Derivation of Ideal Gas Equation MCQ - Practice Questions & Answers

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Ideal Gas Equation is considered one of the most asked concept.
63 Questions around this concept.

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Hydrogen bomb is based on the principle of

nuclear fission

natural radioactivity

nuclear fusion

artificial radioactivity

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What is the density of $N_2$ gas at $227^{\circ} \mathrm{C}$ and 5.00 atm pressure? ( $\mathrm{R}=0.0821$ Latm $K^{-1} \mathrm{~mol}^{-1}$ )

$0.29 \mathrm{~g} / \mathrm{L}$

$1.40 \mathrm{~g} / \mathrm{L}$

$2.81 \mathrm{~g} / \mathrm{L}$

$3.41 \mathrm{~g} / \mathrm{L}$

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Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1 mol litre^-1. [ R = 0.082 litre atm mol^-1 K^-1 ]

at S.T.P

when V = 22.42 L

when T = 12 K

Impossible under any condition

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To an evacuated vessel with a Movable Piston under external pressure of 1 atm.
5 mole at He and 2 moles at untrnoum compound [vapor pressure 68 atm at $\mathrm{O}^{\circ} \mathrm{C}$ ] is introduced considering the ideal Gas behavior.

The total volume (in lite) ot gas at $0^{\circ} \mathrm{C}$ is close to -

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$\begin{aligned} & \text { The Reduced Temperature }=\theta=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{A}}} \\ & \text { The Reduced Pressure }=\Gamma=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{A}}} \\ & \text { The Reduced volume }=\phi=\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{A}}}\end{aligned}$

Hence, it can be said that the Reduced equation of state may be given as $\mathrm{\rightarrow }$

$\left(\frac{\pi}{3}+\frac{1}{\phi}\right)(\phi-1)=\frac{8}{3} \theta$

$\left(\frac{\pi}{4}+\frac{1}{\phi}\right)(3 \theta-1)=\frac{8}{3} \phi$

$\left(\frac{M}{3}+\frac{1}{\phi}\right)(\phi-1)=\frac{8}{3} \theta$

$\left(\frac{r}{5}+\frac{1}{\phi}\right)(4 \phi-1)=\frac{8}{4} \phi$

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Two glass bulbs A and B are connected by a very Small Tube having a stop-cock. Bulb A has a volume $50 \mathrm{~cm}^3$ and contained the gas, mutule bulb B was empty. On opening the stop-cocts. the pressure fall down to $50 \%$ The volume at bulb B must be

50 cm³

150 cm³

100 cm³

250 cm³

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A $5.0 L$ elast contains 32 g at oxygen at $27^{\circ} \mathrm{C}$ [Assume $\mathrm{O}_2$ is betwaing ideally] The pressure involev the elask in bar is[Given $\mathrm{R}=.0831 \mathrm{~L}_{\text {bar }} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]

5.986

4.986

3.896

4.896

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Choose the correct option for the total bossing in a mixture of $6 \mathrm{gO}_2$ and $4 \mathrm{gH}_2$ confined in a total volume of one litre at $\mathrm{O}^{\circ} \mathrm{C}$ is ?

$
{ }_{[ } \mathrm{R}=082 \mathrm{Latm} \mathrm{~mol}^{-1} \mathrm{H}^{-1}, \mathrm{~T}=273 \mathrm{k}_{]}
$

2.602

46.8

25.18

48.8

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The volume occupied by$g \mathrm{~g}$ of water vapour at $300^{\circ} \mathrm{C}$ and 1 bar pressure will be
[$\mathrm{R}=.083{\mathrm{bar} \mathrm{Lk}^{-1} \mathrm{~mol}^{-1}}^{-1}$]

5.37

5.35

5.23

5.20

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A container of volume 2.24 L can with stand a Maximum Pressure of atm at 298 K before exploding The Maximum Amount of oxygen sing) that can be safety put in this container at this temperature is closet to?

2.88

4.2

2.66

1.4

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Concepts Covered - 0

Ideal Gas Equation

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

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

$\begin{array}{l}{V \propto \frac{n T}{P}} \\\\ {V=R \frac{n T}{P}}\end{array}$

$\begin{array}{l}{\text { 'R' is the proportionality constant. On rearranging the above equation we get: }} \\ {\text { PV = nRT }} \\ {\text { This is the ideal gas equation as it is obeyed by the hypothetical gases called ideal gases under all conditions }}\end{array}$

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

$\begin{array}{l}{\text {For a single molecule, gas constant is known as }} \\ {\text {Boltzmann constant }(\mathrm{k}) \text { . }} \\ {\mathrm{k}=\mathrm{R} \mathrm{N}_{0}} \\ {\: \: \: \, =1.38 \times 10^{-3} \mathrm{J} / \mathrm{deg-abs} / \mathrm{molecule}} \\ {\: \: \: \, =1.38 \times 10^{-16} \text { erg/deg-abs/molecule }}\end{array}$

Combined gas law:
The 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 . With the change in either of the variables, all the three variables change to . Then we can write:

Combining the both above equation we get

The above relation is called the combined gas law.

Density and Molar Mass of a Gaseous Substance

$\\\mathrm{Ideal \: gas\: equation\: is \: PV = nRT \quad \quad \quad.......(i)}\\\mathrm{\text {On rearranging the above equation, we get }}\\\\\mathrm{\frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\: \: \: .......(ii)}\\\\\mathrm{n\left(N_{0} . \text { of moles }\right)=\frac{\text { Given mass }(m)}{\text { Molar mass }(M)}\quad....(iii)}\\\\\text {Putting the value of 'n' from equation (iii) in equation (ii), we get: }\\\\\frac{\mathrm{m}}{\mathrm{MV}}=\frac{\mathrm{P}}{\mathrm{RT}}\quad \quad \quad \quad \quad \quad \quad \quad \quad\quad\quad\: \: \: \: .........(iv)\\\\\text {We know that density (d) is mass (m) per unit volume (V) }\\\\\mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad .........(v)\\\\\text {Replacing } \frac{\mathrm{m}}{\mathrm{V}} \text { in eq. (iv) with d(density), then equation (iv) becomes:}\\\\$