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Maxwell's Speed of Distribution Law

If E k is th...

Question

If E $_{k}$ is the average kinetic energy per mole of a gas, then:

P V = \frac{3}{2} E_{k}

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P = \frac{3}{2} V E_{k}

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P V = \frac{2}{3} E_{k}

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3 P V = E_{k}

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Solution

The correct option is D $P V = \frac{2}{3} E_{k}$
The pressure $P$ can be expressed in terms of translational kinetic energy of the molecules, $i . e .$ ,
$P = \frac{2}{3} E$ .
Here, $E = \frac{1}{2} ρ C^{2}$ , which is the kinetic energy per unit volume.
$⟹ P = \frac{2}{3} \times \frac{1}{2} \times \frac{m_{n}}{V} C^{2}$ .
$⟹ P V = \frac{2}{3} E_{k}$ , where $E_{k}$ is the average kinetic energy per mole of a gas.

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Similar questions

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}}

\frac{3}{2} P V

The relation between the gas pressure P and average kinetic energy per unit volume E is

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