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Relating Translational KE and T

Kinetic energ...

Question

Kinetic energy of $1$ mole of an ideal gas is:

\frac{3}{2} R T

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\frac{3}{2} K T

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\frac{R T}{2}

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\frac{2 R}{3}

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Solution

The correct option is A $\frac{3}{2} R T$
Answer $A .$ $\frac{3}{2} R T$
Average kinetic energy of a molecule is $K E_{a v g} = \frac{3}{2} K T$
$K E_{t o t a l} = \frac{3}{2} n N_{A} K T n =$ number of moles.
$= \frac{3}{2} n R T$ as $N_{A} K = R$
$= \frac{3}{2} R T$ for $n = 1 m o l e .$

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E = \frac{3}{2} RT

C_{p}

E = (3 / 2) R T, t h e n C_{p}

E = \frac{3}{2} RT

C_{p}

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})

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