Tag: behaviour of perfect gas and kinetic theory of gases

There will be three degrees of freedom. Two are along x-direction and y-directions due to translation and the last degree of freedom due to angular rotation as the  man climbs up.

For a system of N particles having K independent relations among them, the degrees of freedom of the system is given by 3N-K. 3N  is due to three degrees of freedom associated with each particle if all the particles are independent of each other (i.e K=0) and due to K relation among them, degrees of freedom reduces to 3N-K

Units of R, N and K are $ Joule \times mole^{-1} \times K^{-1} $, $ mole^{-1} $ and $Joule \times  K^{-1}$
So $ R = N \times K $

A monoatomic gas has 3 degrees of freedom for translation motion.
Hence, option A is correct.

At a given temperature, the pressure of a container is determined by the number of times gas molecules strike the container walls. If the gas is compressed to a smaller volume, then the same number of molecules will strike against a smaller surface area; the number of collisions against the container will increase, and, by extension, the pressure will increase as well.
Increasing the kinetic energy of the particles will increase the pressure of the gas.
So, the internal energy of a gas Is the total kinetic energy of randomly moving molecules.
Hence, option C is correct.

Vibrational degree of freedom:
(a) For linear molecule = 3N - 5.
(b) For non-linear molecule = 3N - 6. where N is no. of atoms present in a molecule.

Diatomic, like your oxygen molecule, can move in three dimensions and spin in two directions.

 Monoatomic gas has only 3 degree of freedom (all translational)

$\because \gamma =1+\dfrac { 2 }{ f } $
$\Longrightarrow \dfrac { 2 }{ f } =\gamma -1\Longrightarrow f=\dfrac { 2 }{ \gamma -1 } $