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NEET PHYSICSKinetic TheoryEasy

Question

The temperature at which the RMS speed of atoms in neon gas is equal to the RMS speed of hydrogen molecules at 15C15^\circ\text{C} is: (the atomic mass of neon =20.2 u= 20.2 \text{ u}, molecular mass of hydrogen =2 u= 2 \text{ u})

A

2.9×103 K2.9 \times 10^3 \text{ K}

B

2.9 K2.9 \text{ K}

C

0.15×103 K0.15 \times 10^3 \text{ K}

D

0.29×103 K0.29 \times 10^3 \text{ K}

Step-by-Step Solution

The root mean square (RMS) speed of a gas is given by vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}, where TT is the absolute temperature and MM is the molar mass.

  1. Condition: vrms(Ne)=vrms(H2)v_{rms}(\text{Ne}) = v_{rms}(\text{H}_2) 3RTNeMNe=3RTH2MH2\sqrt{\frac{3RT_{\text{Ne}}}{M_{\text{Ne}}}} = \sqrt{\frac{3RT_{\text{H}_2}}{M_{\text{H}_2}}}

  2. Simplify: Squaring both sides and canceling constants (3R3R): TNeMNe=TH2MH2\frac{T_{\text{Ne}}}{M_{\text{Ne}}} = \frac{T_{\text{H}_2}}{M_{\text{H}_2}} TNe=TH2×MNeMH2T_{\text{Ne}} = T_{\text{H}_2} \times \frac{M_{\text{Ne}}}{M_{\text{H}_2}}

  3. Substitute Values: TH2=15C=15+273=288 KT_{\text{H}_2} = 15^\circ\text{C} = 15 + 273 = 288 \text{ K} MNe=20.2 uM_{\text{Ne}} = 20.2 \text{ u}

  • MH2=2 uM_{\text{H}_2} = 2 \text{ u} TNe=288×20.22=288×10.1T_{\text{Ne}} = 288 \times \frac{20.2}{2} = 288 \times 10.1 TNe2908.8 KT_{\text{Ne}} \approx 2908.8 \text{ K}
  1. Scientific Notation: 2908.8 K2.9×103 K2908.8 \text{ K} \approx 2.9 \times 10^3 \text{ K}

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 Theorytemperaturehydrogenmoleculescirctextcatomic

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