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Heat and Thermodynamics question

2016 · 10 Apr · Shift 1 · Q44
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Heat and Thermodynamics question

2016 · 10 Apr · Shift 1 · Q44

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Which of the following shows the correct relationship between the pressure ‘P’ and density ρ\rhoρ of an ideal gas at constant temperature ?
  1. A
    JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 368 English Option 1
  2. B
    JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 368 English Option 2
  3. C
    JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 368 English Option 3
  4. D
    JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 368 English Option 4
View written solutionFree

Correct answer: D

  1. For an ideal gas, the equation of state is

PV=nRTPV=nRTPV=nRT

  1. Write density in terms of mass and volume:

ρ=mV\rho=\frac{m}{V}ρ=Vm​

Also, number of moles

n=mMn=\frac{m}{M}n=Mm​

where MMM is the molar mass.

  1. Substitute n=mMn=\frac{m}{M}n=Mm​ into the ideal gas equation:

PV=mMRTPV=\frac{m}{M}RTPV=Mm​RT

  1. Rearranging,

P=mV⋅RTMP=\frac{m}{V}\cdot \frac{RT}{M}P=Vm​⋅MRT​

Since

mV=ρ\frac{m}{V}=\rhoVm​=ρ

we get

P=ρRTMP=\rho\frac{RT}{M}P=ρMRT​

  1. At constant temperature TTT, and for a given gas MMM is constant, so

P∝ρP\propto \rhoP∝ρ

Thus, pressure is directly proportional to density. The correct graph/relationship is a straight line passing through the origin.

  1. Therefore, the correct option is the one representing

P∝ρP \propto \rhoP∝ρ

Given the stored correct answer is D, this matches the derived result.

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