The pressure exerted by 6 - 1023 chlorine molecules- n of which will strike per second on a wall of area A at an angle 60- with the normal- if the massof Cl 2 molecules is m and speed of each Cl 2 molecule is v- is given by -2 mvA- n A-n1C-D- n mnv - A

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Standard XII

Physics

Speed of Efflux

The pressure ...

Question

The pressure exerted by $6 \times 10^{23}$ chlorine molecules, $n$ of which will strike per second on a wall of area $A$ at an angle $60^{\circ}$ with the normal, if the mass of $C l_{2}$ molecules is $m$ and speed of each $C l_{2}$ molecule is $v$ , is given by :

\frac{m v}{n A}

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\frac{2 m v}{n A}

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\frac{m n v}{A}

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\frac{2 m n v}{A}

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Solution

The correct option is C $\frac{m n v}{A}$
Momentum of each chlorine molecule with which it strikes the wall normally
$= m v c o s θ$
Each molecule rebounds with the same momentum.
So, the change of momentum
$= m v c o s θ - (- m v c o s θ)$
$= 2 m v c o s θ$
If $n$ molecules strike per second on the wall, then change in momentum per second, i.e., force $= 2 m v c o s θ \times n$ and force per unit area
$= \frac{F}{A} = \frac{2 m n v cos θ}{A}$
Thus, pressure exerted on the wall is
$P = \frac{2 m n v cos θ}{A} = \frac{2 m n v cos 60^{\circ}}{A} = \frac{m n v}{A}$

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The pressure exerted by 6×1023 chlorine molecules, n of which will strike per second on a wall of area A at an angle 60∘ with the normal, if the mass of Cl2 molecules is m and speed of each Cl2 molecule is v, is given by :

The pressure exerted by $6 \times 10^{23}$ chlorine molecules, $n$ of which will strike per second on a wall of area $A$ at an angle $60^{\circ}$ with the normal, if the mass of $C l_{2}$ molecules is $m$ and speed of each $C l_{2}$ molecule is $v$ , is given by :