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Kinetic energy per unit volume of a gas is |
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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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