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Q.
The temperature of the ideal gas is increased from $27^{\circ} C$ to $927^{\circ} C$. The root mean square speed of its molecules becomes
AMUAMU 2002
Solution:
Root mean square velocity $\left(v_{r m s}\right)$ of an ideal gas at temperature $T$ is
$ v_{r m s} =\sqrt{\frac{3 R T}{M}} $
$\therefore \frac{v_{1}}{v_{2}} =\sqrt{\frac{T_{1}}{T_{2}}} $
Given, $T_{1}=27^{\circ} C$
$=273+27=300 \,K$,
$T_{2}=927^{\circ} C$
$T_{2}=927+273=1200 \,K$
$\therefore \frac{v_{1}}{v_{2}}=\sqrt{\frac{300}{1200}}=\frac{1}{2}$
Hence, root mean square velocity will become double.