At what temperature, the root-mean-square velocity of $S{O}_2$ will be the same as that of $C{H}_4$ at ${27}^{\circ }C$?a. $3000K$ b. $1345K$ c. $1200K$ d. $1700K$

Sure! Here are the step by step solutions for finding the root-mean-square velocity of ${\text{CH}}_4$ and ${\text{SO}}_2$ gases at a given temperature: For ${\text{CH}}_4$: 1. Use the formula for root-mean-square velocity: $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant, $T$ is the temperature in Kelvin, and $M$ is the molar mass of the gas. 2. Substitute the values given: $R=8.31\text{J/(mol}\cdot \text{K)}$, $T=300\text{K}$, and $M=16\text{g/mol}$ (the molar mass of ${\text{CH}}_4$). 3. Convert the molar mass to kilograms per mole: $M=16\text{g/mol}=0.016\text{kg/mol}$. 4. Plug in the values and solve for $v_{\text{rms}}$: $\begin{array}{rl}{v}_{\text{rms}}& =\sqrt{\frac{3RT}{M}}\\ & =\sqrt{\frac{3\times 8.31\text{J/(mol}\cdot \text{K)}\times 300\text{K}}{0.016\text{kg/mol}}}\\ & \approx 657\text{m/s}\end{array}$ Therefore, the root-mean-square velocity of ${\text{CH}}_4$ at $300\text{K}$ is approximately $657\text{m/s}$. For ${\text{SO}}_2$: 1. Use the same formula for root-mean-square velocity: $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$, but with the molar mass of ${\text{SO}}_2$, which is $64\text{g/mol}$. 2. Substitute the values given: $R=8.31\text{J/(mol}\cdot \text{K)}$, and $M=64\text{g/mol}$. 3. Solve for $T$ by setting the two expressions for $v_{\text{rms}}$ equal to each other: $\begin{array}{rl}\sqrt{\frac{3RT}{M}}& =\sqrt{\frac{3RT}{M}}\\ \sqrt{\frac{3RT}{16}}& =\sqrt{\frac{3RT}{64}}\\ \frac{3RT}{16}& =\frac{3RT}{64}\\ 64\times \frac{3RT}{16}& =64\times \frac{3RT}{64}\\ 12RT& =3RT\\ T& =\frac{1}{4}{T}_0\end{array}$ where $T_0$ is the temperature at which the root-mean-square velocity of ${\text{CH}}_4$ was calculated (in this case, $T_0=300\text{K}$). 4. Substitute the value of $T$ back into the formula for $v_{\text{rms}}$ and solve: $\begin{array}{rl}{v}_{\text{rms}}& =\sqrt{\frac{3RT}{M}}\\ & =\sqrt{\frac{3\times 8.31\text{J/(mol}\cdot \text{K)}\times 1200\text{K}}{64\text{g/mol}}}\\ & \approx 505\text{m/s}\end{array}$ Therefore, the root-mean-square velocity of ${\text{SO}}_2$ at $1200\text{K}$