States of Matter - Result Question 1

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

(2019 Main, 10 April II)

(a) $v _{mp}$ of $H _2(300 K)$; $v _{mp}$ of $N _2(300 K)$; $v _{mp}$ of $O _2(400 K)$

(b) $v _{mp}$ of $O _2(400 K) ; v _{mp}$ of $N _2(300 K) ; v _{mp}$ of $H _2(300 K)$

(c) $v _{mp}$ of $N _2(300 K)$; $v _{mp}$ of $O _2(400 K)$; $v _{mp}$ of $H _2(300 K)$

(d) $v _{mp}$ of $N _2(300 K) ; v _{mp}$ of $H _2(300 K) ; v _{mp}$ of $O _2(400 K)$

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Answer:

Correct Answer: 1. (d)

Solution:

Key Idea From kinetic gas equation,

Most probable velocity $\left(v _{mp}\right)=\sqrt{\frac{2 R T}{M}}$

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

$v _{mp}=\sqrt{\frac{2 R T}{M}}$, i.e. $v _{mp} \propto \sqrt{\frac{T}{M}}$
Gas $\boldsymbol{M}$ $\boldsymbol{T}(K)$ $\sqrt{\boldsymbol{T} / \boldsymbol{M}}$
$H _2$ 2 300 $\sqrt{300 / 2}=\sqrt{150} \ldots$ III (Highest)
$N _2$ 28 300 $\sqrt{300 / 28}=\sqrt{10.71} \ldots$ I (Lowest)
$O _2$ 32 400 $\sqrt{400 / 32}=\sqrt{12.5} \ldots$ II

So,

I. corresponds to $v _{mp}$ of $N _2(300 K)$

II. corresponds to $v _{mp}$ of $O _2(400 K)$

III. corresponds to $v _{mp}$ of $H _2(300 K)$



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