The molecules of a given mass of gas have RMS velocity of $200 ext{ ms}^{-1}$ at $27^\circ ext{C}$ and $1.0

Question

The molecules of a given mass of gas have RMS velocity of $200 	ext{ ms}^{-1}$ at $27^\circ	ext{C}$ and $1.0 	imes 10^5 	ext{ Nm}^{-2}$ pressure. When the temperature and the pressure of the gas are respectively, $127^\circ	ext{C}$ and $0.05 	imes 10^5 	ext{ Nm}^{-2}$, the RMS velocity of its molecules in $	ext{ms}^{-1}$ is:

Accepted Answer

1.  **Identify the Formula:** The Root Mean Square (RMS) velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass. 
2.  **Analyze Dependencies:** For a given mass of gas (constant $M$), the RMS velocity depends only on temperature ($v_{rms} \propto \sqrt{T}$). It is independent of pressure changes if the gas remains ideal.
3.  **Convert Units:** 
    *   Initial Temperature ($T_1$) = $27^\circ	ext{C} = 27 + 273 = 300 	ext{ K}$.
    *   Final Temperature ($T_2$) = $127^\circ	ext{C} = 127 + 273 = 400 	ext{ K}$.
4.  **Calculate New Velocity:**
    *   $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
    *   $\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$
    *   $v_2 = 200 	imes \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} 	ext{ ms}^{-1}$.

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