21. (JEE Main 2021 (Online) 24th February Evening Shift )

The root mean square speed of molecules of a given mass of a gas at 27 $^{\circ}$ C and 1 atmosphere pressure is 200 ms^{$-$ 1}. The root mean square speed of molecules of the gas at 127 $^{\circ}$ C and 2 atmosphere pressure is $\frac{x}{\sqrt{3}}$ ms^{$-$ 1}. The value of x will be _________.

Correct answer is (400)

Given, T₁ = 27 $^{\circ}$ C = 27 + 273 = 300K, p₁ = 1 atm, v₁ = 200 ms^{$-$ 1}, T₂ = 127 $^{\circ}$ C = 400 K, p₂ = 12 atm, v₂ = ?

As we know that,

Root mean square speed, $v_{r m s} = \sqrt{\frac{3 R T}{m}}$

$∴$ $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$

$\Rightarrow v_{2} = \sqrt{\frac{4}{3}} v_{1} = \frac{2}{\sqrt{3}} \times 200 = \frac{400}{\sqrt{3}}$ ms^{$-$ 1}

$\Rightarrow \frac{x}{\sqrt{3}} = \frac{400}{\sqrt{3}} \Rightarrow x = 400$

22. (JEE Main 2020 (Online) 5th September Evening Slot )

Nitrogen gas is at 300^oC temperature. The temperature (in K) at which the rms speed of a H₂ molecule would be equal to the rms speed of a nitrogen molecule, is _______.
(Molar mass of N₂ gas 28 g).

Correct answer is (40 to 41)

V_rms = $\sqrt{\frac{3 R T}{M}}$

V_N₂ = $\sqrt{\frac{3 R (573)}{28}}$

V_H₂ = $\sqrt{\frac{3 R T}{2}}$

Given, V_N₂ = V_H₂

$\sqrt{\frac{3 R T}{2}}$ = $\sqrt{\frac{3 R (573)}{28}}$

$\Rightarrow$ $\frac{T}{2} = \frac{573}{28}$

$\Rightarrow$ T = 41 K

23. (JEE Main 2019 (Online) 9th April Morning Slot )

For a given gas at 1 atm pressure, rms speed of the molecule is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be :

(A) 100 m/s

(B) 100 $\sqrt{5}$ m/s

(C) 80 $\sqrt{5}$ m/s

(D) 80 m/s

Correct answer is (B)

$V_{r m s} = \sqrt{\frac{3 R T}{M_{w}}}$

$\Rightarrow V_{r m s} \propto \sqrt{T}$

Now, $\frac{v}{200} = \sqrt{\frac{500}{400}}$

$\Rightarrow \frac{v}{200} = \frac{\sqrt{5}}{2}$

$\Rightarrow v = 100 \sqrt{5}$ m/s

24. (JEE Main 2019 (Online) 8th April Evening Slot )

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to :
[Boltzmann Constant k_B = 1.38 × 10^–23 J/K Avogadro Number N_A = 6.02 × 10²⁶ /kg Radius of Earth : 6.4 × 10⁶ m Gravitational acceleration on Earth = 10ms^–2]

(A) 3 × 10⁵ K

(B) 10⁴ K

(C) 650 K

(D) 800 K

Correct answer is (B)

$V_{r m s} = \sqrt{\frac{3 R T}{M}} = 11.2 \times 10^{3} m / s$

$\Rightarrow$ $T = \frac{M}{3 R} \times {(11.2 \times 10^{3})}^{2}$

= $\frac{2 \times 10^{- 3}}{3 \times 8.3} \times 125.44 \times 10^{6} = 10^{4} K$

25. (JEE Main 2019 (Online) 9th January Evening Slot )

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27^oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]

(A) 0.9 kJ

(B) 6 kJ

(C) 10 kJ

(D) 14 kJ

Correct answer is (C)

We know,

V_rms $\propto$ $\sqrt{T}$

So, to make V_rms double we have to make temperature 4 times.

$∴$ Final temperature = 300 $\times$ 4 = 1200 K

As N₂ gas present in the closed vessel

So it is a isochoric process.

$∴$ Q = _nC_v $Δ$ T

= $\frac{15}{28} \times (\frac{5}{2} R) (1200 - 300)$

= 10000 J

= 10 kJ

Kinetic Theory of Gases