Q4 / 90 · JEE Main 2019 (10 Apr Shift 2) · Chemistry

Points $I, I I$ and $I I I$ in the following plot respectively correspond to ( $V_{m p} :$ most probable velocity)

A
$V_{m p}$ of $N_{2} (300 K);$ $V_{m p}$ of $H_{2} (300 K); V_{m p}$ of $O_{2} (400 K);$
B
$V_{m p}$ of $H_{2} (300 K);$ $V_{m p}$ of $N_{2} (300 K); V_{m p}$ of $O_{2} (400 K);$
Correct answer
C
$V_{m p}$ of $N_{2} (300 K);$ $V_{m p}$ of $O_{2} (400 K); V_{m p}$ of $H_{2} (300 K);$
D
$V_{m p}$ of $O_{2} (400 K);$ $V_{m p}$ of $N_{2} (300 K); V_{m p}$ of $H_{2} (300 K);$

Solution

$V_{mp} = \sqrt{\frac{2 R T}{M}} \propto \sqrt{\frac{T}{M}}$

For $N_{2}$ at $300 K,$ $V_{mp} \propto \sqrt{\frac{T}{M}} \propto \sqrt{\frac{300}{28}} = 3.2$

For $O_{2}$ at $300 K,$ $V_{mp} \propto \sqrt{\frac{T}{M}} \propto \sqrt{\frac{400}{32}} = 3.53$

For $H_{2}$ at $300 K,$ $V_{mp} \propto \sqrt{\frac{T}{M}} \propto \sqrt{\frac{300}{2}} = 12.24$

$V_{mp}$ of $N_{2} < U_{m p}$ of $O_{2} < U_{m p}$ of $H_{2}$

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

Points I, II and III in the following plot respectively correspond to (Vmp: most probable velocity)

Solution

Relevant Questions

Points $I, I I$ and $I I I$ in the following plot respectively correspond to ( $V_{m p} :$ most probable velocity)