Question

At what temperature will the oxygen molecules have the same root mean square speed as
hydrogen molecules at 200 K
(1) 800 K
(2) 1600 K
(3) 2400 K
(4) 3200 K​

Answer 1

Answer:

Formulas used:

RMS velocity for gases is given by the following:

$v \: rms \: = \sqrt{ \frac{3RT}{m} }$

R => gas constant , T => temperature and m => molar mass of gas.

Calculation:

As per the question :

V rms(oxygen) = V rms(hydrogen)

=> √{(3RT)/32} = √{(3R 200)/2}

Squaring on both sides :

=> 3RT/32 = 3R (200)/2

=> T/32 = 100

=> T = 3200 K

So final answer is 3200 K.

Option D) is correct.

Answer 2

Answer:

• At Temperature (T₁) = 3200 K the Root mean square velocity will be same.

Given:

1. Temperature of Hydrogen (T₂) = 200 K.

Explanation:

$\rule{300}{1.5}$

From the Formula we Know,

$\large\bigstar \: {\boxed{\tt V_{rms} = \sqrt{\dfrac{3RT}{M}}}}$

$\bold{Here}\begin{cases}\text{R Denotes Gas Constant} \\ \text{T Denotes Temperature} \\ \text{M Denotes Molar Mass}\end{cases}$

$\large{\boxed{\tt V_{rms} = \sqrt{\dfrac{3RT}{M}}}}$

Now,

$\large{\tt \longmapsto (V_{rms})_{Oxygen} = (V_{rms})_{Hydrogen}}$

Substituting the values,

$\large{\tt \longmapsto \sqrt{\dfrac{3RT_1}{M_1}} = \sqrt{\dfrac{3RT_2}{M_2}}}$

Substituting the values,

$\large{\tt \longmapsto \sqrt{\dfrac{3R \times T_1}{32}} = \sqrt{\dfrac{3RT_2}{2}}}$

• M₁ = 32 Grams (Molecular Form)
• M₂ = 2 Grams (Molecular Form)

Squaring on both sides,

$\large{\tt \longmapsto \Bigg(\sqrt{\dfrac{3R \times T_1}{32}}\Bigg)^2 = \Bigg(\sqrt{\dfrac{3RT_2}{2}}\Bigg)^2}$

$\large{\tt \longmapsto \dfrac{3R \times T_1}{32} = \dfrac{3R \times T_2}{2}}$

$\large{\tt \longmapsto \dfrac{3R \times T_1}{32} = \dfrac{3R \times 200}{2}}$

• T₂ = 200 K.

Now,

$\large{\tt \longmapsto \dfrac{\cancel{3R} \times T_1}{32} = \dfrac{\cancel{3R} \times 200}{2}}$

$\large{\tt \longmapsto \dfrac{T_1}{32} = \dfrac{200}{2}}$

$\large{\tt \longmapsto \dfrac{T_1}{32} = \cancel{\dfrac{200}{2}}}$

$\large{\tt \longmapsto \dfrac{T_1}{32} = 100}$

$\large{\tt \longmapsto T_1 = 100 \times 32}$

$\large\longmapsto{\underline{\boxed{\red{\tt T_1 = 3200 \: K}}}}$

∴ At Temperature (T₁) = 3200 K the Root mean square velocity will be same.

$\rule{300}{1.5}$