Browse Questions

The temperature at which the r.m.s. velocity of oxygen molecules equal that of nitrogen molecules at $\;100^{0}C\;$ is nearly :

$(a)\;426.3K\qquad(b)\;456.3K\qquad(c)\;436.3K\qquad(d)\;446.3K$

Can you answer this question?

Answer : $\;426.3K$
Explanation :
$\sqrt{\large\frac{3RT_{1}}{M_{1}}}=\sqrt{\large\frac{3RT_{2}}{M_{2}}}$
$T_{1}=\large\frac{M_{1}}{M_{2}}\;.T_{2}=\large\frac{32}{28} \times 373 = 4263 K$
answered Mar 26, 2014 by