Molecular velocities of two gases at the same temperature (T) are u_1 and u_2 . Their masses are m_1 and m_2 respectively. Which of the following expressions is correct at temperature T ?

Correct answer: (D) m_1 u_1^2=m_2 u_2^2. We know u × √(T/m) but as T is same for both gases Hence u ∝ 1/(√m) ∴ fracu_1u_2=√fracm_2~m_1 or fracu_1^2u_2^2=fracm_2~m_1 or m_1 u_1^2=m_2 u_2^2

Q7 / 155 · WBJEE 2021 · Chemistry

Molecular velocities of two gases at the same temperature $(\mathrm{T})$ are $\mathrm{u}_{1}$ and $\mathrm{u}_{2} .$ Their masses are $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ respectively. Which of the following expressions is correct at temperature $\mathrm{T} ?$

A
$\quad \frac{\mathrm{m}_{1}}{\mathrm{u}_{1}^{2}}=\frac{\mathrm{m}_{2}}{\mathrm{u}_{2}^{2}}$
B
$\mathrm{m}_{1} \mathrm{u}_{1}=\mathrm{m}_{2} \mathrm{u}_{2}$
C
$\frac{\mathrm{m}_{1}}{\mathrm{u}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{u}_{2}}$
Correct answer
D
$\mathrm{m}_{1} \mathrm{u}_{1}{ }^{2}=\mathrm{m}_{2} \mathrm{u}_{2}{ }^{2}$

Solution

We know $u \times \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ but as $\mathrm{T}$ is same for both gases Hence u $\propto \frac{1}{\sqrt{\mathrm{m}}}$ $\therefore \frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}=\sqrt{\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}}$ or $\frac{\mathrm{u}_{1}^{2}}{\mathrm{u}_{2}^{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ or $\mathrm{m}_{1} \mathrm{u}_{1}^{2}=\mathrm{m}_{2} \mathrm{u}_{2}^{2}$

Hence, the correct answer is option: 'D'.

Molecular velocities of two gases at the same temperature $(\mathrm{T})$ are $\mathrm{u}_{1}$ and $\mathrm{u}_{2} .$ Their masses are $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ respectively. Which of the following expressions is correct at temperature $\mathrm{T} ?$

Solution

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