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# Three moles of an argon gas are at a temperature of 275 K. Calculate the root-mean-square (rms) speed of an atom in the gas.

Solution

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We have to find the value of root-mean-square speed of an atom in the gas.

$\textbf{Given values:}$

$T=275 \: \text{K}$

$R=8.314 \: \text{J}/\text{mol} \cdot \text{K}$

$M=39.95 \: \text{g} \rightarrow 0.03995 \: \text{kg}$

As we know, the root-mean-square speed of the molecules of a gas is given by:

\begin{align*} v_{rms}&=\sqrt{\frac{3 k_B T}{m}}=\sqrt{\frac{3RT}{M}}\tag{Equation 1.}\\ \end{align*}

Now, we have to substitute values in equation 1 and solve for $v_{rms}$:

\begin{align*} v_{rms}&=\sqrt{\frac{3RT}{M}}\tag{Equation 1.}\\ v_{rms}&=\sqrt{\frac{3(8.314 \: \text{J}/\text{mol} \cdot \text{K})(275 \: \text{K})}{ 0.03995 \: \text{kg}}} \tag{Put values in equation.}\\ v_{rms}&=414.35596 \: \text{m}/\text{s}\\ \end{align*}

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