ABC26PH0715 · Gaseous State

Subject: Physical Chemistry · Chapter: Gaseous State · Topic: Gaseous State – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Hard

The Maxwell probability distribution of molecular speeds for a gas is:
\[ \mathrm{F(v)\,dv} = 4\pi \mathrm{v}^{2}\left(\dfrac{\mathrm{m}}{2\pi \mathrm{kT}}\right)^{3/2} \exp\left(-\dfrac{\mathrm{mv}^{2}}{2\mathrm{kT}}\right) \mathrm{dv} \]
where 'v' is the speed, 'm' the mass of the gas molecule and k the Boltzmann constant. (i) Use F(v) to show that the most probable speed $\mathrm{v_{mp}}$ is given by the expression.
\[ \mathrm{v_{mp}} = \left(\dfrac{2\mathrm{RT}}{\mathrm{M}}\right)^{1/2} \]
(ii) Use $\mathrm{R = 8\,J\,K^{-1}\,mol^{-1}}$ in the above expression to calculate the $\mathrm{v_{mp}}$ for $\mathrm{CH_4(g)}$ at $127^{\circ}\mathrm{C}$.
Answer
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Explanation
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