At what temperature (K) will the r.m.s. veloc | vedclass

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(D) The $r.m.s.$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$.Given,$v_{rms} = v_e = 11.2 	imes 10^3 \ m/s$.The molar mass of hyd

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$10075$

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(D) The $r.m.s.$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$.Given,$v_{rms} = v_e = 11.2 	imes 10^3 \ m/s$.The molar mass of hydrogen $(H_2)$ is $M_w = 2 	imes 10^{-3} \ kg/mol$.Equating the two: $\sqrt{\frac{3RT}{M_w}} = 11.2 	imes 10^3$.Squaring both sides: $\frac{3RT}{M_w} = (11.2 	imes 10^3)^2$.$T = \frac{(11.2 	imes 10^3)^2 	imes M_w}{3R}$.Substituting $R = 8.314 \ J/(mol \cdot K)$ and $M_w = 2 	imes 10^{-3} \ kg/mol$:$T = \frac{125.44 	imes 10^6 	imes 2 	imes 10^{-3}}{3 	imes 8.314} \approx \frac{250.88 	imes 10^3}{24.942} \approx 10058 \ K$.Rounding to the nearest provided option,$T \approx 10075 \ K$.

At what temperature $(K)$ will the $r.m.s.$ velocity of hydrogen gas be equal to the escape velocity from the Earth?

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