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Kinetic Energy Of Ideal Gas MCQ - Practice Questions with Answers

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

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  • Kinetic energy of ideal gas is considered one of the most asked concept.

  • 41 Questions around this concept.

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 A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 00C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than radius of the earth. (kB is Boltzmann constant)                    

 

An HCl molecule has rotational, translational and vibrational motions. If the RMS velocity of HCl molecules in its gaseous phase is \overline{v}m is its mass and k_{B} is Boltzmann constant, then its temperature will be :

When a flask containing Helium and Argon in the Ratio of 4:3 by mass. The Temperature of the mixture is 27^{\circ}C. Find the ratio of average kinetic Energy per molecule of two gases.

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What is the Energy of diatomic gas whose pressure =8 \times 10^4 \mathrm{~N} / \mathrm{m}^2 and density of the gas = 4\, kg/m^{3}?

 

What is the value of thermal velocity ms-1 of the Helium atom at room temperature?
{\left[K_B=1.4 \times 10^{-23}\right.} \left.\mathrm{J\,K}^{-1}, m_{\mathrm{ru}}=7 \times 10^{-27} \mathrm{~kg}\right] \\

A cylinder contains N moles of diatonic gas at Temperature T. Heat is supplied such that temperature remains constant but n  moles of gas are converted to monoatomic gas. What will be the change in Kinetic Energy?

 

A gas has volume V and pressure P. The total translational kinetic energy of all the molecules of the gas is -

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The graph which represents the variation of the mean kinetic energy of molecules with temperature C is -

At 27^{0}C temperature, the kinetic energy of an ideal gas is E_{1}. If the temperature is increased 327^{0}C, then the kinetic energy will be - 

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The translational kinetic energy of 1gm a molecule of a gas, at temperature, 300K is:

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Kinetic energy of ideal gas

The kinetic energy of ideal gas-

In ideal gases, the molecules are considered as point particles. The point particles can have only translational motion and thus only
translational energy. So for an ideal gas, the internal energy can only be translational kinetic energy.

Hence kinetic energy (or internal energy) of   n mole ideal gas

E=\frac{1}{2} nM v_{m s}^{2}=\frac{1}{2} nM \times \frac{3 R T}{M}=\frac{3}{2} nR T

 1.  kinetic energy of 1 molecule 

        E= \frac{3}{2} k T

        where   k = Boltzmann’s constant

and  k=1.38 \times 10^{-23} J / K

  i.e  Kinetic energy per molecule of gas does not depends upon the mass of the molecule but only depends upon the temperature of the gas.

2. kinetic energy of 1 mole ideal gas

E= \frac{3}{2} R T

i.e Kinetic energy per mole of gas depends only upon the temperature of the gas.

3. At T = 0, E = 0 i.e. at absolute zero the molecular motion stops.

  • The relation between pressure and kinetic energy

As we know P=\frac{1}{3} \frac{m N}{V} v_{m s}^{2}=\frac{1}{3} \frac{M}{V} v_{m s}^{2} \Rightarrow P=\frac{1}{3} \rho v_{m s}^{2}  ......(1)

And K.E. per unit volume= E=\frac{1}{2}\left(\frac{M}{V}\right) v_{m s}^{2}=\frac{1}{2} \rho v_{m s}^{2}...... (2)

So from equation (1) and (2), we can say that P=\frac{2}{3}E

i.e. the pressure exerted by an ideal gas is numerically equal to the two-third of the mean kinetic energy of translation per unit volume of the gas.

  • Law of Equipartition of Energy-

According to this law, for any system in thermal equilibrium, the total energy is equally distributed among its various degrees of freedom.

I.e Each degree of freedom is associated with energy E= \frac{1}{2} k T

1. At a given temperature T, all ideal gas molecules will have the same average translational kinetic energy as \frac{3}{2}kT

2. Different energies of a system of the degree of freedom f are as follows

    \begin{array}{l}{\text { (i) Total energy associated with each molecule }=\frac{f}{2} k T} \\ \\ {\text { (ii) Total energy associated with } N \text { molecules }=\frac{f}{2} N k T} \\ \\ {\text { (iii) Total energy associated with } 1 \text { mole }=\frac{f}{2} R T} \\ \\ {\text { (iv) Total energy associated with } n \text { mole }=\frac{nf}{2} R T} \end{array}

 

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Kinetic energy of ideal gas

Physics Part II Textbook for Class XI

Page No. : 325

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