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NEET PHYSICSKinetic TheoryMedium

Question

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is:

A

2/3

B

3/4

C

2

D

1/2

Step-by-Step Solution

According to the Ideal Gas Equation, PV=nRTPV = nRT. The number of moles nn can be written as mass (m)Molar Mass (M)\frac{\text{mass } (m)}{\text{Molar Mass } (M)}. Substituting this into the gas equation: PV=mMRTPV = \frac{m}{M}RT. Rearranging to introduce density (ρ=mV\rho = \frac{m}{V}): P=mVRTM=ρRTMP = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M}. Solving for Molar Mass (MM): M=ρRTPM = \frac{\rho RT}{P}.

Given conditions:

  1. Temperature is constant (TA=TBT_A = T_B).
  2. Pressure of A is twice that of B: PA=2PBPBPA=12P_A = 2P_B \Rightarrow \frac{P_B}{P_A} = \frac{1}{2}.
  3. Density of A is 1.5 times that of B: ρA=1.5ρBρAρB=1.5=32\rho_A = 1.5 \rho_B \Rightarrow \frac{\rho_A}{\rho_B} = 1.5 = \frac{3}{2}.

Calculation: Ratio of molecular weights MAMB=ρART/PAρBRT/PB=(ρAρB)×(PBPA)\frac{M_A}{M_B} = \frac{\rho_A RT / P_A}{\rho_B RT / P_B} = \left( \frac{\rho_A}{\rho_B} \right) \times \left( \frac{P_B}{P_A} \right). MAMB=32×12=34\frac{M_A}{M_B} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}.

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 Theoryvesselsseparatelycontaintemperaturepressure

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