Question

Let the most probable velocity of hydrogen molecules at a temperature $t^{0}C$ is $V_0$. Suppose all the molecules dissociate into atoms when the temperature is raised to ${(2t+273)}^{o}C$ then what is the new r.m.s velocity?

Answer 1

The average velocity of gas particles is found using the root mean square velocity formula.

${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{\frac{1}{2}}$

where,

${\mu }_{rms}$= root mean square velocity in m/s

R= ideal gas constant= 8.3145J/K

T= absolute temperature in kelvin

M= mass of a mole of the gas in kg

Initial temp. t degree celcius

$T_0=(t+273K)$

and when temperature is increased to (final) =$(2t+273)$ degree celcius

$T_f=(2t+273)+273=2\times (t+273)=2\times T_0$

Now as we know =${\left(\frac{3RT}{M}\right)}^{\frac{1}{2}}$

or it is directly proportional to $V\alpha \sqrt{T}$

$V_0=k\sqrt{T_0}$ where K is constant K=$\left( \dfrac { 3RT }{ M } \right) ^{ \dfrac { 1 }{ 2 } }$

and at $T_f$

$V_f=k\sqrt{T_f};since{T}_f=2T_0$

$\frac{V_f}{V_0}=\sqrt{\left(\frac{T_f}{T_0}\right)}=\sqrt{2}$

$V_f=\sqrt{2\times V_0}$

Hence the answer is not given in the option