do not react with each other.
The pressure of an ideal gas is determined by its collisions with the container, not collisions with molecules of other substances since there are no other collisions. A gas will expand to fill the container it is in without affecting the pressure of another gas.
So it can be concluded that the pressure of a certain gas is based on the number of moles of that gas and the volume and temperature of the system. Since the gases in a mixture of gases are in one container, the Volume (V) and Temperature (T) for the different gases are the same as well.
Each gas exerts its own pressure on the system, which can be added up to find the total pressure of the mixture of gases in a container. This is shown by the equation
Ptotal=PA+PB+..
The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
PressureTotal = PressureGas 1 + PressureGas 2 + PressureGas 3 + … PressureGas n
An alternative of this equation can be used to determine the partial pressure of an individual gas in the mixture.
If the total pressure is known and the moles of each component gas are known, the partial pressure can be computed using the formula:
Px = PTotal ( nx / nTotal )
where:
Px = partial pressure of gas x PTotal = total pressure of all gases nx = number of moles of gas x nTotal = number of moles of all gases
This relationship applies to ideal gases but can be used in real gases with very little error.
Questions on Ideal Gas Law
Example 1
A mixture of nitrogen, oxygen and helium is filled in a cylinder. This mixture of these three gases exerts a total pressure of 12 atm on the wall of a cylinder. If the partial pressures of nitrogen and oxygen are 2 atm and 3 atm respectively, then calculate the partial pressure of helium.
Solution:
Given data:
Total pressure of the gas, PTotal = 12 atm
Partial pressure of the nitrogen gas, P1 = 2 atm
Partial pressure of the oxygen gas, P2 = 3 atm
Partial pressure of the helium gas, P3 = ?
Using the formula of dalton’s law,
PTotal = P1 + P2 + P3 + …. Pn
PTotal = P1 +P2 + P3 (Because, a mixture is made up of three gases)
12 = 2 + 3 + P3
P3 = 12 – 5
P3 = 7 atm
Therefore, the partial pressure of the helium gas is 7 atm.
Example 2
A mixture of hydrogen gas and oxygen gas exerts a total pressure of 1.5 atm on the walls of its container. If the partial pressure of hydrogen is 1 atm, find the mole fraction of oxygen in the mixture.
Given, Phydrogen = 1 atm, Ptotal = 1.5 atm
Applying Dalton’s law formula, Ptotal = Phydrogen + Poxygen
Therefore, Poxygen = 0.5 atm
Now, the mole fraction of oxygen, Xoxygen = (Poxygen/Ptotal) = 0.5/1.5 = 0.33
Therefore, the mole fraction of oxygen in the mixture is 0.33
Example 3
Nitrogen is collected over water at 40.0 °C. What is the partial pressure of nitrogen if the total pressure is 99.42 kPa?
Solution:
1) The vapor pressure of water at 40.0 °C is looked up and found to be 7.38 kPa.
2) Dalton’s Law of Partial Pressure is used:
Ptot = PN2 + PH2O
99.42 kPa = PN2 + 7.38 kPa
PN2 = 92.04 kPa
Example 4
Five gases filled in the container exerts a 24 atm of total pressure on the walls of the container. The partial pressures of the four gases are 2 atm, 5 atm, 6 atm and 8 atm respectively. Calculate the partial pressure of the fifth gas.
Solution:
Given data:
Total pressure of the gas, PTotal = 24 atm
Partial pressure of the gas 1, P1 = 2 atm
Partial pressure of the gas 2, P2 = 5 atm
Partial pressure of the gas 3, P3 = 6 atm
Partial pressure of the gas 4, P4 = 8 atm
Partial pressure of the gas 5, P5 = ?
Using the formula of dalton’s law,
PTotal = P1 + P2 + P3 + …. Pn
24 = 2 +5 + 6 + 8 + P5 (Because, a mixture is made up of five gases)
24 = 21 + P5
P5 = 24 – 21
P5 = 3 atm
Therefore, the partial pressure of the fifth gas is 3 atm.