11. (JEE Main 2021 (Online) 18th March Morning Shift )

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (k_B is Boltzmann constant)

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

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

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

(D) $k_{B} T$

Correct answer is (A)

Energy associated with each digress of freedom is $\frac{1}{2} k_{B} T$ .

12. (JEE Main 2021 (Online) 17th March Morning Shift )

Two ideal polyatomic gases at temperatures T₁ and T₂ are mixed so that there is no loss of energy. If F₁ and F₂, m₁ and m₂, n₁ and n₂ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :

(A) $\frac{n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2}}{F_{1} + F_{2}}$

(B) $\frac{n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2}}{n_{1} F_{1} + n_{2} F_{2}}$

(C) $\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

(D) $\frac{n_{1} F_{1} T_{1} + n_{2} F_{2} T_{2}}{n_{1} + n_{2}}$

Correct answer is (B)

Initial internal energy = Final internal energy

$\frac{F_{1}}{2} n_{1} R T_{1} + \frac{F_{2}}{2} n_{2} R T_{2} = \frac{F_{1}}{2} n_{1} R T + \frac{F_{2}}{2} n_{2} R T$

$\Rightarrow$ $T = \frac{F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2}}{F_{1} n_{1} + F_{2} n_{2}}$

13. (JEE Main 2021 (Online) 26th February Evening Shift )

The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U $=$ 3PV + 4. The gas is :

(A) either monoatomic or diatomic.

(B) monoatomic only.

(C) polyatomic only.

(D) diatomic only.

Correct answer is (C)

U = 3PV + 4

$\Rightarrow$ $\frac{n f}{2}$ RT = 3PV + 4

$\Rightarrow$ $\frac{f}{2}$ PV = 3PV + 4

$\Rightarrow$ f = 6 + $\frac{8}{P V}$

Since degree of freedom is more than 6 therefore gas is polyatomic.

14. (JEE Main 2021 (Online) 25th February Evening Shift )

Given below are two statements :

Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.

Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.

In the light of the above statements, choose the correct answer from the options given below :

(A) Both Statement I and Statement II are true

(B) Both Statement I and Statement II are false

(C) Statement I is true but Statement II is false.

(D) Statement I is false but Statement II is true.

Correct answer is (C)

The translational kinetic energy & rotational kinetic energy both obey Maxwell's distribution independent of each other.

T.K.E. of diatomic molecules = $\frac{3}{2} k T$

R.K.E. of diatomic molecules = $\frac{2}{2} k T$

So statement I is true but statement II is false.

15. (JEE Main 2021 (Online) 1st September Evening Shift )

The average translational kinetic energy of N₂ gas molecules at ............. $^{\circ}$ C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k_B = 1.38 $\times$ 10^{$-$ 23} J/K) (Fill the nearest integer).

Correct answer is (500)

Given, the average translational kinetic energy of dinitrogen (N₂) = Kinetic energy of an electron .... (i)

Translational kinetic energy of dinitrogen (N₂)

$K E = \frac{3}{2} K_{B} T$

Here, T = temperature of the gas,

and K_B = Boltzmann constant.

Kinetic energy of an electron = eV

Given, the potential differential of an electron, V = 0.1 V

Substituting the values in the Eq. (i), we get

$\frac{3}{2} K_{B} T = e V$

$\Rightarrow \frac{3}{2} \times 1.38 \times 10^{- 23} \times T = 1.6 \times 10^{- 19} \times (0.1)$

$T = 773 K = 773 - 273^{\circ} C = 500^{\circ} C$

Kinetic Theory of Gases