Question

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

• A. $\sqrt{\frac{2}{3}}{V}_0$
• B. $\sqrt{3(2+273/t)}{V}_0$
• C. $2\sqrt{3}{V}_0$
• D. $\sqrt{6}{V}_0$

Answer 1

Answer: (D)

$\sqrt{6}{V}_0$

Solution:- (D) $\sqrt{6}{V}_{\circ }$

As we know that,

$V_{\left(\text{Most probable}\right)}=\sqrt{\frac{2RT}{M}}.....\left(1\right)$

At $T=t℃=(t+273)K$

$M=2g$

$V_{\left(\text{M.P.}\right)}=V_{\circ }$

$\therefore V_{\left(\text{M.P.}\right)}=\sqrt{\frac{2R(t+273)}{2}}=\sqrt{R(t+273)}$

$\Rightarrow V_{\circ }=\sqrt{R(t+273)}$

At $T=(2t+273)℃=2(t+273)K$

Since the molecule is dissociated into atom.

$M=1g$

$\therefore V_{\left(\text{M.P.}\right)}=\sqrt{\frac{2R\cdot 2(t+273)}{1}}=2\sqrt{R(t+273)}=2V_{\circ }$

Now,

$V_{\left(\text{r.m.s.}\right)}=\sqrt{\frac{3RT}{M}}.....\left(3\right)$

From equation $\left(1\right)\&\left(3\right)$, we get

$V_{\left(\text{r.m.s.}\right)}=\sqrt{\frac{3}{2}}\times V_{\left(\text{M.P.}\right)}$

$\therefore V_{\left(\text{r.m.s.}\right)}=\sqrt{\frac{3}{2}}\times 2V_{\circ }=\sqrt{6}{V}_{\circ }$