6. (JEE Main 2022 (Online) 25th July Morning Shift )

Following statements are given :

(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.

(B) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.

(C) The average kinetic energy of a gas molecule decreases with increase in volume.

(D) Pressure of a gas increases with increase in temperature at constant pressure.

(E) The volume of gas decreases with increase in temperature.

Choose the correct answer from the options given below :

(A) (A) and (D) only

(B) (A), (B) and (D) only

(C) (B) and (D) only

(D) (A), (B) and (E) only

Correct answer is (A)

Because KE $\propto$ T so A is correct, B is incorrect, statement C cannot be said, statement D is contradicting itself, statement E is incorrect (Isothermal process)

So no answer correct (Bonus)

If the statement of D would have been.

"Pressure of gas increases with increase in temperature at constant volume, "then statement D would have been correct, so in that case answer would have been 'A'.

7. (JEE Main 2022 (Online) 26th June Evening Shift )

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27 $^{\circ}$ C. The ratio of their average kinetic energy per molecule respectively will be :

(A) 3 : 2

(B) 9 : 4

(C) 2 : 3

(D) 1 : 1

Correct answer is (D)

$K E_{a v g} = \frac{3}{2} k T$ (At lower temperature)

As temperature is same for both the gases.

$\Rightarrow$ Both gases will have same average kinetic energy.

$\Rightarrow \frac{{(K E_{a v g})}_{\arg o n}}{{(K E_{a v g})}_{o x y g e n}} = \frac{1}{1}$

8. (JEE Main 2022 (Online) 28th June Morning Shift )

The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be _____________ J. (Given R = 8.31 J/mol.K)

Correct answer is (7479)

$U = 2 (\frac{3}{2} R) 300$

$= 3 \times 8.31 \times 300$

$= 7479$ J

9. (JEE Main 2021 (Online) 27th July Morning Shift )

The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2 $\times$ 10^{$-$ 9} J per molecules is :

(A) 0.75 $\times$ 10¹¹

(B) 3 $\times$ 10¹¹

(C) 1.5 $\times$ 10¹¹

(D) 6 $\times$ 10¹¹

Correct answer is (C)

KE = $\frac{3}{2} k T$

PV = $\frac{N}{N_{A}} R T$

N = $\frac{P V}{k T}$

= N = 1.5 $\times$ 10¹¹

10. (JEE Main 2021 (Online) 22th July Evening Shift )

What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T?

(A) $\frac{3}{2} k_{B} T$

(B) $k_{B} T$

(C) $\frac{2}{3} k_{B} T$

(D) $\frac{1}{2} k_{B} T$

Correct answer is (A)

For a monoatomic ideal gas, the average kinetic energy per molecule is determined by the equipartition theorem. This theorem states that the energy is equally distributed among all the available degrees of freedom.

A monoatomic gas has three translational degrees of freedom, corresponding to motion in the x, y, and z directions. Each degree of freedom contributes an average energy of $\frac{1}{2} k_{B} T$ , where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature.

So for a monoatomic gas with three translational degrees of freedom, the average energy per molecule is: $\frac{3}{2} k_{B} T .$

Kinetic Theory of Gases