SATHEE: States of Matter - Result Question 97

States of Matter - Result Question 97

####1. Points I, II and III in the following plot respectively correspond to ( $v_{m p} :$ most probable velocity)

(2019 Main, 10 April II)

(a) $v_{m p}$ of $H_{2} (300 K)$ ; $v_{m p}$ of $N_{2} (300 K)$ ; $v_{m p}$ of $O_{2} (400 K)$

(b) $v_{m p}$ of $O_{2} (400 K); v_{m p}$ of $N_{2} (300 K); v_{m p}$ of $H_{2} (300 K)$

(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 $N_{2} (300 K); v_{m p}$ of $H_{2} (300 K); v_{m p}$ of $O_{2} (400 K)$

Show Answer

Answer:

Correct Answer: 1. (d)

Solution:

Key Idea From kinetic gas equation,

Most probable velocity $(v_{m p}) = \sqrt{\frac{2 R T}{M}}$

where, $R =$ gas constant, $T =$ temperature, $M =$ molecular mass

		$v_{m p} = \sqrt{\frac{2 R T}{M}}$ , i.e. $v_{m p} \propto \sqrt{\frac{T}{M}}$
Gas	$M$	$T (K)$	$\sqrt{T / M}$
$H_{2}$	2	300	$\sqrt{300 / 2} = \sqrt{150} \dots$ III (Highest)
$N_{2}$	28	300	$\sqrt{300 / 28} = \sqrt{10.71} \dots$ I (Lowest)
$O_{2}$	32	400	$\sqrt{400 / 32} = \sqrt{12.5} \dots$ II

So,

I. corresponds to $v_{m p}$ of $N_{2} (300 K)$

II. corresponds to $v_{m p}$ of $O_{2} (400 K)$

III. corresponds to $v_{m p}$ of $H_{2} (300 K)$

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