Essential points to be remembered in kinetic theory of gases were discussed in the post dated 13^{th} March 2008. Questions on kinetic theory of gases were discussed subsequently. You can access all posts related to kinetic theory of gases by clicking on the label, ‘kinetic theory’ below this post. To access older posts you need to click on the ‘older posts’ button.

Today we will discuss a few more typical multiple choice questions on kinetic theory of gases:

(1) The root mean square (R.M.S.) speed *v *of the molecules of an ideal gas is given by the expressions,

*v =* √(3*RT/M** *) and

*v =* √(3*kT/m** *) where *R* is universal gas constant, *T* is the absolute (Kelvin) temperature, *M* is the molar mass, *k* is Boltzman’s constant and* m* is the molecular mass. The R.M.S. speed of oxygen molecules (O_{2}) at temperature *T*_{1} is *v*_{1}. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).

(a) *v*_{1}/2

(b) *v*_{1}

(c) √2 *v*_{1}

(d) 2*v*_{1}

(e) 4*v*_{1}

We have *v*_{1}* =* √(3*RT*_{1}*/M** *) or

*v*_{1}* =* √(3*kT*_{1}*/m** *)

On dissociation the molar mass as well as the molecular mass gets halved. Using the second equation, the R.M.S. speed *v* after dissociation is given by

*v = *√[3*k*×2*T*_{1}*/ *(*m/*2* *)] = 2√(3*kT*_{1}*/m** *) = 2*v*_{1}_{}

(2) Four moles of an ideal diatomic gas is heated at constant volume from 20º C to 30º C. The molar specific heat of the gas at constant pressure (*C*_{p}) is 30.3 Jmol^{–1}K^{–1} and the universal gas constant (*R*) is 8.3 Jmol^{–1}K^{–1}. The increase in internal energy of the gas is

(a) 80.3 J

(b) 303 J

(c) 332 J

(d) 880 J

(e) 1212 J* * * *

The increase in internal energy is *MC*_{v}∆*T* where *M* is the mass of the sample of the gas, *C*_{v }is the specific heat at constant volume and ∆*T* is the rise in temperature of the gas. If we use the *molar specific heat* of the gas at constant volume for *C*_{v}, the *number of moles* in the sample of the gas is to be used in the place of *M*.

Now, *C*_{v} = *C*_{p }–* R* = 30.3 – 8.3 = 22 Jmol^{–1}K^{–1}.

Therefore, the increase in internal energy of the gas is 4×22×10 = 880 J.

(3) In the case of real gases, the equation of state, *PV = RT* (where *P*,* V *and* T* are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure *P* and the measured volume *V*. The corrections for *P* and *V* arise respectively due to

(a) intermolecular attraction and the size of molecules

(b) size of molecules and expansion of the container

(c) expansion of the container and intermolecular attraction

(d) kinetic energy of molecules and collision of molecules

(e) intermolecular attraction and collision of molecules

In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for *P* arises due to intermolecular attraction.

The entire volume *V* of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.

The correct option is (a).

(4) Gases exert pressure on the walls of the container because the gas molecules

(a) collide one another

(b) exert intermolecular attraction

(c) possess momentum

(d) expand on absorbing heat

(e) exert repulsive force

Because of the momentum of the gas molecules, they collide with the walls of the containing vessel and momentum transfer takes place, resulting in a force on the walls. Pressure is force per unit area. The basic reason for the pressure is the momentum of the gas molecules [Option (c)].

Now, see similar questions with solution here.

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