1.

Kinetic energy per unit volume of a gas is

Answer» According to the kinetic theory of an ideal gas, the pressure p exerted by the gas is
`p=(1)/(3)rhoc^(2)_(rms)=(1)/(3)(m)/(V)c^(2)_(rms)`
`:. pV=(1)/(3)mc^(2)_(rms)`
where `c_(rms)` is the speed rms speed (root-mean-square speed) of the gas molecules, m, V and `rho` are the mass, volume and denstiy of the gas, respectively. If there are n moles of the gas and M is the molar mass,
m=nM, so that `pV=(1)/(3)nMc^(2)_(rms)" "`.....(1)
The equation of state of an ideal gas is `pV=nRT " "`....(2)
where T is the absolute temperature of the gas and R is the molar gas constant.
From Eqs.(1) and (2), we get,
`(1)/(3)nMc^(2)_(rms)=nRT`
`:. Mc^(2)_(rms)=3RT" "`....(3)
`:. (1)/(2)Mc^(2)_(rms)=(3)/(2)RT" "`....(4)
where the term on the left-hand side is the kinetic energy of one mole of the gas.
`:.` Kinetic energy per mole of the gas `=(3)/(2)RT" "`.....(5)
Kinetic energy per molecule of the gas `=("kinetic energy per mole")/("Avogadro constant")`
`=(3RT)/(2N_(A))`
`=(3)/(2)k_(B)T" "`....(6)
where the Boltzmann constant, `k_(B)=(R )/(N_(A))`.


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