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The molecules of a given mass of gas have rms velocity of 200 m s1200 \text{ m s}^{-1} at 27C27^{\circ}\text{C} and 1.0×105 N m21.0 \times 10^5 \text{ N m}^{-2} pressure. When the temperature and pressure of the gas are increased to, respectively, 127C127^{\circ}\text{C} and 0.05×105 N m20.05 \times 10^5 \text{ N m}^{-2}, the rms velocity of its molecules in m s1\text{m s}^{-1} will become:

A

4003\frac{400}{\sqrt{3}}

B

10023\frac{100\sqrt{2}}{3}

C

1003\frac{100}{3}

D

1002100\sqrt{2}

Step-by-Step Solution

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

Given:

  1. Initial state: v1=200 m s1v_1 = 200 \text{ m s}^{-1}, T1=27C=300 KT_1 = 27^{\circ}\text{C} = 300 \text{ K}.
  2. Final state: T2=127C=400 KT_2 = 127^{\circ}\text{C} = 400 \text{ K}.

Calculation: Using the proportionality vrmsTv_{rms} \propto \sqrt{T}: v2v1=T2T1\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} v2200=400300=43=23\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} v2=200×23=4003 m s1v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \text{ m s}^{-1}.

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