JEE PHYSICS MCQ: KINETIC THEORY OF GASES

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The average translational kinetic energy of air molecules is

0.040 \mathrm{eV}\,\, \left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)

. Calculate the temperature of the air. Boltzmann constant

k=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}

View answer and explanation

Correct option: C

310 K

Option C (310 K) is correct because the average translational kinetic energy is directly proportional to the temperature, and the calculation yields approximately 310 K.

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Assertion: The pressure exerted by an enclosed ideal gas does not depend on the shape of the container.
Reason: The pressure of an ideal gas depends on the number moles, temperature and volume of the enclosed If

c_s

is the velocity of sound in air and

c

is the rms velocity, then Four particles have velocities

1,0,2,3 \mathrm{~m} / \mathrm{s}

. The root mean square of the particles is (in

\mathrm{m} / \mathrm{s}

) The translational degrees of freedom

(f_t)

and rotational degrees of freedom

(f_r)

{CH}_4

molecule are: On any planet, the presence of atmosphere implies (

{C_{rms}} =

= root mean square velocity of molecules and

{V_e} =

escape velocity) At a given temperature, the pressure of an ideal gas of density

\rho

is proportional to