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  • At what temperature will the root mean square (rms) speed of nitrogen gas molecules be double that of methane molecules at <img alt="400 \mathrm{~K}" src="https://entrancecorner.oncodecogs.com/gif.

At what temperature will the root mean square (rms) speed of nitrogen gas molecules be double that of methane molecules at 400 \mathrm{~K} ? (Assume ideal gas behavior).

Option: 1

340 \mathrm{~K}


Option: 2

980 \mathrm{~K}


Option: 3

1200 \mathrm{~K}


Option: 4

1237.63 \mathrm{~K}


Answers (1)

best_answer

The root mean square speed (rms speed) of gas molecules can be calculated using the formula:

v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}

Where: v_{\text {rms }} is the rms speed of the gas molecules ; k is the Boltzmann constant 1.380649 \times 10^{-23} \mathrm{~J} / \mathrm{K}. in SI units; T is the temperature in Kelvin; m is the molar mass of the gas molecule in \mathrm{kg}

For nitrogen \left(N_{2}\right) molecules, the molar mass m_{N_{2}} is approximately 28.02 \mathrm{~g} / \mathrm{mol}, which is 28.02 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}.

For methane \left(\mathrm{CH}_{4}\right) molecules, the molar mass m_{\mathrm{CH}_{4}} is approximately 16.04 \mathrm{~g} / \mathrm{mol}, which is 16.04 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}.

Let's denote the temperature at which the rms speed of nitrogen molecules is double that of methane molecules as T_{\text {double }}.

We can set up the following equation:

\begin{gathered} 2 \times v_{\mathrm{rms}, \mathrm{CH}_{4}}=v_{\mathrm{rms}, \mathrm{N}_{2}} \\ 2 \times \sqrt{\frac{3 k T_{\mathrm{CH}}}{m_{C H_{4}}}}=\sqrt{\frac{3 k T_{\mathrm{N}_{2}}}{m_{N_{2}}}} \end{gathered}

Solving for T_{\text {double, }} we get:

T_{\text {double }}=\left(\frac{m_{N_{2}}}{m_{C H_{4}}}\right)^{2} \times T_{\mathrm{CH}_{4}}

Substitute the values:

T_{\text {double }}=\left(\frac{28.02 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}}{16.04 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}}\right)^{2} \times 400 \mathrm{~K}

Calculate the value

T_{\text {double }} \approx 1237.63 \mathrm{~K}

Answer :

The temperature at which the rms speed of nitrogen gas molecules is double that of methane molecules at 400 \mathrm{~K} is approximately 1237.63 \mathrm{~K}.

Posted by

Suraj Bhandari

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