The r.m.s. speed of oxygen molecules at 47^{c | vedclass

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(C) The formula for r.m.s. speed is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $V_{rms, O_2} = V_{rms, H_2}$.Temperature of oxygen $T_{O_2}

Vedclass · Accepted Answer

$-253$

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(C) The formula for r.m.s. speed is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $V_{rms, O_2} = V_{rms, H_2}$.Temperature of oxygen $T_{O_2} = 273 + 47 = 320 \ K$.Equating the speeds: $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.Squaring both sides and simplifying: $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.Substituting the values: $\frac{320}{32} = \frac{T_{H_2}}{2}$.$10 = \frac{T_{H_2}}{2} \implies T_{H_2} = 20 \ K$.Converting to Celsius: $T(^{\circ}C) = 20 - 273 = -253^{\circ}C$.

The r.m.s. speed of oxygen molecules at $47^{\circ} C$ is equal to that of the hydrogen molecules kept at . . . . . . ${ }^{\circ} C$. (Mass of oxygen molecule/mass of hydrogen molecule $= 32 / 2$)

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