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NEET PHYSICSKinetic TheoryEasy

Question

At 10C10^\circ\text{C} the value of the density of a fixed mass of an ideal gas divided by its pressure is xx. At 110C110^\circ\text{C} this ratio is:

A

xx

B

383283x\frac{383}{283}x

C

10110x\frac{10}{110}x

D

283383x\frac{283}{383}x

Step-by-Step Solution

According to the Ideal Gas Equation, PV=nRTPV = nRT. Since n=mMn = \frac{m}{M} (mass / Molar mass) and density ρ=mV\rho = \frac{m}{V}, we can substitute to find the relationship between density and pressure: P=mVRTM=ρRTMP = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M} Rearranging for the ratio of density to pressure: ρP=MRT\frac{\rho}{P} = \frac{M}{RT} For a fixed mass of a specific gas, the Molar mass (MM) is constant. Thus, the ratio is inversely proportional to the absolute temperature: ρP1T\frac{\rho}{P} \propto \frac{1}{T}

  1. Initial Condition: T1=10C=10+273=283 KT_1 = 10^\circ\text{C} = 10 + 273 = 283 \text{ K}. x=k283(where k is a constant)x = \frac{k}{283} \quad \text{(where } k \text{ is a constant)}
  2. Final Condition: T2=110C=110+273=383 KT_2 = 110^\circ\text{C} = 110 + 273 = 383 \text{ K}. x=k383x' = \frac{k}{383}
  3. Calculate Ratio: xx=k/383k/283=283383\frac{x'}{x} = \frac{k / 383}{k / 283} = \frac{283}{383} x=283383xx' = \frac{283}{383}x

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.

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