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

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,

$\mathrm{V} \propto \mathrm{n}$ (P and T constant )

According to Boyle's Law,

$V \propto 1 / P$ (T and n constant).

According to Charles' Law,

$V \propto T$ (P and n constant). Combining the three laws, we get:

$V \propto n T P V=R n T P$

' R ' is the proportionality constant. On rearranging the above equation, we get the following:
PV=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 Constant R or S: Molar gas constant or universal gas constant. Values of R=0.0821lit, atm, K⁻¹, mol⁻¹$=8.314$ joule $K^{-1} \mathrm{~mol}^{-1}=8.314 \times 10^7 \mathrm{ergK}^{-1} \mathrm{~mol}^{-1}=2 \mathrm{calK}^{-1} \mathrm{~mol}^{-1}$

For a single molecule, the gas constant is known as the Boltzmann constant ( k ) and unit $\left(\mathrm{m}^2 \mathrm{kgs}^{-2} \mathrm{~K}^{-1}\right)$
$k=RN0=1.38×10−3 J/deg− abs / molecule =1.38×10−16erg/ deg-abs / molecule$

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 T₁, P₁, and V₁. With the change in either of the variables, all three variables change to T₂, P₂, and V₂. Then we can write:

$P_1 V_1 T_1=n R$ and $P_2 V_2 T_2=n R$
Combining the above equations, we get

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

Related Topics link

• NMR Spectroscopy
• Proton Neutron Discovery
• Protons
• Proton Neutron Discovery
• Atomic number mass number

Density and Molar Mass of a Gaseous Substance

Ideal gas equation is PV=nRT
On rearranging the above equation, we get
$\mathrm{nV}=\mathrm{PRT}$

n (No. of moles )= Given mass (m) Molar mass (M)

Putting the value of ' n ' from equation (iii) in equation (ii), we get:
$\mathrm{mMV}=\mathrm{PRT}$

We know that density (d) is mass (m) per unit volume (V)
d=mV

Replacing mV in the equation. (iv) with d (density), then equation (iv) becomes:

$\mathrm{dM}=\mathrm{PRT}$

Rearranging the above equation, we get M=dRTP

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

Some Solved Examples

Example 1: A refrigeration tank holding 5 liters of gas with molecular formula C₂Cl₂ F4 at 298 K and 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, n=3×50.082×298=0.613moles

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,$\mathrm{P}=0.17 \times 0.082 \times 2985=0.83 \mathrm{~atm}$

Hence, the correct answer is option (1)

Example 2: Two identical flasks contain gases A and B at the same temperature. If the density of A=3 g/dm³ and that of B=1.5 g/dm^3, and the molar mass of A = 12 and B, the ratio of pressure exerted by gases is :

1)P_AP_B=2

2)P_AP_B=1

3) P_AP_B=4

4)P_AP_B=3

Solution

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

PM=dRT

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

P_A=3RTM_A

P_B=1.5RTM_B

P_AP_B=2M_BM_A=2×2M_AM_A= 4

Example 3: Which of the following statements is correct?
(a) Equal masses of all gases occupy the same volume at 57 P.
(b) An wat ideal gas cannot be liquefied.
(c) the kinetic energy of gaseous molecules is servo at .
(d) Gases having very low critical temperatures often show nan ideal behavior at room temperature.

1) a and c only

2) b and c only

3) a and d only

4) (correct) b and d only

Solution

(a) It is not equal masses but equal moles of g axes.
(b) An ideal gas does not exhibit molecular attractions.
(c) KE is zero at 0 K and not at $0^{\circ} \mathrm{C}$
(d) The room temperature is much larger than the critical temperature. Options (b) and (d) are correct.

Hence, the answer is option (4).

Example 4: An open vessel at $27^{\circ} \mathrm{C}$ is heated until two fifth of the ain in it has escaped from the vessel. Assuming that the volume of the Vessel remains constant, the temperature at which the vessel has been heated in
1) 700 K
2) (correct) 500 K
3) 750 K
4) $500^{\circ} \mathrm{C}$

Solution

given, temperature $\left(\mathrm{T}_1\right)=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$
volume of vessel = constant
press cure in vessel $=$ constant
the volume of ain reduced by $\frac{2}{5}$ so the remaining volume of ain is $\frac{3}{5}$. Let, at $T_1$ the volume of air inside the vessel be $n$ so at $T_2$ the volume of air will be $\frac{3}{5} n$.
Now, a s p and vane constant, 10

$\mathrm{n} \cdot \mathrm{~T}_1=\frac{3}{5} \mathrm{nT}_2-(1)$

Putting the value of $\mathrm{T}_1$ in equation (1) we get,

$\begin{aligned}
& \mathrm{n}=300=\frac{3}{5} \mathrm{n} \cdot \mathrm{~T}_2 \\
& \Rightarrow \mathrm{~T}_2=300 \times \frac{5}{3}=500 \mathrm{k}
\end{aligned}$
Hence, the answer is option (2).

Example 5: A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of at and ban Sf molar volume of is 250 L under such conditions, what is the percentage of sodium bicarbonate in each tablet?

1) (correct) 8.4

2) 0.84

3) 16.8

4) 33.6

Solution

$\begin{aligned}
& 2 \mathrm{NaHCO}_3+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow 2 \mathrm{CO}_2+\mathrm{Na}_2 \mathrm{C}_4 \mathrm{O}_4+\mathrm{H}_2 \mathrm{O} \\
& 2 \mathrm{~mol} \\
& 1 \mathrm{~mol}
\end{aligned}$

⇒ In the reaction, several moles of $\mathrm{CO}_2$ produced.

$\begin{aligned}
\mathrm{n} & =\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1 \mathrm{bar} \times 0.25 \times 10^{-3} \mathrm{~L}}{0.082 \mathrm{LatmK}^{-1} \mathrm{~mol}^{-1} \times 298.15 \mathrm{~K}} \\
& =1.08 \times 10^{-5} \mathrm{~mol}
\end{aligned}$
Number of moles of $\mathrm{NaHCO}_3=\frac{\text { Wt.ofNaHCO }}{3}$
Molecular mass of $\mathrm{NaHCO}_3$
Hence, the answer is option (1).

Practice more questions with the link given below