The temperature of an ideal gas is increased from $200 \mathrm{~K}$ to $800 \mathrm{~K}$. If r.m.s. speed of gas at $200 \mathrm{~K}$ is $v_{0}$. Then, r.m.s. speed of the gas at $800 \mathrm{~K}$ will be:

<p>The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:</p> <p>$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$</p> <p>where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.</p> <p>In this case, we are given that the initial temperature is $200 \, K$ and the final temperature is $800 \, K$. Let the r.m.s speed at $200 \, K$ be $v_0$, then:</p> <p>$v_0 = \sqrt{\frac{3R \cdot 200}{M}}$</p> <p>Now, we want to find the r.m.s speed at $800 \, K$, let&#39;s call this $v_1$:</p> <p>$v_1 = \sqrt{\frac{3R \cdot 800}{M}}$</p> <p>Now, divide $v_1$ by $v_0$:</p> <p>$$\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}$$</p> <p>Simplify the expression:</p> <p>$\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$</p> <p>So, $v_1 = 2v_0$.</p> <p>Hence, the r.m.s speed of the gas at $800 \, K$ will be 2 times the r.m.s speed of the gas at $200 \, K$.</p>