Question

Assuming the expression for pressure P exerted by ideal gas, prove that
the kinetic energy per unit volume is 1.5 P.
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Answer 1

kinetic energy of ideal gas per unit volume is 1.5 times of pressure exerted by it.

it is given that pressure exerted by ideal gas is P.

we know, kinetic energy = 1/2 mv²

where m is mass of object.

now for ideal gas, m is mass of each molecule , V is rms velocity of gaseous molecule.

from Maxwell's equation,

rms velocity, V = √{3P/ρ}

so, kinetic energy of ideal gas , K.E = 1/2 × m × {√3P/ρ}² = 3mP/2ρ

as we know, ρ = m / V [ where m is mass of gaseous molecule and V is volume of gaseous molecule]

so, K.E = 3mP/2m/V = 3PV/2

⇒K.E/V = kinetic energy per unit volume = 3P/2 = 1.5 P [ hence proved ]

Answer 2

$\huge\bold\purple{Answer:-}$

it is given that pressure exerted by ideal gas is P.

we know, kinetic energy = 1/2 mv²

where m is mass of object.

now for ideal gas, m is mass of each molecule , V is rms velocity of gaseous molecule.

from Maxwell's equation,

rms velocity, V = √{3P/ρ}

so, kinetic energy of ideal gas , K.E = 1/2 × m × {√3P/ρ}² = 3mP/2ρ

as we know, ρ = m / V [ where m is mass of gaseous molecule and V is volume of gaseous molecule]

so, K.E = 3mP/2m/V = 3PV/2

⇒K.E/V = kinetic energy per unit volume = 3P/2 = 1.5 P [ hence proved ]