A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f1(v)dv denote the fraction of molecules with speed between v and (v + dv) with f2(v)dv, similarly for oxygen. Then

Question

A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f1(v)dv denote the fraction of molecules with speed between v and (v + dv) with f2(v)dv, similarly for oxygen. Then (A) f1(v) + f2(v) = f(v) obeys the Maxwell's distribution law (B) f1(v), f2(v) will obey the Maxwell's distribution law separately (C) Neither f1(v) nor f2(v) will obey the Maxwell's distribution law (D) f2(v) and f1(v) will be the same

Tardigrade Admin · Accepted Answer

The Maxwell-Boltzmann speed distribution function (Nv = (dN/dv)) depends on the mass of the gas molecule. [Here, dN is the number of molecules with speeds between v and (v + dv)]. The masses of hydrogen and oxygen molecules are different.

Q. A vessel of volume $V$ contains a mixture of $1$ mole of hydrogen and $1$ mole of oxygen (both considered as ideal). Let $f_{1} (v) d v$ denote the fraction of molecules with speed between $v$ and $(v + d v)$ with $f_{2} (v) d v$ , similarly for oxygen. Then

$f_{1} (v) + f_{2} (v) = f (v)$ obeys the Maxwell's distribution law

$f_{1} (v)$ , $f_{2} (v)$ will obey the Maxwell's distribution law separately

Neither $f_{1} (v)$ nor $f_{2} (v)$ will obey the Maxwell's distribution law

$f_{2} (v)$ and $f_{1} (v)$ will be the same

The Maxwell-Boltzmann speed distribution function $(N_{v} = \frac{d N}{d v})$ depends on the mass of the gas molecule. [Here, $d N$ is the number of molecules with speeds between $v$ and $(v + d v)$ ]. The masses of hydrogen and oxygen molecules are different.

Q. A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f1​(v)dv denote the fraction of molecules with speed between v and (v+dv) with f2​(v)dv, similarly for oxygen. Then

f1​(v)+f2​(v)=f(v) obeys the Maxwell's distribution law

f1​(v), f2​(v) will obey the Maxwell's distribution law separately

Neither f1​(v) nor f2​(v) will obey the Maxwell's distribution law

f2​(v) and f1​(v) will be the same