KE per unit volume is E. The pressure exerted by the gas is given by:

\frac{E}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 E}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 E}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{E}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{2 E}{3}$
Pressure exerted by gas is given by $P = \frac{1}{3} ρ v^{2}$
where $v$ is the velocity of gas particles, $ρ$ is the density of gas.
Kinetic energy per unit volume $= E = \frac{\frac{1}{2} m v^{2}}{V} = \frac{1}{2} ρ v^{2}$
Thus $P = \frac{2}{3} (\frac{1}{2} ρ v^{2}) = \frac{2 E}{3}$

Suggest Corrections

Similar questions

Internal energy and pressure for unit volume of gas are related by

A.)P=2/3E. B.)P=3/2E. C.)P=1/3E. D.)P=2E

E

Given standard $E^{⊖}$ :

$F e^{3 +} + 3 e^{-} \to F e$ ; $E^{⊖} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e$ ; $E^{⊖} = - 0.440 V$

The $E^{⊖}$ of $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

E^{\circ}

E^{\circ}

V_{(a q)}^{+ 2} + 2 e^{-} \to V, E^{\circ} = - 1.2 V

F e_{(a q)}^{+ 3} + 3 e^{-} \to F e, E^{\circ} = - 0.05 V

A u_{(a q)}^{+ 3} + 3 e^{-} \to A u, E^{\circ} = + 1.50 V

H g_{(a q)}^{+ 2} + 2 e^{-} \to H g, E^{\circ} = + 0.86 V

N O_{3}^{-}

Given standard electrode potentials:

$F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V$

The standard electrode potential $E^{o}$ for $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

Join BYJU'S Learning Program