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Derivation of Ideal Gas Equation MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Ideal Gas Equation is considered one of the most asked concept.

  • 64 Questions around this concept.

Solve by difficulty

Hydrogen bomb is based on the principle of

What is the density of N2 gas at 227C and 5.00 atm pressure? ( R=0.0821 Latm K1 mol1 )

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 ]

An open vessel at 27C is heated until two fifth of the air (assumed as an ideal gas ) 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 is: (in K)

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 OC ] is introduced considering the ideal Gas behavior.

The total volume (in lite) ot gas at 0C is close to -

 The Reduced Temperature =θ=TTA The Reduced Pressure =Γ=PPA The Reduced volume =ϕ=VVA

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

Two glass bulbs A and B are connected by a very Small Tube having a stop-cock. Bulb A has a volume 50 cm3 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

A 5.0L elast contains 32 g at oxygen at 27C [Assume O2 is betwaing ideally] The pressure involev the elask in bar is[Given R=.0831 Lbar K1 mol1 ]

 

 

Choose the correct option for the total bossing in a mixture of 6gO2 and 4gH2 confined in a total volume of one litre at OC is ?

[R=082Latm mol1H1, T=273k]
 

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The volume occupied byg g of water vapour at 300C and 1 bar pressure will be
[R=.083barLk1 mol11]

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:}\\\\

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

\text { Rearanging the above equation, we get } \mathrm{M}=\frac{\mathrm{dRT}}{\mathrm{P}}

\text {The above equation gives the relation between the density and molar mass of a gaseous substance. }

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