The r.m.s. speed of hydrogen at S.T.P. is u t | vedclass

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(C) At $S.T.P.$, the initial temperature $T_1 = 273 	ext{ K}$.Given that the gas is heated at constant pressure, according to Charles's Law, $V \prop

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) At $S.T.P.$, the initial temperature $T_1 = 273 	ext{ K}$.Given that the gas is heated at constant pressure, according to Charles's Law, $V \propto T$, so $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.Given $V_2 = 3V_1$, we have $T_2 = 3T_1 = 3 	imes 273 	ext{ K} = 819 	ext{ K}$.The r.m.s. speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{3T_1}{T_1}} = \sqrt{3}$.Therefore, $v_2 = \sqrt{3} u 	ext{ m/s}$.Thus, the final temperature is $819 	ext{ K}$ and the r.m.s. speed is $\sqrt{3} u 	ext{ m/s}$.

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

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