Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$\frac{400}{\sqrt{3}}$

$\frac{100\sqrt{2}}{3}$

$\frac{100}{3}$

$100\sqrt{2}$

Step-by-Step Solution

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.
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.
Convert Units: Initial Temperature ( $T_1$ ) = $27^\circ\text{C} = 27 + 273 = 300 \text{ K}$ . Final Temperature ( $T_2$ ) = $127^\circ\text{C} = 127 + 273 = 400 \text{ K}$ .
Calculate New Velocity: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$ $\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$

$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \text{ ms}^{-1}$ .

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