Derivation of Ideal Gas Equation - Definition, Formula, Limitations, FAQs

The ideal gas equation is usually called the ideal gas law since it is quite fundamental in the course of learning gases; it contains in one mathematical expression the basic principles relating to pressure, volume, temperature, and number of moles of gas. This formula is always written as PV=nRT, where P = pressure, V = volume, n = number of moles, R = universal gas constant, and T = Temperature in Kelvin. It is an integrated law that has Boyle's Law, Charles's Law, Gay-Lussac's Law, and all such laws combined into a single equation.

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This Story also Contains

1. Ideal Gas Equation
2. Some Solved Examples
3. Summary

The ideal gas equation assumes that gases are composed of a large number of small particles moving constantly and randomly without forces between them, except those of elastic collision. This model simplifies complex behaviors of real gases into simpler calculations and predictions for many conditions. While the ideal gas law is usually an excellent approximation for most gases under most conditions, the accuracy of the approximation decreases at extremely high pressures or extremely low temperatures, where real gases deviate significantly from ideality. The applications of the ideal gas equation are very important in most fields of science and engineering. Such applications are core in areas of chemistry, physics, and meteorology, and explain and predict the behavior of gases under different conditions. It also applies to industrial processes, including the storage of gases, chemical reaction kinetics, and thermodynamic cycles.

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Ideal Gas Equation

The ideal gas equation is an equation that is followed by the ideal gases. A gas that would obey Boyle's and Charles's laws under all the conditions of temperature and pressure is called an ideal gas.
As discussed the behavior of gases is described by certain laws such as Avogadro's Law, Boyle's Law, and Charles' Law.

According to Avogadro's Law; $\mathrm{V} \propto \mathrm{n}(\mathrm{P}$ and T constant $)$ According to Boyle's Law; $\mathrm{V} \propto \frac{1}{P}$ ( T and n constant) According to Charles' Law; $\mathrm{V} \propto \mathrm{T}(\mathrm{P}$ and n constant) Combining the three laws; we get:

$\begin{aligned} & V \propto \frac{n T}{P} \\ & V=R \frac{n T}{P}\end{aligned}$

' $R$ ' is the proportionality constant. On rearranging the above equation we get the following:
$$
\mathrm{PV}=\mathrm{nRT}
$$

This is the ideal gas equation as it is obeyed by the hypothetical gases called ideal gases under all conditions

Universal Gas Constant or Ideal Gas ConstantR or S : Molar gas constant or universal gas constant Values of $\mathrm{R}=0.0821 \mathrm{lit}$, atm, $\mathrm{K}^{-1}, \mathrm{~mol}^{-1}$
$
\begin{aligned}
& =8.314 \text { joule } \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\
& =8.314 \times 10^7 \mathrm{erg} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\
& =2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$

For a single molecule, the gas constant is known as Boltzmann constant ( k ) and unit $\left(\mathrm{m}^2 \mathrm{kgs}^{-2} \mathrm{~K}^{-1}\right)$
$
\begin{aligned}
\mathrm{k} & =\mathrm{RN}_0 \\
& =1.38 \times 10^{-3} \mathrm{~J} / \mathrm{deg}-\text { abs } / \text { molecule } \\
& =1.38 \times 10^{-16} \mathrm{erg} / \text { deg-abs } / \text { molecule }
\end{aligned}
$

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Combined gas law
Boyle's and Charles' Law can be combined to give a relationship between the three variables P, V, and T. The initial temperature, pressure, and volume of a gas are $T_1, P_1$ and $V_1$. With the change in either of the variables, all three variables change to $T_2, P_2$, and $V_2$. Then we can write:

$\frac{P_1 V_1}{T_1}=n R$ and $\frac{P_2 V_2}{T_2}=n R$
Combining the above equations we get

$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}=n R$
The above relation is called the combined gas law.

Density and Molar Mass of a Gaseous Substance

Ideal gas equation is $\mathrm{PV}=\mathrm{nRT}$ $\qquad$
On rearranging the above equation, we get
$$
\begin{aligned}
& \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}} \\
& \mathrm{n}\left(\mathrm{N}_0 . \text { of moles }\right)=\frac{\text { Given mass }(\mathrm{m})}{\text { Molar mass }(M)}
\end{aligned}
$$
$\qquad$
Putting the value of ' $n$ ' from equation (iii) in equation (ii), we get:
$$
\frac{\mathrm{m}}{\mathrm{MV}}=\frac{\mathrm{P}}{\mathrm{RT}}
$$
$\qquad$
We know that density (d) is mass (m) per unit volume (V)
$$
\mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}
$$

Replacing $\frac{\mathrm{m}}{\mathrm{V}}$ in eq. (iv) with d (density), then equation (iv) becomes:

$\frac{\mathrm{d}}{\mathrm{M}}=\frac{\mathrm{P}}{\mathrm{RT}}$

Rearranging the above equation, we get $M=\frac{d R T}{P}$

The above equation gives the relation between the density and molar mass of a gaseous substance

Recommended topic video on(Derivation of Ideal Gas Equation)

Some Solved Examples

Example 1: A refrigeration tank holding 5 liters of gas with molecular formula $\mathrm{C}_2 \mathrm{Cl}_2 \mathrm{~F}_4$ at 298 Kand 3 atm pressure developed a leak. When the leak was discovered and repaired, the tank had lost 76 g of the gas. What was the pressure of the gas remaining in the tank at 298 K?

1) 0.83

2)1.23

3)0.67

4)2.23

Solution

First, understand the question.

there is a tank holding 5 liter of gas with molecular formula C₂Cl₂ F4 at 298K at pressure 3 atm

after leaking 76 g of gas lost

We have to find the pressure of the gas remaining in the tank at 298K.

Now,

given some gas is lost in grams, so we need to find the initial quantity in grams from PV=nRT and n = weight/molar mass

then we will have the quantity in grams of remaining gas.

Then we have to convert it into moles and find p from PV=nRT.

According to the ideal gas equation, we have:

pV = nRT

Thus, $\mathrm{n}=\frac{3 \times 5}{0.082 \times 298}=0.613 \mathrm{moles}$

Thus, the total weight of the gas originally present = 0.613 x 171g
= 105g
Now, 76g of the gas is already lost, thus the remaining gas:
= 105 - 76 = 29g
Thus, total moles of the gas remaining = 29/171 = 0.17 moles
Again, according to the ideal gas equation, we have:
PV = nRT

Thus, $\mathrm{P}=\frac{0.17 \times 0.082 \times 298}{5}=0.83 \mathrm{~atm}$

Therefore, Option(1) is correct.

Example 2: Two identical flasks contain gases A and B at the same temperature. If density of $A=3 \mathrm{~g} / \mathrm{dm}^3$ and that of $B=1.5 \mathrm{~g} / \mathrm{dm}^3$and the molar mass of $A=\frac{1}{2}$ of $B$ , the ratio of pressure exerted by gases is :

1)$\frac{P_A}{P_B}=2$

2)$\frac{P_A}{P_B}=1$

3) $\frac{P_A}{P_B}=4$

4)$\frac{P_A}{P_B}=3$

Solution

As we learned in Ideal Gas Law in terms of density -

$P M=d R T$

- wherein
where
d - density of gas
P - Pressure
R - Gas Constant
T - Temperature
M - Molar Mass

$\begin{aligned} & P_A=\frac{3 R T}{M_A} ; P_B=\frac{1.5 R T}{M_B} \\ & \frac{P_A}{P_B}=\frac{2 M_B}{M_A}=\frac{2 \times 2 M_A}{M_A}=4\end{aligned}$

Example 3: $\mathrm{NaClO}_3$ is used, even in spacecraft, to produce $O_2$. The daily consumption of pure $O_2$ by a person is 492L at $1 \mathrm{~atm}, 300 \mathrm{~K}$ . How much amount of $\mathrm{NaClO}_3$ , in grams, is required to produce $O_2$ for the daily consumption of a person at $1 \mathrm{~atm}, 300 \mathrm{~K}$ ? _______.

$\mathrm{NaClO}_3(\mathrm{~s})+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{O}_2(\mathrm{~g})+\mathrm{NaCl}(\mathrm{s})+\mathrm{FeO}(\mathrm{s})$

$R=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

1) 2130

2)2140

3)2456

4)2458

Solution

moles of $\mathrm{NaClO}_3$ = moles of $\mathrm{O}_2$

moles of $\mathrm{O}_2=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1 \times 492}{0.082 \times 300}=20 \mathrm{ml}$

Mass of $\mathrm{NaClO}_3=20 \times 106.5=2130 \mathrm{~g}$

Example 4: The volume occupied by 4.75 g of acetylene gas at 50^oC and 740 mmHg pressure is _______L. (Rounded off to the nearest integer)

[Given R = 0.0826 L atm K^-1 mol ^-1]

1) 5

2)5.2

3)6

4)7

Solution

$\begin{aligned} & \mathrm{T}=50^{\circ} \mathrm{C}=323.15 \mathrm{~K} \\ & \mathrm{P}=740 \mathrm{~mm} \text { of } \mathrm{Hg}=\frac{740}{760} \mathrm{~atm} \\ & \mathrm{~V}=? \\ & \text { moles }(\mathrm{n})=\frac{4.75}{26} \\ & \mathrm{~V}=\frac{4.75}{26} \times \frac{0.0821 \times 323.15}{740} \times 760 \\ & \mathrm{~V}=4.97 \simeq 5 \mathrm{Lit}\end{aligned}$

Answer ; 5

Example 5: A car tyre is filled with nitrogen gas at 35 psi and $27^{\circ} \mathrm{C}$. It will burst of pressure exceeds 40 psi. The temperature in ${ }^{\circ} \mathrm{C}$ at which the car tyre will burst is ______ (Rounded off to the nearest integer)

1) 70

2)81

3)85

4)78

Solution

T₁ = 27^oC = 273 + 27 K = 300 K

$\begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} \\ & \frac{35}{300}=\frac{40}{\mathrm{~T}_2} \\ & \mathrm{~T}_2=\frac{40 \times 300}{35} \\ & \mathrm{~T}_2=342.86 \mathrm{~K} \\ & \mathrm{~T}_2=69.85^{\circ} \mathrm{C} \simeq 70^{\circ} \mathrm{C}\end{aligned}$

Hence, the answer is (70).

Summary

The ideal gas equation is PV = nRT, a very important formula in gas dynamics that links the pressure, volume, temperature, and quantity of the gas in a system. It combines Boyle's Law, Charles' Law, Gay-Lussac's Law, and all these relationships into one easy-to-use equation in one model, which will help predict and understand gasses under various conditions. In this equation, P is pressure, V is volume, n is the number of moles of gas, R is the universal constant for gases, and T is the temperature in Kelvin. The ideal gas equation is based on the assumption that the particles of the gas are always in constant, random motion and that their interaction with one another is only through elastic collisions, thus exerting negligible intermolecular forces. This idealization serves to simplify the really complex nature of real gases and hence makes the calculations easier and predictions more at hand. However, the formula becomes much less accurate for real gases under very high pressure or extremely low-temperature conditions, where ideal behavior deviates. The ideal gas equation has immense importance that cuts across a number of scientific and industrial fields. This includes being one of the basics in chemistry when determining the reactions of gases, in physics when studying thermodynamics and kinetic theory, and in meteorology when modeling in the atmosphere. Industrial applications of this equation range from storing gases to the manufacture of chemicals, through designing engines and refrigeration systems.

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