The temperature at which the rms speed of mol | vedclass

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(C) The rms speed of molecules in a gas is given by the formula $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$,where $k_B$ is the Boltzmann constant,$T$ is the

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(C) The rms speed of molecules in a gas is given by the formula $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$,where $k_B$ is the Boltzmann constant,$T$ is the absolute temperature in Kelvin,and $m$ is the mass of the molecule.From this relation,we can see that $v_{rms} \propto \sqrt{T}$.Given the initial temperature $T_1 = 27^{\circ} C = (27 + 273) K = 300 K$.Let the initial rms speed be $v_1$. We want the final rms speed $v_2 = 2v_1$ at temperature $T_2$.Using the proportionality $v_{rms} \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values,we get $2 = \sqrt{\frac{T_2}{300}}$.Squaring both sides,$4 = \frac{T_2}{300}$,which gives $T_2 = 1200 K$.Converting back to Celsius,$T_2 = (1200 - 273)^{\circ} C = 927^{\circ} C$.

The temperature at which the rms speed of molecules in hydrogen gas will be double of its initial value at $27^{\circ} C$ is (in $^{\circ} C$)

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