The kinetic theory of gases

explains the behaviour of an ideal gas.
describes the motion of a single atom or molecule.
relates the temperature of the gas with K.E. of atoms of the gas
All of the above

The ratio of principal molar heat capacities of a gas is maximum for

adiatomic gas
a monatomic gas
apolyatomic gas having linear molecules.
apolyatomic gas having non-linear molecules.

The correct statement of the law of equipartition of energy is

the total energy of a gas is equally divided among all the molecules.
The gas possess equal energies in all the three directions x,y and z-axis.
the total energy of a gas is equally divided between kinetic and potential energies.
the total kinetic energy of a gas molecules is equally divided among translational and rotational kinetic energies.

The internal energy of an ideal gas is

the sum of total kinetic and potential energies.
the total translational kinetic energy.
the total kinetic energy of randomly moving molecules.
the total kinetic energy of gas molecules.

The total degree of freedom of a CO₂ gas molecule is

In kinetic theory of gases, it is assumed that molecules

have same mass but can have different volume
have same volume but mass can bedifferent
have different mass as well as volume
have same mass but negligible volume.

Gases exert pressure on the walls of the container because the gas molecules

possess momentum
collide with each other
have finite volume
obey gas laws.

Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T.The mass of a molecule is ‘m’ Neglecting vibrational energy terms, the false statement is

a molecule can have a speed greater than \(\sqrt{2}\)v
v is proportional to \(\sqrt{T}\)
the average rotational K.E. of a molecule is ¹⁄₄ mv²
the average K.E. of a molecule is ⁵⁄₆mv²

If masses of all molecule of a gas are halved and their speed doubled then the ratio of initial and final pressure will be

2 : 1
1 : 2
4 : 1
1 : 4

Solution

Since P=\(\frac{1}{3}\frac{mNv^{2}rms}{V}\)

since m is halved & speed is doubled so pressure become twice.

On a particular day, the relative humidity is 100% and the room temperature is 30°C, then the dew point is

70ºC
30ºC
100ºC
0ºC

Solution

Relative humidity (R.H)=\(\frac{S.V.P\: at\: dew\: point}{S.V.P\: at\: room\: temp.}\)

since here R.H is 100%

⇒room temp. = dew point =30°C

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