The r.m.s. velocity of hydrogen at S.T.P. is | vedclass

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(A) At $S.T.P.$,the initial temperature $T_1 = 273 \ K$.Since the process occurs at constant pressure,according to Charles's Law,$V \propto T$,which i

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$819 \ K, \sqrt{3} u \ m/s$

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(A) At $S.T.P.$,the initial temperature $T_1 = 273 \ K$.Since the process occurs at constant pressure,according to Charles's Law,$V \propto T$,which implies $\frac{V_2}{V_1} = \frac{T_2}{T_1}$.Given $V_2 = 3V_1$,we have $\frac{3V_1}{V_1} = \frac{T_2}{T_1} \Rightarrow T_2 = 3T_1$.Therefore,$T_2 = 3 	imes 273 = 819 \ K$.The r.m.s. speed is given by $V_{rms} = \sqrt{\frac{3RT}{M_0}}$,which means $V_{rms} \propto \sqrt{T}$.Thus,$\frac{V_{rms}'}{V_{rms}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{3T_1}{T_1}} = \sqrt{3}$.Given $V_{rms} = u$,the new r.m.s. speed is $V_{rms}' = \sqrt{3} u \ m/s$.

Column $- I$	Column $- II$
$(A)$ Root mean square speed of gas molecules	$(P)$ $\frac{1}{3} n m \bar{v}^{2}$
$(B)$ Pressure exerted by ideal gas	$(Q)$ $\sqrt{\frac{3 RT}{M}}$
$(C)$ Average kinetic energy of a molecule	$(R)$ $\frac{5}{2} RT$
$(D)$ Total internal energy of $1$ mole of a diatomic gas	$(S)$ $\frac{3}{2} k_{B} T$

The r.m.s. velocity of hydrogen at $S.T.P.$ is $u \ m/s$. If the gas is heated at constant pressure until its volume becomes three times,then the final temperature of the gas and the r.m.s. speed are respectively:

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