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NEET PHYSICSKinetic TheoryMedium

Question

The molecules of a given mass of gas have RMS velocity of 200 ms1200 \text{ ms}^{-1} at 27C27^\circ\text{C} and 1.0×105 Nm21.0 \times 10^5 \text{ Nm}^{-2} pressure. When the temperature and the pressure of the gas are respectively, 127C127^\circ\text{C} and 0.05×105 Nm20.05 \times 10^5 \text{ Nm}^{-2}, the RMS velocity of its molecules in ms1\text{ms}^{-1} is:

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

  1. Identify the Formula: The Root Mean Square (RMS) velocity of gas molecules is given by vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}, where RR is the gas constant, TT is the absolute temperature, and MM is the molar mass.
  2. Analyze Dependencies: For a given mass of gas (constant MM), the RMS velocity depends only on temperature (vrmsTv_{rms} \propto \sqrt{T}). It is independent of pressure changes if the gas remains ideal.
  3. Convert Units: Initial Temperature (T1T_1) = 27C=27+273=300 K27^\circ\text{C} = 27 + 273 = 300 \text{ K}. Final Temperature (T2T_2) = 127C=127+273=400 K127^\circ\text{C} = 127 + 273 = 400 \text{ K}.
  4. Calculate New Velocity: 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 ms1v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \text{ ms}^{-1}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Kinetic Theory. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSKinetic Theorymoleculesvelocitycirctextcpressuretemperature

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