The Maxwell probability distribution of molecular speeds for a gas is: F(v)dv = 4π v² (m/(2π kT))3/2exp… · ABC26PH0715

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:
$$F(v)dv = 4\pi v^2\left(\frac{m}{2\pi kT}\right)^{3/2}\exp\left(-\frac{mv^2}{2kT}\right)dv$$
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 $v_{mp}$ is given by the expression.
$$v_{mp} = \left(\frac{2RT}{M}\right)^{1/2}$$
(ii) Use $R = 8\,\mathrm{J\,K^{-1}\,mol^{-1}}$ in the above expression to calculate the $v_{mp}$ for $\mathrm{CH_4(g)}$ at 127°C. IIT JAM 2012
Answer
SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation
No explanation was printed for this question.

Study loop for Gaseous State

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app