Ideal Gas Equation

Ideal Gas Equation

The equation which relates the pressure (P), volume (V) and temperature (T) of the given state of an ideal gas is known as an ideal gas equation or equation of state.

From Boyle’s law, we get $V\propto \frac{1}{P}...(1)$

and From Charle’s Law, we get $V\propto T_{\dots (2)}$

And from Avogadro’s Law, we get $V\propto n\dots .(3)$

And from equation (1), (2), (3)

we can write

$\begin{array}{rl}& V\propto \frac{nT}{P}\\ & \text{or}\frac{PV}{nT}=\text{constant}\\ & \text{or}\frac{PV}{nT}=R(\text{where}R\text{is proportionality constant})\\ & \text{or}PV=nRT\end{array}$

So, Ideal Gas Equation is given as

P V = n R T

where

T= Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

Universal gas constant (R)

At S.T.P. the value of the universal gas constant is the same for all gases. And its value is given as

$R=8.31\frac{\text{Joule}}{\text{mole}\times \text{Kelvin}}=2\frac{\mathrm{cal}}{\text{mole}\times \text{Kelvin}}$

And its Dimension is : $\left[M{L}^2{T}^{-2}{\theta }^{-1}\right]$

Boltzman's constant (k)

It is represented by the per mole gas constant.

$k=\frac{R}{N}=\frac{8.31}{6.023\times {10}^{23}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Specific gas constant (r)

It is represented by per gram gas constant.

i.e. $r=\frac{R}{M}$

It's unit is $\frac{\text{Joule}}{\mathrm{gm}\times \text{kelvin}}$

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Solved Examples Based on the Ideal gas equation

Example 1: Which of the following shows the correct relationship between the pressure ‘P’ and density ρ of an ideal gas at constant temperature?

1)

2)

3)

4)

Solution:

Ideal gas equation

P V=n R T
wherein

T = Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

For an ideal gas

$PM=\rho RT$

Hence, the graph between $\mathrm{P}\mathrm{\&}\rho$ at constant temperature is straight line passing through origin.

Hence, the answer is option (4).

Example 2: ‘ n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be :

1) $\frac{9P_0{V}_0}{4nR}$
2) $\frac{3P_0{V}_0}{2nR}$
3) $\frac{9P_0{V}_0}{2nR}$
4) $\frac{9P_0{V}_0}{nR}$

Solution:

At any point between A & B, we can write a relation between P & V by using the equation of the straight-line

$\begin{array}{rl}& V-V_0=\frac{2V_0-V_0}{P_0-2P_0}(P-2P_0)\\ & V-V_0=\frac{-V_0}{P_0}(P-2P_0)\\ & P\left(\frac{-V_0}{P_0}\right)+2V_0=V-V_0\\ & P=\frac{-P_0}{V_0}(V-3V_0)\end{array}$

From the ideal gas equation

$\begin{array}{rl}& \mathrm{PV}=\mathrm{nRT}\\ & \Rightarrow \frac{nRT}{V}=\frac{-P_0}{V_0}(V-3V_0)\\ & T=\frac{-P_0}{nR{V}_0}(V^2-3V_{0}V)\end{array}$

$\text{For temperature to be maximum at any point}\frac{dT}{dV}=0$

$\begin{array}{rl}& \Rightarrow 2V-3V_0=0\\ & \therefore V=\frac{3V_0}{2}\\ & \therefore T_{\text{max}}=\frac{-P_0}{nR{V}_0}(\frac{9}{4}{V}_0^2-\frac{9}{2}{V}_0^2)=-\frac{P_0}{nR{V}_0}\cdot \frac{-9}{4}{V}_0^2=\frac{9}{4}\frac{P_0{V}_0}{nR}\end{array}$

Hence, the answer is option option (1).

Example 3: One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

1) $\frac{25}{16}\frac{p_0{v}_0}{R}$
2) $\frac{25}{8}\frac{p_0{v}_0}{R}$
3) $\frac{25}{4}\frac{p_0{v}_0}{R}$
4) $\frac{5}{8}\frac{p_0{v}_0}{R}$

Solution:

Ideal gas equation

$PV = nRT$
wherein

T= Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

equation of the given path is

$\begin{array}{rl}& p-3p_0=-\frac{2p_0}{v_0}(v-v_0)\\ & p=\frac{-2p_0}{v_0}v+5p_0\\ & PV=RT\text{or}P=RT/V\\ & \therefore T=(-2p_0/R{v}_0)v^2+\frac{5p_0}{R}v\end{array}$

for T to be maximum

$\begin{array}{rl}& dT/dv=0\\ & \frac{-2p_0}{R{v}_0}2v+5p_0/R=0\\ & v=5v_0/4\\ & T_{max}=\frac{25p_0{v}_0}{8R}\end{array}$

Hence, the answer is option (2)

Example 4: This is true for an ideal gas.

(1) Molecule of gas is identical spherical rigid and perfectly elastic point mass.

(2) There is always some attractive and repulsive force acting between gas molecules.

(3) The density of a gas is constant at all points of the container molecules

1) only 1

2) 1 and 3

3) 2 and 3

4) only 3

Solution:

For ideal gas

1. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses (It means that when they collide with each other, then there is no loss of energy during collision).
2. The density of gas does not change at any point in the container.
3. No attractive or repulsive force acts between gas molecules.

Hence, the answer is the option 2

Example 5: The temperature of an open room of volume 30 ${\mathrm{m}}^3$ increases from ${17}^{\circ }\mathrm{C}$ to ${27}^{\circ }\mathrm{C}$ due to the sunshine. The atmospheric pressure in the room remains $1\times {10}^5\mathrm{Pa}$. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f-n_i$ will be :

1) −1.61×10²³

2) 1.38×10²³

3) 2.5×10²⁵

4) −2.5×10²⁵

Solution:

$\begin{array}{rl}& \mathrm{PV}=\mathrm{nRT}\\ & \Rightarrow n_i=\frac{PV}{R{T}_i},n_f=\frac{PV}{R{T}_f}\\ & \Rightarrow n_f-n_i=\frac{PV}{R}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{{10}^5\times 30}{8.31}\times (\frac{1}{300}-\frac{1}{290})\\ & n_f-n_i=\frac{{10}^5\times 30}{8.31}\times \frac{-10}{300\times 290}=\frac{-{10}^5}{290\times 8.31}\end{array}$
change in the Number of molecules
$=\frac{-{10}^5\times 6.023\times {10}^{23}}{290\times 8.31}=-2.5\times {10}^{25}$

Hence, the answer is the option (4).

Summary

Not all gases can be represented by the ideal gas equation; only ideal gases can. It cannot be used for gases that are exceptionally large, exhibit strong intermolecular forces, collide in an inelastic manner, or lack kinetic energy proportional to the ideal gas's absolute temperature. Thus, there are restrictions on the Ideal Gas Equation. Because the molecular mass and the intermolecular forces are important under these circumstances, it is ineffective at low temperatures, high densities, and extremely high pressures. Heavy gases, such as refrigerants, and gases with extremely high intermolecular forces, such as water vapour, cannot be treated with it.