QJEEM23Ph1654: JEE Main 2023 Physics - Class 11 JEEM23Ph1654 Solution: The temperature of an ideal gas is increased from to . If…

The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:

$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

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:

$v_0 = \sqrt{\frac{3R \cdot 200}{M}}$

Now, we want to find the r.m.s speed at $800 \, K$ , let's call this $v_1$ :

$v_1 = \sqrt{\frac{3R \cdot 800}{M}}$

Now, divide $v_1$ by $v_0$ :

$\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}$

Simplify the expression:

$\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$

So, $v_1 = 2v_0$ .

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 $200 \, K$ .