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Introduction

The average k...

Question

The average kinetic energy $(K . E ._{a v g .})$ per gram mole of gas is :

\frac{3}{2} R T

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\frac{1}{2} R T

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\frac{2}{3} R T

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\frac{2}{3 R T}

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Solution

The correct option is A $\frac{3}{2} R T$
The average kinetic energy for a molecule $= \frac{3}{2} k T$
Per gram mole:

$K . E_{a v g .} = \frac{3}{2} k N_{A} T$ ------ (1)

where,
$N_{A}$ is the Avogadro's number.

$\frac{R}{N_{A}} = k$

Hence,

$k N_{A} = R$ ------ (2)

Substituting (2) in (1) we have

$K . E ._{a v g .} = \frac{3}{2} R T$

Option $(A)$ is correct.

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Similar questions

M

T

X

P V = \frac{1}{3} (m N_{A}) u^{2} = \frac{1}{3} M u^{2}

m N_{A} = M

N_{A} = Avogadro's number

u = root mean square velocity

Translational kinetic energy of n mole \frac{1}{2} M u^{2} = \frac{3}{2} (P V) = \frac{3}{2} (n R T)

Average translational kinetic energy per molecule = \frac{3}{2} (\frac{R T}{N_{A}}) = \frac{3}{2} (K T)

u_{r m s} = \sqrt{\frac{3 P V}{M}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{d}}

u_{a v} = \sqrt{\frac{8 R T}{π M}}

10^{- 12}

27^{\circ} C

(Given R = 8 J K^{- 1} m o l^{- 1}, N_{A} = 6 \times 10^{23})

P V = \frac{1}{3} (m N_{A}) u^{2} = \frac{1}{3} M u^{2}

m N_{A} = M

N_{A} = Avogadro's number

u = root mean square velocity

Translational kinetic energy of n mole \frac{1}{2} M u^{2} = \frac{3}{2} (P V) = \frac{3}{2} (n R T)

Average translational kinetic energy per molecule = \frac{3}{2} (\frac{R T}{N_{A}}) = \frac{3}{2} (K T)

u_{r m s} = \sqrt{\frac{3 P V}{M}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{d}}

u_{a v} = \sqrt{\frac{8 R T}{π M}}

List-I	List-II
(I) Translation kinetic energy	1. $\frac{3}{2} P$
(II) Rotational kinetic energy of $C O_{2}$	2. 15/13
(III) Translation kinetic energy per unit volume	3. 7/5
(IV) $λ$ for $C O_{2}$ at very high temperature	4. Function of T only 5. RT 6. $\frac{3}{2} R T$

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