The molecules of a given mass of gas have rms velocity of $200 ext{ m s}^{-1}$ at $27^{\circ} ext{C}$ and $

Question

The molecules of a given mass of gas have rms velocity of $200 	ext{ m s}^{-1}$ at $27^{\circ}	ext{C}$ and $1.0 	imes 10^5 	ext{ N m}^{-2}$ pressure. When the temperature and pressure of the gas are increased to, respectively, $127^{\circ}	ext{C}$ and $0.05 	imes 10^5 	ext{ N m}^{-2}$, the rms velocity of its molecules in $	ext{m s}^{-1}$ will become:

Accepted Answer

The root mean square (rms) velocity ($v_{rms}$) of ideal gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass . The pressure ($P$) does not explicitly affect $v_{rms}$ if the temperature is known, as $v_{rms}$ depends solely on $T$ for a given gas.

**Given:**
1. Initial state: $v_1 = 200 	ext{ m s}^{-1}$, $T_1 = 27^{\circ}	ext{C} = 300 	ext{ K}$.
2. Final state: $T_2 = 127^{\circ}	ext{C} = 400 	ext{ K}$.

**Calculation:**
Using the proportionality $v_{rms} \propto \sqrt{T}$:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
$\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$
$v_2 = 200 	imes \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} 	ext{ m s}^{-1}$.

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